cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A000045 Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155
Offset: 0

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Author

N. J. A. Sloane, 1964

Keywords

Comments

D. E. Knuth writes: "Before Fibonacci wrote his work, the sequence F_{n} had already been discussed by Indian scholars, who had long been interested in rhythmic patterns that are formed from one-beat and two-beat notes. The number of such rhythms having n beats altogether is F_{n+1}; therefore both Gopāla (before 1135) and Hemachandra (c. 1150) mentioned the numbers 1, 2, 3, 5, 8, 13, 21, ... explicitly." (TAOCP Vol. 1, 2nd ed.) - Peter Luschny, Jan 11 2015
In keeping with historical accounts (see the references by P. Singh and S. Kak), the generalized Fibonacci sequence a, b, a + b, a + 2b, 2a + 3b, 3a + 5b, ... can also be described as the Gopala-Hemachandra numbers H(n) = H(n-1) + H(n-2), with F(n) = H(n) for a = b = 1, and Lucas sequence L(n) = H(n) for a = 2, b = 1. - Lekraj Beedassy, Jan 11 2015
Susantha Goonatilake writes: "[T]his sequence was well known in South Asia and used in the metrical sciences. Its development is attributed in part to Pingala (200 BC), later being associated with Virahanka (circa 700 AD), Gopala (circa 1135), and Hemachandra (circa 1150)—all of whom lived and worked prior to Fibonacci." (Toward a Global Science: Mining Civilizational Knowledge, p. 126) - Russ Cox, Sep 08 2021
Also sometimes called Hemachandra numbers.
Also sometimes called Lamé's sequence.
For a photograph of "Fibonacci"'s 1202 book, see the Leonardo of Pisa link below.
F(n+2) = number of binary sequences of length n that have no consecutive 0's.
F(n+2) = number of subsets of {1,2,...,n} that contain no consecutive integers.
F(n+1) = number of tilings of a 2 X n rectangle by 2 X 1 dominoes.
F(n+1) = number of matchings (i.e., Hosoya index) in a path graph on n vertices: F(5)=5 because the matchings of the path graph on the vertices A, B, C, D are the empty set, {AB}, {BC}, {CD} and {AB, CD}. - Emeric Deutsch, Jun 18 2001
F(n) = number of compositions of n+1 with no part equal to 1. [Cayley, Grimaldi]
Positive terms are the solutions to z = 2*x*y^4 + (x^2)*y^3 - 2*(x^3)*y^2 - y^5 - (x^4)*y + 2*y for x,y >= 0 (Ribenboim, page 193). When x=F(n), y=F(n + 1) and z > 0 then z=F(n + 1).
For Fibonacci search see Knuth, Vol. 3; Horowitz and Sahni; etc.
F(n) is the diagonal sum of the entries in Pascal's triangle at 45 degrees slope. - Amarnath Murthy, Dec 29 2001 (i.e., row sums of A030528, R. J. Mathar, Oct 28 2021)
F(n+1) is the number of perfect matchings in ladder graph L_n = P_2 X P_n. - Sharon Sela (sharonsela(AT)hotmail.com), May 19 2002
F(n+1) = number of (3412,132)-, (3412,213)- and (3412,321)-avoiding involutions in S_n.
This is also the Horadam sequence (0,1,1,1). - Ross La Haye, Aug 18 2003
An INVERT transform of A019590. INVERT([1,1,2,3,5,8,...]) gives A000129. INVERT([1,2,3,5,8,13,21,...]) gives A028859. - Antti Karttunen, Dec 12 2003
Number of meaningful differential operations of the k-th order on the space R^3. - Branko Malesevic, Mar 02 2004
F(n) = number of compositions of n-1 with no part greater than 2. Example: F(4) = 3 because we have 3 = 1+1+1 = 1+2 = 2+1.
F(n) = number of compositions of n into odd parts; e.g., F(6) counts 1+1+1+1+1+1, 1+1+1+3, 1+1+3+1, 1+3+1+1, 1+5, 3+1+1+1, 3+3, 5+1. - Clark Kimberling, Jun 22 2004
F(n) = number of binary words of length n beginning with 0 and having all runlengths odd; e.g., F(6) counts 010101, 010111, 010001, 011101, 011111, 000101, 000111, 000001. - Clark Kimberling, Jun 22 2004
The number of sequences (s(0),s(1),...,s(n)) such that 0 < s(i) < 5, |s(i)-s(i-1)|=1 and s(0)=1 is F(n+1); e.g., F(5+1) = 8 corresponds to 121212, 121232, 121234, 123212, 123232, 123234, 123432, 123434. - Clark Kimberling, Jun 22 2004 [corrected by Neven Juric, Jan 09 2009]
Likewise F(6+1) = 13 corresponds to these thirteen sequences with seven numbers: 1212121, 1212123, 1212321, 1212323, 1212343, 1232121, 1232123, 1232321, 1232323, 1232343, 1234321, 1234323, 1234343. - Neven Juric, Jan 09 2008
A relationship between F(n) and the Mandelbrot set is discussed in the link "Le nombre d'or dans l'ensemble de Mandelbrot" (in French). - Gerald McGarvey, Sep 19 2004
For n > 0, the continued fraction for F(2n-1)*phi = [F(2n); L(2n-1), L(2n-1), L(2n-1), ...] and the continued fraction for F(2n)*phi = [F(2n+1)-1; 1, L(2n)-2, 1, L(2n)-2, ...]. Also true: F(2n)*phi = [F(2n+1); -L(2n), L(2n), -L(2n), L(2n), ...] where L(i) is the i-th Lucas number (A000204). - Clark Kimberling, Nov 28 2004 [corrected by Hieronymus Fischer, Oct 20 2010]
For any nonzero number k, the continued fraction [4,4,...,4,k], which is n 4's and a single k, equals (F(3n) + k*F(3n+3))/(F(3n-3) + k*F(3n)). - Greg Dresden, Aug 07 2019
F(n+1) (for n >= 1) = number of permutations p of 1,2,3,...,n such that |k-p(k)| <= 1 for k=1,2,...,n. (For <= 2 and <= 3, see A002524 and A002526.) - Clark Kimberling, Nov 28 2004
The ratios F(n+1)/F(n) for n > 0 are the convergents to the simple continued fraction expansion of the golden section. - Jonathan Sondow, Dec 19 2004
Lengths of successive words (starting with a) under the substitution: {a -> ab, b -> a}. - Jeroen F.J. Laros, Jan 22 2005
The Fibonacci sequence, like any additive sequence, naturally tends to be geometric with common ratio not a rational power of 10; consequently, for a sufficiently large number of terms, Benford's law of first significant digit (i.e., first digit 1 <= d <= 9 occurring with probability log_10(d+1) - log_10(d)) holds. - Lekraj Beedassy, Apr 29 2005 (See Brown-Duncan, 1970. - N. J. A. Sloane, Feb 12 2017)
F(n+2) = Sum_{k=0..n} binomial(floor((n+k)/2),k), row sums of A046854. - Paul Barry, Mar 11 2003
Number of order ideals of the "zig-zag" poset. See vol. 1, ch. 3, prob. 23 of Stanley. - Mitch Harris, Dec 27 2005
F(n+1)/F(n) is also the Farey fraction sequence (see A097545 for explanation) for the golden ratio, which is the only number whose Farey fractions and continued fractions are the same. - Joshua Zucker, May 08 2006
a(n+2) is the number of paths through 2 plates of glass with n reflections (reflections occurring at plate/plate or plate/air interfaces). Cf. A006356-A006359. - Mitch Harris, Jul 06 2006
F(n+1) equals the number of downsets (i.e., decreasing subsets) of an n-element fence, i.e., an ordered set of height 1 on {1,2,...,n} with 1 > 2 < 3 > 4 < ... n and no other comparabilities. Alternatively, F(n+1) equals the number of subsets A of {1,2,...,n} with the property that, if an odd k is in A, then the adjacent elements of {1,2,...,n} belong to A, i.e., both k - 1 and k + 1 are in A (provided they are in {1,2,...,n}). - Brian Davey, Aug 25 2006
Number of Kekulé structures in polyphenanthrenes. See the paper by Lukovits and Janezic for details. - Parthasarathy Nambi, Aug 22 2006
Inverse: With phi = (sqrt(5) + 1)/2, round(log_phi(sqrt((sqrt(5) a(n) + sqrt(5 a(n)^2 - 4))(sqrt(5) a(n) + sqrt(5 a(n)^2 + 4)))/2)) = n for n >= 3, obtained by rounding the arithmetic mean of the inverses given in A001519 and A001906. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 19 2007
A result of Jacobi from 1848 states that every symmetric matrix over a p.i.d. is congruent to a triple-diagonal matrix. Consider the maximal number T(n) of summands in the determinant of an n X n triple-diagonal matrix. This is the same as the number of summands in such a determinant in which the main-, sub- and superdiagonal elements are all nonzero. By expanding on the first row we see that the sequence of T(n)'s is the Fibonacci sequence without the initial stammer on the 1's. - Larry Gerstein (gerstein(AT)math.ucsb.edu), Mar 30 2007
Suppose psi=log(phi). We get the representation F(n)=(2/sqrt(5))*sinh(n*psi) if n is even; F(n)=(2/sqrt(5))*cosh(n*psi) if n is odd. There is a similar representation for Lucas numbers (A000032). Many Fibonacci formulas now easily follow from appropriate sinh and cosh formulas. For example: the de Moivre theorem (cosh(x)+sinh(x))^m = cosh(mx)+sinh(mx) produces L(n)^2 + 5F(n)^2 = 2L(2n) and L(n)F(n) = F(2n) (setting x=n*psi and m=2). - Hieronymus Fischer, Apr 18 2007
Inverse: floor(log_phi(sqrt(5)*F(n)) + 1/2) = n, for n > 1. Also for n > 0, floor((1/2)*log_phi(5*F(n)*F(n+1))) = n. Extension valid for integer n, except n=0,-1: floor((1/2)*sign(F(n)*F(n+1))*log_phi|5*F(n)*F(n+1)|) = n (where sign(x) = sign of x). - Hieronymus Fischer, May 02 2007
F(n+2) = the number of Khalimsky-continuous functions with a two-point codomain. - Shiva Samieinia (shiva(AT)math.su.se), Oct 04 2007
This is a_1(n) in the Doroslovacki reference.
Let phi = A001622 then phi^n = (1/phi)*a(n) + a(n+1). - Gary W. Adamson, Dec 15 2007
The sequence of first differences, F(n+1)-F(n), is essentially the same sequence: 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... - Colm Mulcahy, Mar 03 2008
Equals row sums of triangle A144152. - Gary W. Adamson, Sep 12 2008
Except for the initial term, the numerator of the convergents to the recursion x = 1/(x+1). - Cino Hilliard, Sep 15 2008
F(n) is the number of possible binary sequences of length n that obey the sequential construction rule: if last symbol is 0, add the complement (1); else add 0 or 1. Here 0,1 are metasymbols for any 2-valued symbol set. This rule has obvious similarities to JFJ Laros's rule, but is based on addition rather than substitution and creates a tree rather than a single sequence. - Ross Drewe, Oct 05 2008
F(n) = Product_{k=1..(n-1)/2} (1 + 4*cos^2 k*Pi/n), where terms = roots to the Fibonacci product polynomials, A152063. - Gary W. Adamson, Nov 22 2008
Fp == 5^((p-1)/2) mod p, p = prime [Schroeder, p. 90]. - Gary W. Adamson & Alexander R. Povolotsky, Feb 21 2009
A000032(n)^2 - 5*F(n)^2 = 4*(-1)^n. - Gary W. Adamson, Mar 11 2009
Output of Kasteleyn's formula for the number of perfect matchings of an m X n grid specializes to the Fibonacci sequence for m=2. - Sarah-Marie Belcastro, Jul 04 2009
(F(n),F(n+4)) satisfies the Diophantine equation: X^2 + Y^2 - 7XY = 9*(-1)^n. - Mohamed Bouhamida, Sep 06 2009
(F(n),F(n+2)) satisfies the Diophantine equation: X^2 + Y^2 - 3XY = (-1)^n. - Mohamed Bouhamida, Sep 08 2009
a(n+2) = A083662(A131577(n)). - Reinhard Zumkeller, Sep 26 2009
Difference between number of closed walks of length n+1 from a node on a pentagon and number of walks of length n+1 between two adjacent nodes on a pentagon. - Henry Bottomley, Feb 10 2010
F(n+1) = number of Motzkin paths of length n having exactly one weak ascent. A Motzkin path of length n is a lattice path from (0,0) to (n,0) consisting of U=(1,1), D=(1,-1) and H=(1,0) steps and never going below the x-axis. A weak ascent in a Motzkin path is a maximal sequence of consecutive U and H steps. Example: a(5)=5 because we have (HHHH), (HHU)D, (HUH)D, (UHH)D, and (UU)DD (the unique weak ascent is shown between parentheses; see A114690). - Emeric Deutsch, Mar 11 2010
(F(n-1) + F(n+1))^2 - 5*F(n-2)*F(n+2) = 9*(-1)^n. - Mohamed Bouhamida, Mar 31 2010
From the Pinter and Ziegler reference's abstract: authors "show that essentially the Fibonacci sequence is the unique binary recurrence which contains infinitely many three-term arithmetic progressions. A criterion for general linear recurrences having infinitely many three-term arithmetic progressions is also given." - Jonathan Vos Post, May 22 2010
F(n+1) = number of paths of length n starting at initial node on the path graph P_4. - Johannes W. Meijer, May 27 2010
F(k) = number of cyclotomic polynomials in denominator of generating function for number of ways to place k nonattacking queens on an n X n board. - Vaclav Kotesovec, Jun 07 2010
As n->oo, (a(n)/a(n-1) - a(n-1)/a(n)) tends to 1.0. Example: a(12)/a(11) - a(11)/a(12) = 144/89 - 89/144 = 0.99992197.... - Gary W. Adamson, Jul 16 2010
From Hieronymus Fischer, Oct 20 2010: (Start)
Fibonacci numbers are those numbers m such that m*phi is closer to an integer than k*phi for all k, 1 <= k < m. More formally: a(0)=0, a(1)=1, a(2)=1, a(n+1) = minimal m > a(n) such that m*phi is closer to an integer than a(n)*phi.
For all numbers 1 <= k < F(n), the inequality |k*phi-round(k*phi)| > |F(n)*phi-round(F(n)*phi)| holds.
F(n)*phi - round(F(n)*phi) = -((-phi)^(-n)), for n > 1.
Fract(1/2 + F(n)*phi) = 1/2 -(-phi)^(-n), for n > 1.
Fract(F(n)*phi) = (1/2)*(1 + (-1)^n) - (-phi)^(-n), n > 1.
Inverse: n = -log_phi |1/2 - fract(1/2 + F(n)*phi)|.
(End)
F(A001177(n)*k) mod n = 0, for any integer k. - Gary Detlefs, Nov 27 2010
F(n+k)^2 - F(n)^2 = F(k)*F(2n+k), for even k. - Gary Detlefs, Dec 04 2010
F(n+k)^2 + F(n)^2 = F(k)*F(2n+k), for odd k. - Gary Detlefs, Dec 04 2010
F(n) = round(phi*F(n-1)) for n > 1. - Joseph P. Shoulak, Jan 13 2012
For n > 0: a(n) = length of n-th row in Wythoff array A003603. - Reinhard Zumkeller, Jan 26 2012
From Bridget Tenner, Feb 22 2012: (Start)
The number of free permutations of [n].
The number of permutations of [n] for which s_k in supp(w) implies s_{k+-1} not in supp(w).
The number of permutations of [n] in which every decomposition into length(w) reflections is actually composed of simple reflections. (End)
The sequence F(n+1)^(1/n) is increasing. The sequence F(n+2)^(1/n) is decreasing. - Thomas Ordowski, Apr 19 2012
Two conjectures: For n > 1, F(n+2)^2 mod F(n+1)^2 = F(n)*F(n+1) - (-1)^n. For n > 0, (F(2n) + F(2n+2))^2 = F(4n+3) + Sum_{k = 2..2n} F(2k). - Alex Ratushnyak, May 06 2012
From Ravi Kumar Davala, Jan 30 2014: (Start)
Proof of Ratushnyak's first conjecture: For n > 1, F(n+2)^2 - F(n)*F(n+1) + (-1)^n = 2*F(n+1)^2.
Consider: F(n+2)^2 - F(n)*F(n+1) - 2*F(n+1)^2
= F(n+2)^2 - F(n+1)^2 - F(n+1)^2 - F(n)*F(n+1)
= (F(n+2) + F(n+1))*(F(n+2) - F(n+1)) - F(n+1)*(F(n+1) + F(n))
= F(n+3)*F(n) - F(n+1)*F(n+2) = -(-1)^n.
Proof of second conjecture: L(n) stands for Lucas number sequence from A000032.
Consider the fact that
L(2n+1)^2 = L(4n+2) - 2
(F(2n) + F(2n+2))^2 = F(4n+1) + F(4n+3) - 2
(F(2n) + F(2n+2))^2 = (Sum_{k = 2..2n} F(2k)) + F(4n+3).
(End)
The relationship: INVERT transform of (1,1,0,0,0,...) = (1, 2, 3, 5, 8, ...), while the INVERT transform of (1,0,1,0,1,0,1,...) = (1, 1, 2, 3, 5, 8, ...) is equivalent to: The numbers of compositions using parts 1 and 2 is equivalent to the numbers of compositions using parts == 1 (mod 2) (i.e., the odd integers). Generally, the numbers of compositions using parts 1 and k is equivalent to the numbers of compositions of (n+1) using parts 1 mod k. Cf. A000930 for k = 3 and A003269 for k = 4. Example: for k = 2, n = 4 we have the compositions (22; 211, 121; 112; 1111) = 5; but using parts 1 and 3 we have for n = 5: (311, 131, 113, 11111, 5) = 5. - Gary W. Adamson, Jul 05 2012
The sequence F(n) is the binomial transformation of the alternating sequence (-1)^(n-1)*F(n), whereas the sequence F(n+1) is the binomial transformation of the alternating sequence (-1)^n*F(n-1). Both of these facts follow easily from the equalities a(n;1)=F(n+1) and b(n;1)=F(n) where a(n;d) and b(n;d) are so-called "delta-Fibonacci" numbers as defined in comments to A014445 (see also the papers of Witula et al.). - Roman Witula, Jul 24 2012
F(n) is the number of different (n-1)-digit binary numbers such that all substrings of length > 1 have at least one digit equal to 1. Example: for n = 5 there are 8 binary numbers with n - 1 = 4 digits (1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111), only the F(n) = 5 numbers 1010, 1011, 1101, 1110 and 1111 have the desired property. - Hieronymus Fischer, Nov 30 2012
For positive n, F(n+1) equals the determinant of the n X n tridiagonal matrix with 1's along the main diagonal, i's along the superdiagonal and along the subdiagonal where i = sqrt(-1). Example: Det([1,i,0,0; i,1,i,0; 0,i,1,i; 0,0,i,1]) = F(4+1) = 5. - Philippe Deléham, Feb 24 2013
For n >= 1, number of compositions of n where there is a drop between every second pair of parts, starting with the first and second part; see example. Also, a(n+1) is the number of compositions where there is a drop between every second pair of parts, starting with the second and third part; see example. - Joerg Arndt, May 21 2013 [see the Hopkins/Tangboonduangjit reference for a proof, see also the Checa reference for alternative proofs and statistics]
Central terms of triangles in A162741 and A208245, n > 0. - Reinhard Zumkeller, Jul 28 2013
For n >= 4, F(n-1) is the number of simple permutations in the geometric grid class given in A226433. - Jay Pantone, Sep 08 2013
a(n) are the pentagon (not pentagonal) numbers because the algebraic degree 2 number rho(5) = 2*cos(Pi/5) = phi (golden section), the length ratio diagonal/side in a pentagon, has minimal polynomial C(5,x) = x^2 - x - 1 (see A187360, n=5), hence rho(5)^n = a(n-1)*1 + a(n)*rho(5), n >= 0, in the power basis of the algebraic number field Q(rho(5)). One needs a(-1) = 1 here. See also the P. Steinbach reference under A049310. - Wolfdieter Lang, Oct 01 2013
A010056(a(n)) = 1. - Reinhard Zumkeller, Oct 10 2013
Define F(-n) to be F(n) for n odd and -F(n) for n even. Then for all n and k, F(n+2k)^2 - F(n)^2 = F(n+k)*( F(n+3k) - F(n-k) ). - Charlie Marion, Dec 20 2013
( F(n), F(n+2k) ) satisfies the Diophantine equation: X^2 + Y^2 - L(2k)*X*Y = F(4k)^2*(-1)^n. This generalizes Bouhamida's comments dated Sep 06 2009 and Sep 08 2009. - Charlie Marion, Jan 07 2014
For any prime p there is an infinite periodic subsequence within F(n) divisible by p, that begins at index n = 0 with value 0, and its first nonzero term at n = A001602(i), and period k = A001602(i). Also see A236479. - Richard R. Forberg, Jan 26 2014
Range of row n of the circular Pascal array of order 5. - Shaun V. Ault, May 30 2014 [orig. Kicey-Klimko 2011, and observations by Glen Whitehead; more general work found in Ault-Kicey 2014]
Nonnegative range of the quintic polynomial 2*y - y^5 + 2*x*y^4 + x^2*y^3 - 2*x^3*y^2 - x^4*y with x, y >= 0, see Jones 1975. - Charles R Greathouse IV, Jun 01 2014
The expression round(1/(F(k+1)/F(n) + F(k)/F(n+1))), for n > 0, yields a Fibonacci sequence with k-1 leading zeros (with rounding 0.5 to 0). - Richard R. Forberg, Aug 04 2014
Conjecture: For n > 0, F(n) is the number of all admissible residue classes for which specific finite subsequences of the Collatz 3n + 1 function consists of n+2 terms. This has been verified for 0 < n < 51. For details see Links. - Mike Winkler, Oct 03 2014
a(4)=3 and a(6)=8 are the only Fibonacci numbers that are of the form prime+1. - Emmanuel Vantieghem, Oct 02 2014
a(1)=1=a(2), a(3)=2 are the only Fibonacci numbers that are of the form prime-1. - Emmanuel Vantieghem, Jun 07 2015
Any consecutive pair (m, k) of the Fibonacci sequence a(n) illustrates a fair equivalence between m miles and k kilometers. For instance, 8 miles ~ 13 km; 13 miles ~ 21 km. - Lekraj Beedassy, Oct 06 2014
a(n+1) counts closed walks on K_2, containing one loop on the other vertex. Equivalently the (1,1)entry of A^(n+1) where the adjacency matrix of digraph is A=(0,1; 1,1). - _David Neil McGrath, Oct 29 2014
a(n-1) counts closed walks on the graph G(1-vertex;l-loop,2-loop). - David Neil McGrath, Nov 26 2014
From Tom Copeland, Nov 02 2014: (Start)
Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x) = [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1-x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].
Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. A057078).
BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)].
Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).
Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1].
(End)
F(n+1) equals the number of binary words of length n avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Russell Jay Hendel, Apr 12 2015: (Start)
We prove Conjecture 1 of Rashid listed in the Formula section.
We use the following notation: F(n)=A000045(n), the Fibonacci numbers, and L(n) = A000032(n), the Lucas numbers. The fundamental Fibonacci-Lucas recursion asserts that G(n) = G(n-1) + G(n-2), with "L" or "F" replacing "G".
We need the following prerequisites which we label (A), (B), (C), (D). The prerequisites are formulas in the Koshy book listed in the References section. (A) F(m-1) + F(m+1) = L(m) (Koshy, p. 97, #32), (B) L(2m) + 2*(-1)^m = L(m)^2 (Koshy p. 97, #41), (C) F(m+k)*F(m-k) = (-1)^n*F(k)^2 (Koshy, p. 113, #24, Tagiuri's identity), and (D) F(n)^2 + F(n+1)^2 = F(2n+1) (Koshy, p. 97, #30).
We must also prove (E), L(n+2)*F(n-1) = F(2n+1)+2*(-1)^n. To prove (E), first note that by (A), proof of (E) is equivalent to proving that F(n+1)*F(n-1) + F(n+3)*F(n-1) = F(2n+1) + 2*(-1)^n. But by (C) with k=1, we have F(n+1)*F(n-1) = F(n)^2 + (-1)^n. Applying (C) again with k=2 and m=n+1, we have F(n+3)*F(n-1) = F(n+1) + (-1)^n. Adding these two applications of (C) together and using (D) we have F(n+1)*F(n-1) + F(n+3)*F(n-1) = F(n)^2 + F(n+1)^2 + 2*(-1)^n = F(2n+1) + 2(-1)^n, completing the proof of (E).
We now prove Conjecture 1. By (A) and the Fibonacci-Lucas recursion, we have F(2n+1) + F(2n+2) + F(2n+3) + F(2n+4) = (F(2n+1) + F(2n+3)) + (F(2n+2) + F(2n+4)) = L(2n+2) +L(2n+3) = L(2n+4). But then by (B), with m=2n+4, we have sqrt(L(2n+4) + 2(-1)^n) = L(n+2). Finally by (E), we have L(n+2)*F(n-1) = F(2n+1) + 2*(-1)^n. Dividing both sides by F(n-1), we have (F(2n+1) + 2*(-1)^n)/F(n-1) = L(n+2) = sqrt(F(2n+1) + F(2n+2) + F(2n+3) + F(2n+4) + 2(-1)^n), as required.
(End)
In Fibonacci's Liber Abaci the rabbit problem appears in the translation of L. E. Sigler on pp. 404-405, and a remark [27] on p. 637. - Wolfdieter Lang, Apr 17 2015
a(n) counts partially ordered partitions of (n-1) into parts 1,2,3 where only the order of adjacent 1's and 2's are unimportant. (See example.) - David Neil McGrath, Jul 27 2015
F(n) divides F(n*k). Proved by Marjorie Bicknell and Verner E Hoggatt Jr. - Juhani Heino, Aug 24 2015
F(n) is the number of UDU-equivalence classes of ballot paths of length n. Two ballot paths of length n with steps U = (1,1), D = (1,-1) are UDU-equivalent whenever the positions of UDU are the same in both paths. - Kostas Manes, Aug 25 2015
Cassini's identity F(2n+1) * F(2n+3) = F(2n+2)^2 + 1 is the basis for a geometrical paradox (or dissection fallacy) in A262342. - Jonathan Sondow, Oct 23 2015
For n >= 4, F(n) is the number of up-down words on alphabet {1,2,3} of length n-2. - Ran Pan, Nov 23 2015
F(n+2) is the number of terms in p(n), where p(n)/q(n) is the n-th convergent of the formal infinite continued fraction [a(0),a(1),...]; e.g., p(3) = a(0)a(1)a(2)a(3) + a(0)a(1) + a(0)a(3) + a(2)a(3) + 1 has F(5) terms. Also, F(n+1) is the number of terms in q(n). - Clark Kimberling, Dec 23 2015
F(n+1) (for n >= 1) is the permanent of an n X n matrix M with M(i,j)=1 if |i-j| <= 1 and 0 otherwise. - Dmitry Efimov, Jan 08 2016
A trapezoid has three sides of lengths in order F(n), F(n+2), F(n). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*F(n+1). For a trapezoid with lengths of sides in order F(n+2), F(n), F(n+2), the fourth side will be F(n+3). - J. M. Bergot, Mar 17 2016
(1) Join two triangles with lengths of sides L(n), F(n+3), L(n+2) and F(n+2), L(n+1), L(n+2) (where L(n)=A000032(n)) along the common side of length L(n+2) to create an irregular quadrilateral. Its area is approximately 5*F(2*n-1) - (F(2*n-7) - F(2*n-13))/5. (2) Join two triangles with lengths of sides L(n), F(n+2), F(n+3) and L(n+1), F(n+1), F(n+3) along the common side F(n+3) to form an irregular quadrilateral. Its area is approximately 4*F(2*n-1) - 2*(F(2*n-7) + F(2*n-18)). - J. M. Bergot, Apr 06 2016
From Clark Kimberling, Jun 13 2016: (Start)
Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*.
Let g(n) be the set of nodes in the n-th generation, so that g(0) = {0}, g(1) = {1}, g(2) = {2, x}, g(3) = {3, 2x, x+1, x^2}, etc.
Let T(r) be the tree obtained by substituting r for x.
If a positive integer N is not a square and r = sqrt(N), then the number of (not necessarily distinct) integers in g(n) is A000045(n), for n >= 1. See A274142. (End)
Consider the partitions of n, with all summands initially listed in nonincreasing order. Freeze all the 1's in place and then allow all the other summands to change their order, without displacing any of the 1's. The resulting number of arrangements is a(n+1). - Gregory L. Simay, Jun 14 2016
Limit of the matrix power M^k shown in A163733, Sep 14 2016, as k->infinity results in a single column vector equal to the Fibonacci sequence. - Gary W. Adamson, Sep 19 2016
F(n) and Lucas numbers L(n), being related by the formulas F(n) = (F(n-1) + L(n-1))/2 and L(n) = 2 F(n+1) - F(n), are a typical pair of "autosequences" (see the link to OEIS Wiki). - Jean-François Alcover, Jun 10 2017
Also the number of independent vertex sets and vertex covers in the (n-2)-path graph. - Eric W. Weisstein, Sep 22 2017
Shifted numbers of {UD, DU, FD, DF}-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are P-equivalent iff the positions of pattern P are identical in these paths. - Sergey Kirgizov, Apr 08 2018
For n > 0, F(n) = the number of Markov equivalence classes with skeleton the path on n nodes. See Theorem 2.1 in the article by A. Radhakrishnan et al. below. - Liam Solus, Aug 23 2018
For n >= 2, also: number of terms in A032858 (every other base-3 digit is strictly smaller than its neighbors) with n-2 digits in base 3. - M. F. Hasler, Oct 05 2018
F(n+1) is the number of fixed points of the Foata transformation on S_n. - Kevin Long, Oct 17 2018
F(n+2) is the dimension of the Hecke algebra of type A_n with independent parameters (0,1,0,1,...) or (1,0,1,0,...). See Corollary 1.5 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019
The sequence is the second INVERT transform of (1, -1, 2, -3, 5, -8, 13, ...) and is the first sequence in an infinite set of successive INVERT transforms generated from (1, 0, 1, 0, 1, ...). Refer to the array shown in A073133. - Gary W. Adamson, Jul 16 2019
From Kai Wang, Dec 16 2019: (Start)
F(n*k)/F(k) = Sum_{i=0..n-1; j=0..n-1; i+2*j=n-1} (-1)^(j*(k-1))*L(k)^i*((i+j)!/(i!*j!)).
F((2*m+1)*k)/F(k) = Sum_{i=0..m-1} (-1)^(i*k)*L((2*m-2*i)*k) + (-1)^(m*k).
F(2*m*k)/F(k) = Sum_{i=0..m-1} (-1)^(i*k)*L((2*m-2*i-1)*k).
F(m+s)*F(n+r) - F(m+r)*F(n+s) = (-1)^(n+s)*F(m-n)*F(r-s).
F(m+r)*F(n+s) + F(m+s)*F(n+r) = (2*L(m+n+r+s) - (-1)^(n+s)*L(m-n)*L(r-s))/5.
L(m+r)*L(n+s) - 5*F(m+s)*F(n+r) = (-1)^(n+s)*L(m-n)*L(r-s).
L(m+r)*L(n+s) + 5*F(m+s)*F(n+r) = 2*L(m+n+r+s) + (-1)^(n+s)*5*F(m-n)*F(r-s).
L(m+r)*L(n+s) - L(m+s)*L(n+r) = (-1)^(n+s)*5*F(m-n)*F(r-s). (End)
F(n+1) is the number of permutations in S_n whose principal order ideals in the weak order are Boolean lattices. - Bridget Tenner, Jan 16 2020
F(n+1) is the number of permutations w in S_n that form Boolean intervals [s, w] in the weak order for every simple reflection s in the support of w. - Bridget Tenner, Jan 16 2020
F(n+1) is the number of subsets of {1,2,.,.,n} in which all differences between successive elements of subsets are odd. For example, for n = 6, F(7) = 13 and the 13 subsets are {6}, {1,6}, {3,6}, {5,6}, {2,3,6}, {2,5,6}, {4,5,6}, {1,2,3,6}, {1,2,5,6}, {1,4,5,6}, {3,4,5,6}, {2,3,4,5,6}, {1,2,3,4,5,6}. For even differences between elements see Comment in A016116. - Enrique Navarrete, Jul 01 2020
F(n) is the number of subsets of {1,2,...,n} in which the smallest element of the subset equals the size of the subset (this type of subset is sometimes called extraordinary). For example, F(6) = 8 and the subsets are {1}, {2,3}, {2,4}, {2,5}, {3,4,5}, {2,6}, {3,4,6}, {3,5,6}. It is easy to see that these subsets follow the Fibonacci recursion F(n) = F(n-1) + F(n-2) since we get F(n) such subsets by keeping all F(n-1) subsets from the previous stage (in the example, the F(5)=5 subsets that don't include 6), and by adding one to all elements and appending an additional element n to each subset in F(n-2) subsets (in the example, by applying this to the F(4)=3 subsets {1}, {2,3}, {2,4} we obtain {2,6}, {3,4,6}, {3,5,6}). - Enrique Navarrete, Sep 28 2020
Named "série de Fibonacci" by Lucas (1877) after the Italian mathematician Fibonacci (Leonardo Bonacci, c. 1170 - c. 1240/50). In 1876 he named the sequence "série de Lamé" after the French mathematician Gabriel Lamé (1795 - 1870). - Amiram Eldar, Apr 16 2021
F(n) is the number of edge coverings of the path with n edges. - M. Farrokhi D. G., Sep 30 2021
LCM(F(m), F(n)) is a Fibonacci number if and only if either F(m) divides F(n) or F(n) divides F(m). - M. Farrokhi D. G., Sep 30 2021
Every nonunit positive rational number has at most one representation as the quotient of two Fibonacci numbers. - M. Farrokhi D. G., Sep 30 2021
The infinite sum F(n)/10^(n-1) for all natural numbers n is equal to 100/89. More generally, the sum of F(n)/(k^(n-1)) for all natural numbers n is equal to k^2/(k^2-k-1). Jonatan Djurachkovitch, Dec 31 2023
For n >= 1, number of compositions (c(1),c(2),...,c(k)) of n where c(1), c(3), c(5), ... are 1. To obtain such compositions K(n) of length n increase all parts c(2) by one in all of K(n-1) and prepend two parts 1 in all of K(n-2). - Joerg Arndt, Jan 05 2024
Cohn (1964) proved that a(12) = 12^2 is the only square in the sequence greater than a(1) = 1. - M. F. Hasler, Dec 18 2024
Product_{i=n-2..n+2} F(i) = F(n)^5 - F(n). For example, (F(4)F(5)F(6)F(7)F(8))=(3 * 5 * 8 * 13 * 21) = 8^5 - 8. - Jules Beauchamp, Apr 28 2025
F(n) is even iff n is a multiple of 3. - Stefano Spezia, Jul 06 2025

Examples

			For x = 0,1,2,3,4, x=1/(x+1) = 1, 1/2, 2/3, 3/5, 5/8. These fractions have numerators 1,1,2,3,5, which are the 2nd to 6th terms of the sequence. - _Cino Hilliard_, Sep 15 2008
From _Joerg Arndt_, May 21 2013: (Start)
There are a(7)=13 compositions of 7 where there is a drop between every second pair of parts, starting with the first and second part:
01:  [ 2 1 2 1 1 ]
02:  [ 2 1 3 1 ]
03:  [ 2 1 4 ]
04:  [ 3 1 2 1 ]
05:  [ 3 1 3 ]
06:  [ 3 2 2 ]
07:  [ 4 1 2 ]
08:  [ 4 2 1 ]
09:  [ 4 3 ]
10:  [ 5 1 1 ]
11:  [ 5 2 ]
12:  [ 6 1 ]
13:  [ 7 ]
There are abs(a(6+1))=13 compositions of 6 where there is no rise between every second pair of parts, starting with the second and third part:
01:  [ 1 2 1 2 ]
02:  [ 1 3 1 1 ]
03:  [ 1 3 2 ]
04:  [ 1 4 1 ]
05:  [ 1 5 ]
06:  [ 2 2 1 1 ]
07:  [ 2 3 1 ]
08:  [ 2 4 ]
09:  [ 3 2 1 ]
10:  [ 3 3 ]
11:  [ 4 2 ]
12:  [ 5 1 ]
13:  [ 6 ]
(End)
Partially ordered partitions of (n-1) into parts 1,2,3 where only the order of the adjacent 1's and 2's are unimportant. E.g., a(8)=21. These are (331),(313),(133),(322),(232),(223),(3211),(2311),(1321),(2131),(1132),(2113),(31111),(13111),(11311),(11131),(11113),(2221),(22111),(211111),(1111111). - _David Neil McGrath_, Jul 25 2015
Consider the partitions of 7 with summands initially listed in nonincreasing order. Keep the 1's frozen in position (indicated by "[]") and then allow the other summands to otherwise vary their order: 7; 6,[1]; 5,2; 2,5; 4,3; 3,4; 5,[1,1], 4,2,[1]; 2,4,[1]; 3,3,[1]; 3,3,2; 3,2,3; 2,3,3; 4,[1,1,1]; 3,2,[1,1]; 2,3,[1,1]; 2,2,2,[1]; 3,[1,1,1,1]; 2,2,[1,1,1]; 2,[1,1,1,1,1]; [1,1,1,1,1,1,1]. There are 21 = a(7+1) arrangements in all. - _Gregory L. Simay_, Jun 14 2016

References

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  • Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem II, Missouri Journal of Mathematical Sciences, Vol. 16, No. 1, Winter 2004, pp. 12-17.
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  • R. B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 84.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 4.
  • Marjorie Bicknell and Verner E Hoggatt, Fibonacci's Problem Book, Fibonacci Association, San Jose, Calif., 1974.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pages 24 (Ex. 18), 489, 541.
  • A. Cayley, Theorems in Trigonometry and on Partitions, Messenger of Mathematics, 5 (1876), pp. 164, 188 = Mathematical Papers Vol. 10, n. 634, p. 16.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 84, 111-124, 202-203.
  • B. A. Davey and H. A. Priestley, Introduction to Lattices and Order (2nd edition), CUP, 2002. (See Exercise 1.15.)
  • B. Davis, 'The law of first digits' in 'Science Today' (subsequently renamed '2001') March 1980 p. 55, Times of India, Mumbai.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.2.
  • R. P. Grimaldi, Compositions without the summand 1, Proceedings Thirty-second Southeastern International Conference on Combinatorics, Graph Theory and Computing (Baton Rouge, LA, 2001). Congr. Numer. 152 (2001), 33-43.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.5 The Fibonacci and Related Sequences, pp. 286-288.
  • H. Halberstam and K. F. Roth, Sequences, Oxford, 1966; see Appendix.
  • S. Happersett, "Mathematical meditations", Journal of Mathematics and the Arts, 1 (2007), 29 - 33.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954; see esp. p. 148.
  • V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers. Houghton, Boston, MA, 1969.
  • E. Horowitz and S. Sahni, Fundamentals of Data Structures, Computer Science Press, 1976; p. 338.
  • M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 63.
  • C. Kicey and K. Klimko, Some geometry of Pascal's triangle, Pi Mu Epsilon Journal, 13(4):229-245 (2011).
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 78; Vol. 3, Section 6.2.1.
  • Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.
  • Leonardo of Pisa [Leonardo Pisano], Liber Abaci [The Book of Calculation], 1202.
  • D. Litchfield, D. Goldenheim and C. H. Dietrich, Euclid, Fibonacci and Sketchpad, Math. Teacher, 90 (1997).
  • Lukovits et al., Nanotubes: Number of Kekulé structures and aromaticity, J. Chem. Inf. Comput. Sci, (2003), vol. 43, 609-614. See eq. 2 on page 610.
  • I. Lukovits and D. Janezic, "Enumeration of conjugated circuits in nanotubes", J. Chem. Inf. Comput. Sci., vol. 44, 410-414 (2004). See Table 1, second column.
  • B. Malesevic: Some combinatorial aspects of differential operation composition on the space R^n, Univ. Beograd, Publ. Elektrotehn. Fak., Ser. Mat. 9 (1998), 29-33.
  • G. Mantel, Resten van wederkeerige Reeksen, Nieuw Archief v. Wiskunde, 2nd series, I (1894), 172-184.
  • C. N. Menhinick, The Fibonacci Resonance and other new Golden Ratio discoveries, Onperson, (2015), pages 200-206.
  • S. Mneimneh, Fibonacci in The Curriculum: Not Just a Bad Recurrence, in Proceeding SIGCSE '15 Proceedings of the 46th ACM Technical Symposium on Computer Science Education, Pages 253-258.
  • Hilary I. Okagbue, Muminu O. Adamu, Sheila A. Bishop, Abiodun A. Opanuga, Digit and Iterative Digit Sum of Fibonacci numbers, their identities and powers, International Journal of Applied Engineering Research ISSN 0973-4562 Volume 11, Number 6 (2016) pp 4623-4627.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 49.
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  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 55-58, 255-260.
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  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
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  • S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.
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  • N. N. Vorobiev, Fibonacci Numbers, Birkhauser (Basel; Boston) 2002.
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Links

Crossrefs

Cf. A001622 (phi), A039834 (signed Fibonacci numbers), A001519 (F(2n-1)), A001906 (F(2n)), A001690 (complement), A000213, A000288, A000322, A000383, A060455, A030186, A020695, A020701, A060280 (inv. Eul. Trans), A071679, A099731, A100492, A094216, A094638, A000108, A101399, A101400, A001611, A000071, A157725, A001911, A157726, A006327, A157727, A157728, A157729, A167616, A059929, A144152, A152063, A114690, A003893, A000032, A060441, A000930, A003269, A000957, A057078, A007317, A091867, A104597, A249548, A262342, A001060, A022095, A072649, A163733, A073133, A166861 (Euler Transform), A337009 (Multiset Transf.).
First row of arrays A103323, A172236, A234357. Second row of arrays A099390, A048887, and A092921 (k-generalized Fibonacci numbers).
Cf. also A001175 (Pisano periods), A001177 (Entry points), A001176 (number of zeros in a fundamental period).
Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074.
Fibonacci-Cayley triangle: A327992.
Boustrophedon transforms: A000738, A000744.
Powers: A103323, A105317, A254719.
Numbers of prime factors: A022307 and A038575.
Cf. A061446 (primitive part of Fibonacci numbers), A000010 (comments on product formulas).
Number of digits of F(n): A020909 (base 2), A020911 (base 3), A020912 (base 4), A020913 (base 5), A060384 (base 10), A261585 (base 60).

Programs

  • Axiom
    [fibonacci(n) for n in 0..50]
  • GAP
    Fib:=[0,1];; for n in [3..10^3] do Fib[n]:=Fib[n-1]+Fib[n-2]; od; Fib; # Muniru A Asiru, Sep 03 2017
  • Haskell
    -- Based on code from http://www.haskell.org/haskellwiki/The_Fibonacci_sequence
-- which also has other versions.
fib :: Int -> Integer
fib n = fibs !! n
    where
        fibs = 0 : 1 : zipWith (+) fibs (tail fibs)
{- Example of use: map fib [0..38] Gerald McGarvey, Sep 29 2009 -}
  • Julia
    function fib(n)
   F = BigInt[1 1; 1 0]
   Fn = F^n
   Fn[2, 1]
end
println([fib(n) for n in 0:38]) # Peter Luschny, Feb 23 2017
  • Julia
    # faster
function fibrec(n::Int)
    n == 0 && return (BigInt(0), BigInt(1))
    a, b = fibrec(div(n, 2))
    c = a * (b * 2 - a)
    d = a * a + b * b
    iseven(n) ? (c, d) : (d, c + d)
end
fibonacci(n::Int) = fibrec(n)[1]
println([fibonacci(n) for n in 0:40]) # Peter Luschny, Apr 03 2022
  • Magma
    [Fibonacci(n): n in [0..38]];
  • Maple
    A000045 := proc(n) combinat[fibonacci](n); end;
ZL:=[S, {a = Atom, b = Atom, S = Prod(X,Sequence(Prod(X,b))), X = Sequence(b,card >= 1)}, unlabelled]: seq(combstruct[count](ZL, size=n), n=0..38); # Zerinvary Lajos, Apr 04 2008
spec := [B, {B=Sequence(Set(Z, card>1))}, unlabeled ]: seq(combstruct[count](spec, size=n), n=1..39); # Zerinvary Lajos, Apr 04 2008
# The following Maple command isFib(n) yields true or false depending on whether n is a Fibonacci number or not.
with(combinat): isFib := proc(n) local a: a := proc(n) local j: for j while fibonacci(j) <= n do fibonacci(j) end do: fibonacci(j-1) end proc: evalb(a(n) = n) end proc: # Emeric Deutsch, Nov 11 2014
  • Mathematica
    Table[Fibonacci[k], {k, 0, 50}] (* Mohammad K. Azarian, Jul 11 2015 *)
Table[2^n Sqrt @ Product[(Cos[Pi k/(n + 1)]^2 + 1/4), {k, n}] // FullSimplify, {n, 15}]; (* Kasteleyn's formula specialized, Sarah-Marie Belcastro, Jul 04 2009 *)
LinearRecurrence[{1, 1}, {0, 1}, 40] (* Harvey P. Dale, Aug 03 2014 *)
Fibonacci[Range[0, 20]] (* Eric W. Weisstein, Sep 22 2017 *)
CoefficientList[Series[-(x/(-1 + x + x^2)), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 22 2017 *)
  • Maxima
    makelist(fib(n),n,0,100); /* Martin Ettl, Oct 21 2012 */
  • PARI
    a(n) = fibonacci(n)
  • PARI
    a(n) = imag(quadgen(5)^n)
  • PARI
    a(n)=my(phi=quadgen(5));(phi^n-(-1/phi)^n)/(2*phi-1) \\ Charles R Greathouse IV, Jun 17 2012
  • PARI
    is_A000045=A010056 \\ Characteristic function: see there. - M. F. Hasler, Feb 21 2025
  • Python
    # From Jaap Spies, Jan 05 2007, updated by Peter Luschny, Feb 21 2023:
from itertools import islice
def fib_gen():
    x, y = 0, 1
    while True:
        yield x
        x, y = y, x + y
fib_list = lambda n: list(islice(fib_gen(), n))
  • Python
    is_A000045 = A010056 # See there: Characteristic function. Used e.g. in A377092.
A000045 = lambda n: (4<M. F. Hasler, improving old code from 2023, Feb 20 2025
  • Python
    [(i:=-1)+(j:=1)] + [(j:=i+j)+(i:=j-i) for  in range(100)] # _Jwalin Bhatt, Apr 03 2025
  • Sage
    # Demonstration program from Jaap Spies:
a = sloane.A000045; # choose sequence
print(a)            # This returns the name of the sequence.
print(a(38))        # This returns the 38th term of the sequence.
print(a.list(39))   # This returns a list of the first 39 terms.
  • Sage
    a = BinaryRecurrenceSequence(1,1); print([a(n) for n in range(20)])
# Closed form integer formula with F(1) = 0 from Paul Hankin (see link).
F = lambda n: (4<<(n-1)*(n+2))//((4<<2*(n-1))-(2<<(n-1))-1)&((2<<(n-1))-1)
print([F(n) for n in range(20)]) # Peter Luschny, Aug 28 2016
  • Sage
    print(list(fibonacci_sequence(0, 40))) # Bruno Berselli, Jun 26 2014
  • Scala
    def fibonacci(n: BigInt): BigInt = {
  val zero = BigInt(0)
  def fibTail(n: BigInt, a: BigInt, b: BigInt): BigInt = n match {
    case `zero` => a
    case _ => fibTail(n - 1, b, a + b)
  }
  fibTail(n, 0, 1)
} // Based on "Case 3: Tail Recursion" from Carrasquel (2016) link
(0 to 49).map(fibonacci()) // _Alonso del Arte, Apr 13 2019

Formula

G.f.: x / (1 - x - x^2).
G.f.: Sum_{n>=0} x^n * Product_{k=1..n} (k + x)/(1 + k*x). - Paul D. Hanna, Oct 26 2013
F(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)).
Alternatively, F(n) = ((1/2+sqrt(5)/2)^n - (1/2-sqrt(5)/2)^n)/sqrt(5).
F(n) = F(n-1) + F(n-2) = -(-1)^n F(-n).
F(n) = round(phi^n/sqrt(5)).
F(n+1) = Sum_{j=0..floor(n/2)} binomial(n-j, j).
A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all positive integers n and m. - Michael Somos, Jan 03 2017
E.g.f.: (2/sqrt(5))*exp(x/2)*sinh(sqrt(5)*x/2). - Len Smiley, Nov 30 2001
[0 1; 1 1]^n [0 1] = [F(n); F(n+1)]
x | F(n) ==> x | F(kn).
A sufficient condition for F(m) to be divisible by a prime p is (p - 1) divides m, if p == 1 or 4 (mod 5); (p + 1) divides m, if p == 2 or 3 (mod 5); or 5 divides m, if p = 5. (This is essentially Theorem 180 in Hardy and Wright.) - Fred W. Helenius (fredh(AT)ix.netcom.com), Jun 29 2001
a(n)=F(n) has the property: F(n)*F(m) + F(n+1)*F(m+1) = F(n+m+1). - Miklos Kristof, Nov 13 2003
From Kurmang. Aziz. Rashid, Feb 21 2004: (Start)
Conjecture 1: for n >= 2, sqrt(F(2n+1) + F(2n+2) + F(2n+3) + F(2n+4) + 2*(-1)^n) = (F(2n+1) + 2*(-1)^n)/F(n-1). [For a proof see Comments section.]
Conjecture 2: for n >= 0, (F(n+2)*F(n+3)) - (F(n+1)*F(n+4)) + (-1)^n = 0.
[Two more conjectures removed by Peter Luschny, Nov 17 2017]
Theorem 1: for n >= 0, (F(n+3)^ 2 - F(n+1)^ 2)/F(n+2) = (F(n+3)+ F(n+1)).
Theorem 2: for n >= 0, F(n+10) = 11*F(n+5) + F(n).
Theorem 3: for n >= 6, F(n) = 4*F(n-3) + F(n-6). (End)
Conjecture 2 of Rashid is actually a special case of the general law F(n)*F(m) + F(n+1)*F(m+1) = F(n+m+1) (take n <- n+1 and m <- -(n+4) in this law). - Harmel Nestra (harmel.nestra(AT)ut.ee), Apr 22 2005
Conjecture 2 of Rashid Kurmang simplified: F(n)*F(n+3) = F(n+1)*F(n+2)-(-1)^n. Follows from d'Ocagne's identity: m=n+2. - Alex Ratushnyak, May 06 2012
Conjecture: for all c such that 2-phi <= c < 2*(2-phi) we have F(n) = floor(phi*a(n-1)+c) for n > 2. - Gerald McGarvey, Jul 21 2004
For x > phi, Sum_{n>=0} F(n)/x^n = x/(x^2 - x - 1). - Gerald McGarvey, Oct 27 2004
F(n+1) = exponent of the n-th term in the series f(x, 1) determined by the equation f(x, y) = xy + f(xy, x). - Jonathan Sondow, Dec 19 2004
a(n-1) = Sum_{k=0..n} (-1)^k*binomial(n-ceiling(k/2), floor(k/2)). - Benoit Cloitre, May 05 2005
a(n) = Sum_{k=0..n} abs(A108299(n, k)). - Reinhard Zumkeller, Jun 01 2005
a(n) = A001222(A000304(n)).
F(n+1) = Sum_{k=0..n} binomial((n+k)/2, (n-k)/2)(1+(-1)^(n-k))/2. - Paul Barry, Aug 28 2005
Fibonacci(n) = Product_{j=1..ceiling(n/2)-1} (1 + 4(cos(j*Pi/n))^2). [Bicknell and Hoggatt, pp. 47-48.] - Emeric Deutsch, Oct 15 2006
F(n) = 2^-(n-1)*Sum_{k=0..floor((n-1)/2)} binomial(n,2*k+1)*5^k. - Hieronymus Fischer, Feb 07 2006
a(n) = (b(n+1) + b(n-1))/n where {b(n)} is the sequence A001629. - Sergio Falcon, Nov 22 2006
F(n*m) = Sum_{k = 0..m} binomial(m,k)*F(n-1)^k*F(n)^(m-k)*F(m-k). The generating function of F(n*m) (n fixed, m = 0,1,2,...) is G(x) = F(n)*x / ((1 - F(n-1)*x)^2 - F(n)*x*(1 - F(n-1)*x) - (F(n)*x)^2). E.g., F(15) = 610 = F(5*3) = binomial(3,0)* F(4)^0*F(5)^3*F(3) + binomial(3,1)* F(4)^1*F(5)^2*F(2) + binomial(3,2)* F(4)^2*F(5)^1*F(1) + binomial(3,3)* F(4)^3*F(5)^0*F(0) = 1*1*125*2 + 3*3*25*1 + 3*9*5*1 + 1*27*1*0 = 250 + 225 + 135 + 0 = 610. - Miklos Kristof, Feb 12 2007
From Miklos Kristof, Mar 19 2007: (Start)
Let L(n) = A000032(n) = Lucas numbers. Then:
For a >= b and odd b, F(a+b) + F(a-b) = L(a)*F(b).
For a >= b and even b, F(a+b) + F(a-b) = F(a)*L(b).
For a >= b and odd b, F(a+b) - F(a-b) = F(a)*L(b).
For a >= b and even b, F(a+b) - F(a-b) = L(a)*F(b).
F(n+m) + (-1)^m*F(n-m) = F(n)*L(m);
F(n+m) - (-1)^m*F(n-m) = L(n)*F(m);
F(n+m+k) + (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = F(n)*L(m)*L(k);
F(n+m+k) - (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = L(n)*L(m)*F(k);
F(n+m+k) + (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = L(n)*F(m)*L(k);
F(n+m+k) - (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = 5*F(n)*F(m)*F(k). (End)
A corollary to Kristof 2007 is 2*F(a+b) = F(a)*L(b) + L(a)*F(b). - Graeme McRae, Apr 24 2014
For n > m, the sum of the 2m consecutive Fibonacci numbers F(n-m-1) thru F(n+m-2) is F(n)*L(m) if m is odd, and L(n)*F(m) if m is even (see the McRae link). - Graeme McRae, Apr 24 2014.
F(n) = b(n) + (p-1)*Sum_{k=2..n-1} floor(b(k)/p)*F(n-k+1) where b(k) is the digital sum analog of the Fibonacci recurrence, defined by b(k) = ds_p(b(k-1)) + ds_p(b(k-2)), b(0)=0, b(1)=1, ds_p=digital sum base p. Example for base p=10: F(n) = A010077(n) + 9*Sum_{k=2..n-1} A059995(A010077(k))*F(n-k+1). - Hieronymus Fischer, Jul 01 2007
F(n) = b(n)+p*Sum_{k=2..n-1} floor(b(k)/p)*F(n-k+1) where b(k) is the digital product analog of the Fonacci recurrence, defined by b(k) = dp_p(b(k-1)) + dp_p(b(k-2)), b(0)=0, b(1)=1, dp_p=digital product base p. Example for base p=10: F(n) = A074867(n) + 10*Sum_{k=2..n-1} A059995(A074867(k))*F(n-k+1). - Hieronymus Fischer, Jul 01 2007
a(n) = denominator of continued fraction [1,1,1,...] (with n ones); e.g., 2/3 = continued fraction [1,1,1]; where barover[1] = [1,1,1,...] = 0.6180339.... - Gary W. Adamson, Nov 29 2007
F(n + 3) = 2F(n + 2) - F(n), F(n + 4) = 3F(n + 2) - F(n), F(n + 8) = 7F(n + 4) - F(n), F(n + 12) = 18F(n + 6) - F(n). - Paul Curtz, Feb 01 2008
a(2^n) = Product_{i=0..n-2} B(i) where B(i) is A001566. Example 3*7*47 = F(16). - Kenneth J Ramsey, Apr 23 2008
a(n+1) = Sum_{k=0..n} A109466(n,k)*(-1)^(n-k). -Philippe Deléham, Oct 26 2008
a(n) = Sum_{l_1=0..n+1} Sum_{l_2=0..n}...Sum_{l_i=0..n-i}... Sum_{l_n=0..1} delta(l_1,l_2,...,l_i,...,l_n), where delta(l_1,l_2,...,l_i,...,l_n) = 0 if any l_i + l_(i+1) >= 2 for i=1..n-1 and delta(l_1,l_2,...,l_i,...,l_n) = 1 otherwise. - Thomas Wieder, Feb 25 2009
a(n+1) = 2^n sqrt(Product_{k=1..n} cos(k Pi/(n+1))^2+1/4) (Kasteleyn's formula specialized). - Sarah-Marie Belcastro, Jul 04 2009
a(n+1) = Sum_{k=floor(n/2) mod 5} C(n,k) - Sum_{k=floor((n+5)/2) mod 5} C(n,k) = A173125(n) - A173126(n) = |A054877(n)-A052964(n-1)|. - Henry Bottomley, Feb 10 2010
If p[i] = modp(i,2) and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n >= 1, a(n)=det A. - Milan Janjic, May 02 2010
Limit_{k->oo} F(k+n)/F(k) = (L(n) + F(n)*sqrt(5))/2 with the Lucas numbers L(n) = A000032(n). - Johannes W. Meijer, May 27 2010
For n >= 1, F(n) = round(log_2(2^(phi*F(n-1)) + 2^(phi*F(n-2)))), where phi is the golden ratio. - Vladimir Shevelev, Jun 24 2010, Jun 27 2010
For n >= 1, a(n+1) = ceiling(phi*a(n)), if n is even and a(n+1) = floor(phi*a(n)), if n is odd (phi = golden ratio). - Vladimir Shevelev, Jul 01 2010
a(n) = 2*a(n-2) + a(n-3), n > 2. - Gary Detlefs, Sep 08 2010
a(2^n) = Product_{i=0..n-1} A000032(2^i). - Vladimir Shevelev, Nov 28 2010
a(n)^2 - a(n-1)^2 = a(n+1)*a(n-2), see A121646.
a(n) = sqrt((-1)^k*(a(n+k)^2 - a(k)*a(2n+k))), for any k. - Gary Detlefs, Dec 03 2010
F(2*n) = F(n+2)^2 - F(n+1)^2 - 2*F(n)^2. - Richard R. Forberg, Jun 04 2011
From Artur Jasinski, Nov 17 2011: (Start)
(-1)^(n+1) = F(n)^2 + F(n)*F(1+n) - F(1+n)^2.
F(n) = F(n+2) - 1 + (F(n+1))^4 + 2*(F(n+1)^3*F(n+2)) - (F(n+1)*F(n+2))^2 - 2*F(n+1)(F(n+2))^3 + (F(n+2))^4 - F(n+1). (End)
F(n) = 1 + Sum_{x=1..n-2} F(x). - Joseph P. Shoulak, Feb 05 2012
F(n) = 4*F(n-2) - 2*F(n-3) - F(n-6). - Gary Detlefs, Apr 01 2012
F(n) = round(phi^(n+1)/(phi+2)). - Thomas Ordowski, Apr 20 2012
From Sergei N. Gladkovskii, Jun 03 2012: (Start)
G.f.: A(x) = x/(1-x-x^2) = G(0)/sqrt(5) where G(k) = 1 - ((-1)^k)*2^k/(a^k - b*x*a^k*2^k/(b*x*2^k - 2*((-1)^k)*c^k/G(k+1))) and a=3+sqrt(5), b=1+sqrt(5), c=3-sqrt(5); (continued fraction, 3rd kind, 3-step).
Let E(x) be the e.g.f., i.e.,
E(x) = 1*x + (1/2)*x^2 + (1/3)*x^3 + (1/8)*x^4 + (1/24)*x^5 + (1/90)*x^6 + (13/5040)*x^7 + ...; then
E(x) = G(0)/sqrt(5); G(k) = 1 - ((-1)^k)*2^k/(a^k - b*x*a^k*2^k/(b*x*2^k - 2*((-1)^k)*(k+1)*c^k/G(k+1))), where a=3+sqrt(5), b=1+sqrt(5), c=3-sqrt(5); (continued fraction, 3rd kind, 3-step).
(End)
From Hieronymus Fischer, Nov 30 2012: (Start)
F(n) = 1 + Sum_{j_1=1..n-2} 1 + Sum_{j_1=1..n-2} Sum_{j_2=1..j_1-2} 1 + Sum_{j_1=1..n-2} Sum_{j_2=1..j_1-2} Sum_{j_3=1..j_2-2} 1 + ... + Sum_{j_1=1..n-2} Sum_{j_2=1..j_1-2} Sum_{j_3=1..j_2-2} ... Sum_{j_k=1..j_(k-1)-2} 1, where k = floor((n-1)/2).
Example: F(6) = 1 + Sum_{j=1..4} 1 + Sum_{j=1..4} Sum_{k=1..(j-2)} 1 + 0 = 1 + (1 + 1 + 1 + 1) + (1 + (1 + 1)) = 8.
F(n) = Sum_{j=0..k} S(j+1,n-2j), where k = floor((n-1)/2) and the S(j,n) are the n-th j-simplex sums: S(1,n) = 1 is the 1-simplex sum, S(2,n) = Sum_{k=1..n} S(1,k) = 1+1+...+1 = n is the 2-simplex sum, S(3,n) = Sum_{k=1..n} S(2,k) = 1+2+3+...+n is the 3-simplex sum (= triangular numbers = A000217), S(4,n) = Sum_{k=1..n} S(3,k) = 1+3+6+...+n(n+1)/2 is the 4-simplex sum (= tetrahedral numbers = A000292) and so on.
Since S(j,n) = binomial(n-2+j,j-1), the formula above equals the well-known binomial formula, essentially. (End)
G.f.: A(x) = x / (1 - x / (1 - x / (1 + x))). - Michael Somos, Jan 04 2013
Sum_{n >= 1} (-1)^(n-1)/(a(n)*a(n+1)) = 1/phi (phi=golden ratio). - Vladimir Shevelev, Feb 22 2013
From Raul Prisacariu, Oct 29 2023: (Start)
For odd k, Sum_{n >= 1} a(k)^2*(-1)^(n-1)/(a(k*n)*a(k*n+k)) = phi^(-k).
For even k, Sum_{n >= 1} a(k)^2/(a(k*n)*a(k*n+k)) = phi^(-k). (End)
From Vladimir Shevelev, Feb 24 2013: (Start)
(1) Expression a(n+1) via a(n): a(n+1) = (a(n) + sqrt(5*(a(n))^2 + 4*(-1)^n))/2;
(2) Sum_{k=1..n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);
(3) a(n)/a(n+1) = 1/phi + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)
F(n+1) = F(n)/2 + sqrt((-1)^n + 5*F(n)^2/4), n >= 0. F(n+1) = U_n(i/2)/i^n, (U:= Chebyshev polynomial of the 2nd kind, i=sqrt(-1)). - Bill Gosper, Mar 04 2013
G.f.: -Q(0) where Q(k) = 1 - (1+x)/(1 - x/(x - 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Mar 06 2013
G.f.: x - 1 - 1/x + (1/x)/Q(0), where Q(k) = 1 - (k+1)*x/(1 - x/(x - (k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Apr 23 2013
G.f.: x*G(0), where G(k) = 1 + x*(1+x)/(1 - x*(1+x)/(x*(1+x) + 1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 08 2013
G.f.: x^2 - 1 + 2*x^2/(W(0)-2), where W(k) = 1 + 1/(1 - x*(k + x)/( x*(k+1 + x) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 28 2013
G.f.: Q(0) - 1, where Q(k) = 1 + x^2 + (k+2)*x - x*(k+1 + x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013
Let b(n) = b(n-1) + b(n-2), with b(0) = 0, b(1) = phi. Then, for n >= 2, F(n) = floor(b(n-1)) if n is even, F(n) = ceiling(b(n-1)), if n is odd, with convergence. - Richard R. Forberg, Jan 19 2014
a(n) = Sum_{t1*g(1)+t2*g(2)+...+tn*g(n)=n} multinomial(t1+t2+...+tn,t1,t2,...,tn), where g(k)=2*k-1. - Mircea Merca, Feb 27 2014
F(n) = round(sqrt(F(n-1)^2 + F(n)^2 + F(n+1)^2)/2), for n > 0. This rule appears to apply to any sequence of the form a(n) = a(n-1) + a(n-2), for any two values of a(0) and a(1), if n is sufficiently large. - Richard R. Forberg, Jul 27 2014
F(n) = round(2/(1/F(n) + 1/F(n+1) + 1/F(n+2))), for n > 0. This rule also appears to apply to any sequence of the form a(n) = a(n-1) + a(n-2), for any two values of a(0) and a(1), if n is sufficiently large. - Richard R. Forberg, Aug 03 2014
F(n) = round(1/(Sum_{j>=n+2} 1/F(j))). - Richard R. Forberg, Aug 14 2014
a(n) = hypergeometric([-n/2+1/2, -n/2+1], [-n+1], -4) for n >= 2. - Peter Luschny, Sep 19 2014
Limit_{n -> oo} (log F(n+1)/log F(n))^n = e. - Thomas Ordowski, Oct 06 2014
F(n) = (L(n+1)^2 - L(n-1)^2)/(5*L(n)), where L(n) is A000032(n), with a similar inverse relationship. - Richard R. Forberg, Nov 17 2014
Consider the graph G[1-vertex;1-loop,2-loop] in comment above. Construct the power matrix array T(n,j) = [A^*j]*[S^*(j-1)] where A=(1,1,0,...) and S=(0,1,0,...)(A063524). [* is convolution operation] Define S^*0=I with I=(1,0,...). Then T(n,j) counts n-walks containing (j) loops and a(n-1) = Sum_{j=1..n} T(n,j). - David Neil McGrath, Nov 21 2014
Define F(-n) to be F(n) for n odd and -F(n) for n even. Then for all n and k, F(n) = F(k)*F(n-k+3) - F(k-1)*F(n-k+2) - F(k-2)*F(n-k) + (-1)^k*F(n-2k+2). - Charlie Marion, Dec 04 2014
F(n+k)^2 - L(k)*F(n)*F(n+k) + (-1)^k*F(n)^2 = (-1)^n*F(k)^2, if L(k) = A000032(k). - Alexander Samokrutov, Jul 20 2015
F(2*n) = F(n+1)^2 - F(n-1)^2, similar to Koshy (D) and Forberg 2011, but different. - Hermann Stamm-Wilbrandt, Aug 12 2015
F(n+1) = ceiling( (1/phi)*Sum_{k=0..n} F(k) ). - Tom Edgar, Sep 10 2015
a(n) = (L(n-3) + L(n+3))/10 where L(n)=A000032(n). - J. M. Bergot, Nov 25 2015
From Bob Selcoe, Mar 27 2016: (Start)
F(n) = (F(2n+k+1) - F(n+1)*F(n+k+1))/F(n+k), k >= 0.
Thus when k=0: F(n) = sqrt(F(2n+1) - F(n+1)^2).
F(n) = (F(3n) - F(n+1)^3 + F(n-1)^3)^(1/3).
F(n+2k) = binomial transform of any subsequence starting with F(n). Example F(6)=8: 1*8 = F(6)=8; 1*8 + 1*13 = F(8)=21; 1*8 + 2*13 + 1*21 = F(10)=55; 1*8 + 3*13 + 3*21 + 1*34 = F(12)=144, etc. This formula applies to Fibonacci-type sequences with any two seed values for a(0) and a(1) (e.g., Lucas sequence A000032: a(0)=2, a(1)=1).
(End)
F(n) = L(k)*F(n-k) + (-1)^(k+1)*F(n-2k) for all k >= 0, where L(k) = A000032(k). - Anton Zakharov, Aug 02 2016
From Ilya Gutkovskiy, Aug 03 2016: (Start)
a(n) = F_n(1), where F_n(x) are the Fibonacci polynomials.
Inverse binomial transform of A001906.
Number of zeros in substitution system {0 -> 11, 1 -> 1010} at step n from initial string "1" (1 -> 1010 -> 101011101011 -> ...) multiplied by 1/A000079(n). (End)
For n >= 2, a(n) = 2^(n^2+n) - (4^n-2^n-1)*floor(2^(n^2+n)/(4^n-2^n-1)) - 2^n*floor(2^(n^2) - (2^n-1-1/2^n)*floor(2^(n^2+n)/(4^n-2^n-1))). - Benoit Cloitre, Apr 17 2017
f(n+1) = Sum_{j=0..floor(n/2)} Sum_{k=0..j} binomial(n-2j,k)*binomial(j,k). - Tony Foster III, Sep 04 2017
F(n) = Sum_{k=0..floor((n-1)/2)} ( (n-k-1)! / ((n-2k-1)! * k!) ). - Zhandos Mambetaliyev, Nov 08 2017
For x even, F(n) = (F(n+x) + F(n-x))/L(x). For x odd, F(n) = (F(n+x) - F(n-x))/L(x) where n >= x in both cases. Therefore F(n) = F(2*n)/L(n) for n >= 0. - David James Sycamore, May 04 2018
From Isaac Saffold, Jul 19 2018: (Start)
Let [a/p] denote the Legendre symbol. Then, for an odd prime p:
F(p+n) == [5/p]*F([5/p]+n) (mod p), if [5/p] = 1 or -1.
F(p+n) == 3*F(n) (mod p), if [5/p] = 0 (i.e., p = 5).
This is true for negative-indexed terms as well, if this sequence is extended by the negafibonacci numbers (i.e., F(-n) = A039834(n)). (End)
a(n) = A094718(4, n). a(n) = A101220(0, j, n).
a(n) = A090888(0, n+1) = A118654(0, n+1) = A118654(1, n-1) = A109754(0, n) = A109754(1, n-1), for n > 0.
a(n) = (L(n-3) + L(n-2) + L(n-1) + L(n))/5 with L(n)=A000032(n). - Art Baker, Jan 04 2019
F(n) = F(k-1)*F(abs(n-k-2)) + F(k-1)*F(n-k-1) + F(k)*F(abs(n-k-2)) + 2*F(k)*F(n-k-1), for n > k > 0. - Joseph M. Shunia, Aug 12 2019
F(n) = F(n-k+2)*F(k-1) + F(n-k+1)*F(k-2) for all k such that 2 <= k <= n. - Michael Tulskikh, Oct 09 2019
F(n)^2 - F(n+k)*F(n-k) = (-1)^(n+k) * F(k)^2 for 2 <= k <= n [Catalan's identity]. - Hermann Stamm-Wilbrandt, May 07 2021
Sum_{n>=1} 1/a(n) = A079586 is the reciprocal Fibonacci constant. - Gennady Eremin, Aug 06 2021
a(n) = Product_{d|n} b(d) = Product_{k=1..n} b(gcd(n,k))^(1/phi(n/gcd(n,k))) = Product_{k=1..n} b(n/gcd(n,k))^(1/phi(n/gcd(n,k))) where b(n) = A061446(n) = primitive part of a(n), phi(n) = A000010(n). - Richard L. Ollerton, Nov 08 2021
a(n) = 2*i^(1-n)*sin(n*arccos(i/2))/sqrt(5), i=sqrt(-1). - Bill Gosper, May 05 2022
a(n) = i^(n-1)*sin(n*c)/sin(c) = i^(n-1)*sin(c*n)*csc(c), where c = Pi/2 + i*arccsch(2). - Peter Luschny, May 23 2022
F(2n) = Sum_{k=1..n} (k/5)*binomial(2n, n+k), where (k/5) is the Legendre or Jacobi Symbol; F(2n+1)= Sum_{k=1..n} (-(k+2)/5)*binomial(2n+1, n+k), where (-(k+2)/5) is the Legendre or Jacobi Symbol. For example, F(10) = 1*binomial(10,6) - 1*binomial(10,7) - 1*binomial(10,8) + 1*binomial(10,9) + 0*binomial(10,10), F(11) = 1*binomial(11,6) - 1*binomial(11,7) + 0*binomial(11,8) - 1*binomial(11,9) + 1*binomial(11,10) + 1*binomial(11,11). - Yike Li, Aug 21 2022
For n > 0, 1/F(n) = Sum_{k>=1} F(n*k)/(F(n+2)^(k+1)). - Diego Rattaggi, Oct 26 2022
From Andrea Pinos, Dec 02 2022: (Start)
For n == 0 (mod 4): F(n) = F((n+2)/2)*( F(n/2) + F((n/2)-2) ) + 1;
For n == 1 (mod 4): F(n) = F((n-1)/2)*( F((n-1)/2) + F(2+(n-1)/2) ) + 1;
For n == 2 (mod 4): F(n) = F((n-2)/2)*( F(n/2) + F((n/2)+2) ) + 1;
For n == 3 (mod 4): F(n) = F((n-1)/2)*( F((n-1)/2) + F(2+(n-1)/2) ) - 1. (End)
F(n) = Sum_{i=0..n-1} F(i)^2 / F(n-1). - Jules Beauchamp, May 03 2025

A000005 d(n) (also called tau(n) or sigma_0(n)), the number of divisors of n.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 2, 4, 4, 5, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 4, 6, 2, 8, 2, 6, 4, 4, 4, 9, 2, 4, 4, 8, 2, 8, 2, 6, 6, 4, 2, 10, 3, 6, 4, 6, 2, 8, 4, 8, 4, 4, 2, 12, 2, 4, 6, 7, 4, 8, 2, 6, 4, 8, 2, 12, 2, 4, 6, 6, 4, 8, 2, 10, 5, 4, 2, 12, 4, 4, 4, 8, 2, 12, 4, 6, 4, 4, 4, 12, 2, 6, 6, 9, 2, 8, 2, 8
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

If the canonical factorization of n into prime powers is Product p^e(p) then d(n) = Product (e(p) + 1). More generally, for k > 0, sigma_k(n) = Product_p ((p^((e(p)+1)*k))-1)/(p^k-1) is the sum of the k-th powers of the divisors of n.
Number of ways to write n as n = x*y, 1 <= x <= n, 1 <= y <= n. For number of unordered solutions to x*y=n, see A038548.
Note that d(n) is not the number of Pythagorean triangles with radius of the inscribed circle equal to n (that is A078644). For number of primitive Pythagorean triangles having inradius n, see A068068(n).
Number of factors in the factorization of the polynomial x^n-1 over the integers. - T. D. Noe, Apr 16 2003
Also equal to the number of partitions p of n such that all the parts have the same cardinality, i.e., max(p)=min(p). - Giovanni Resta, Feb 06 2006
Equals A127093 as an infinite lower triangular matrix * the harmonic series, [1/1, 1/2, 1/3, ...]. - Gary W. Adamson, May 10 2007
For odd n, this is the number of partitions of n into consecutive integers. Proof: For n = 1, clearly true. For n = 2k + 1, k >= 1, map each (necessarily odd) divisor to such a partition as follows: For 1 and n, map k + (k+1) and n, respectively. For any remaining divisor d <= sqrt(n), map (n/d - (d-1)/2) + ... + (n/d - 1) + (n/d) + (n/d + 1) + ... + (n/d + (d-1)/2) {i.e., n/d plus (d-1)/2 pairs each summing to 2n/d}. For any remaining divisor d > sqrt(n), map ((d-1)/2 - (n/d - 1)) + ... + ((d-1)/2 - 1) + (d-1)/2 + (d+1)/2 + ((d+1)/2 + 1) + ... + ((d+1)/2 + (n/d - 1)) {i.e., n/d pairs each summing to d}. As all such partitions must be of one of the above forms, the 1-to-1 correspondence and proof is complete. - Rick L. Shepherd, Apr 20 2008
Number of subgroups of the cyclic group of order n. - Benoit Jubin, Apr 29 2008
Equals row sums of triangle A143319. - Gary W. Adamson, Aug 07 2008
Equals row sums of triangle A159934, equivalent to generating a(n) by convolving A000005 prefaced with a 1; (1, 1, 2, 2, 3, 2, ...) with the INVERTi transform of A000005, (A159933): (1, 1,-1, 0, -1, 2, ...). Example: a(6) = 4 = (1, 1, 2, 2, 3, 2) dot (2, -1, 0, -1, 1, 1) = (2, -1, 0, -2, 3, 2) = 4. - Gary W. Adamson, Apr 26 2009
Number of times n appears in an n X n multiplication table. - Dominick Cancilla, Aug 02 2010
Number of k >= 0 such that (k^2 + k*n + k)/(k + 1) is an integer. - Juri-Stepan Gerasimov, Oct 25 2015
The only numbers k such that tau(k) >= k/2 are 1,2,3,4,6,8,12. - Michael De Vlieger, Dec 14 2016
a(n) is also the number of partitions of 2*n into equal parts, minus the number of partitions of 2*n into consecutive parts. - Omar E. Pol, May 03 2017
From Tomohiro Yamada, Oct 27 2020: (Start)
Let k(n) = log d(n)*log log n/(log 2 * log n), then lim sup k(n) = 1 (Hardy and Wright, Chapter 18, Theorem 317) and k(n) <= k(6983776800) = 1.537939... (the constant A280235) for every n (Nicolas and Robin, 1983).
There exist infinitely many n such that d(n) = d(n+1) (Heath-Brown, 1984). The number of such integers n <= x is at least c*x/(log log x)^3 (Hildebrand, 1987) but at most O(x/sqrt(log log x)) (Erdős, Carl Pomerance and Sárközy, 1987). (End)
Number of 2D grids of n congruent rectangles with two different side lengths, in a rectangle, modulo rotation (cf. A038548 for squares instead of rectangles). Also number of ways to arrange n identical objects in a rectangle (NOT modulo rotation, cf. A038548 for modulo rotation); cf. A007425 and A140773 for the 3D case. - Manfred Boergens, Jun 08 2021
The constant quoted above from Nicolas and Robin, 6983776800 = 2^5 * 3^3 * 5^2 * 7 * 11 * 13 * 17 * 19, appears arbitrary, but interestingly equals 2 * A095849(36). That second factor is highly composite and deeply composite. - Hal M. Switkay, Aug 08 2025

Examples

			G.f. = x + 2*x^2 + 2*x^3 + 3*x^4 + 2*x^5 + 4*x^6 + 2*x^7 + 4*x^8 + 3*x^9 + ...

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 840.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 38.
  • G. Chrystal, Algebra: An elementary text-book for the higher classes of secondary schools and for colleges, 6th ed, Chelsea Publishing Co., New York 1959 Part II, p. 345, Exercise XXI(16). MR0121327 (22 #12066)
  • G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, Cambridge, University Press, 1940, p. 55.
  • G. H. Hardy and E. M. Wright, revised by D. R. Heath-Brown and J. H. Silverman, An Introduction to the Theory of Numbers, 6th ed., Oxford Univ. Press, 2008.
  • K. Knopp, Theory and Application of Infinite Series, Blackie, London, 1951, p. 451.
  • D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Chap. II. (For inequalities, etc.)
  • S. Ramanujan, Collected Papers, Ed. G. H. Hardy et al., Cambridge 1927; Chelsea, NY, 1962. Has many references to this sequence. - N. J. A. Sloane, Jun 02 2014
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • B. Spearman and K. S. Williams, Handbook of Estimates in the Theory of Numbers, Carleton Math. Lecture Note Series No. 14, 1975; see p. 2.1.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 285.
  • E. C. Titchmarsh, The Theory of Functions, Oxford, 1938, p. 160.
  • Terence Tao, Poincaré's Legacies, Part I, Amer. Math. Soc., 2009, see pp. 31ff for upper bounds on d(n).

Links

Crossrefs

See A002183, A002182 for records. See A000203 for the sum-of-divisors function sigma(n).
For partial sums see A006218.
Cf. A007427 (Dirichlet Inverse), A001227, A001826, A001842, A005237, A005238, A006558, A006601, A019273, A027750, A034296, A039665, A049051, A049820, A051731, A061017, A066446, A091202, A091220, A106737, A115361, A127093, A129372, A129510, A143319, A156552, A159933, A159934, A163280, A183063, A237665, A263730.
Factorizations into given number of factors: writing n = x*y (A038548, unordered, A000005, ordered), n = x*y*z (A034836, unordered, A007425, ordered), n = w*x*y*z (A007426, ordered).
Cf. A000010, A095849.
Cf. A098198 (Dgf at s=2), A183030 (Dgf at s=3), A183031 (Dgf at s=3).

Programs

  • GAP
    List([1..150],n->Tau(n)); # Muniru A Asiru, Mar 05 2019
  • Haskell
    divisors 1 = [1]
divisors n = (1:filter ((==0) . rem n)
               [2..n `div` 2]) ++ [n]
a = length . divisors
-- James Spahlinger, Oct 07 2012
  • Haskell
    a000005 = product . map (+ 1) . a124010_row  -- Reinhard Zumkeller, Jul 12 2013
  • Julia
    function tau(n)
    i = 2; num = 1
    while i * i <= n
        if rem(n, i) == 0
            e = 0
            while rem(n, i) == 0
                e += 1
                n = div(n, i)
            end
            num *= e + 1
        end
        i += 1
    end
    return n > 1 ? num + num : num
end
println([tau(n) for n in 1:104])  # Peter Luschny, Sep 03 2023
  • Magma
    [ NumberOfDivisors(n) : n in [1..100] ]; // Sergei Haller (sergei(AT)sergei-haller.de), Dec 21 2006
  • Maple
    with(numtheory): A000005 := tau; [ seq(tau(n), n=1..100) ];
  • Mathematica
    Table[DivisorSigma[0, n], {n, 100}] (* Enrique Pérez Herrero, Aug 27 2009 *)
CoefficientList[Series[(Log[1 - q] + QPolyGamma[1, q])/(q Log[q]), {q, 0, 100}], q] (* Vladimir Reshetnikov, Apr 23 2013 *)
a[ n_] := SeriesCoefficient[ (QPolyGamma[ 1, q] + Log[1 - q]) / Log[q], {q, 0, Abs@n}]; (* Michael Somos, Apr 25 2013 *)
a[ n_] := SeriesCoefficient[ q/(1 - q)^2 QHypergeometricPFQ[ {q, q}, {q^2, q^2}, q, q^2], {q, 0, Abs@n}]; (* Michael Somos, Mar 05 2014 *)
a[n_] := SeriesCoefficient[q/(1 - q) QHypergeometricPFQ[{q, q}, {q^2}, q, q], {q, 0, Abs@n}] (* Mats Granvik, Apr 15 2015 *)
With[{M=500},CoefficientList[Series[(2x)/(1-x)-Sum[x^k (1-2x^k)/(1-x^k),{k,M}],{x,0,M}],x]] (* Mamuka Jibladze, Aug 31 2018 *)
  • MuPAD
    numlib::tau (n)$ n=1..90 // Zerinvary Lajos, May 13 2008
  • PARI
    {a(n) = if( n==0, 0, numdiv(n))}; /* Michael Somos, Apr 27 2003 */
  • PARI
    {a(n) = n=abs(n); if( n<1, 0, direuler( p=2, n, 1 / (1 - X)^2)[n])}; /* Michael Somos, Apr 27 2003 */
  • PARI
    {a(n)=polcoeff(sum(m=1, n+1, sumdiv(m, d, (-log(1-x^(m/d) +x*O(x^n) ))^d/d!)), n)} \\ Paul D. Hanna, Aug 21 2014
  • Python
    from sympy import divisor_count
for n in range(1, 20): print(divisor_count(n), end=', ') # Stefano Spezia, Nov 05 2018
  • Sage
    [sigma(n, 0) for n in range(1, 105)]  # Zerinvary Lajos, Jun 04 2009

Formula

If n is written as 2^z*3^y*5^x*7^w*11^v*... then a(n)=(z+1)*(y+1)*(x+1)*(w+1)*(v+1)*...
a(n) = 2 iff n is prime.
G.f.: Sum_{n >= 1} a(n) x^n = Sum_{k>0} x^k/(1-x^k). This is usually called THE Lambert series (see Knopp, Titchmarsh).
a(n) = A083888(n) + A083889(n) + A083890(n) + A083891(n) + A083892(n) + A083893(n) + A083894(n) + A083895(n) + A083896(n).
a(n) = A083910(n) + A083911(n) + A083912(n) + A083913(n) + A083914(n) + A083915(n) + A083916(n) + A083917(n) + A083918(n) + A083919(n).
Multiplicative with a(p^e) = e+1. - David W. Wilson, Aug 01 2001
a(n) <= 2 sqrt(n) [see Mitrinovich, p. 39, also A046522].
a(n) is odd iff n is a square. - Reinhard Zumkeller, Dec 29 2001
a(n) = Sum_{k=1..n} f(k, n) where f(k, n) = 1 if k divides n, 0 otherwise (Mobius transform of A000012). Equivalently, f(k, n) = (1/k)*Sum_{l=1..k} z(k, l)^n with z(k, l) the k-th roots of unity. - Ralf Stephan, Dec 25 2002
G.f.: Sum_{k>0} ((-1)^(k+1) * x^(k * (k + 1)/2) / ((1 - x^k) * Product_{i=1..k} (1 - x^i))). - Michael Somos, Apr 27 2003
a(n) = n - Sum_{k=1..n} (ceiling(n/k) - floor(n/k)). - Benoit Cloitre, May 11 2003
a(n) = A032741(n) + 1 = A062011(n)/2 = A054519(n) - A054519(n-1) = A006218(n) - A006218(n-1) = 1 + Sum_{k=1..n-1} A051950(k+1). - Ralf Stephan, Mar 26 2004
G.f.: Sum_{k>0} x^(k^2)*(1+x^k)/(1-x^k). Dirichlet g.f.: zeta(s)^2. - Michael Somos, Apr 05 2003
Sequence = M*V where M = A129372 as an infinite lower triangular matrix and V = ruler sequence A001511 as a vector: [1, 2, 1, 3, 1, 2, 1, 4, ...]. - Gary W. Adamson, Apr 15 2007
Sequence = M*V, where M = A115361 is an infinite lower triangular matrix and V = A001227, the number of odd divisors of n, is a vector: [1, 1, 2, 1, 2, 2, 2, ...]. - Gary W. Adamson, Apr 15 2007
Row sums of triangle A051731. - Gary W. Adamson, Nov 02 2007
Sum_{n>0} a(n)/(n^n) = Sum_{n>0, m>0} 1/(n*m). - Gerald McGarvey, Dec 15 2007
Logarithmic g.f.: Sum_{n>=1} a(n)/n * x^n = -log( Product_{n>=1} (1-x^n)^(1/n) ). - Joerg Arndt, May 03 2008
a(n) = Sum_{k=1..n} (floor(n/k) - floor((n-1)/k)). - Enrique Pérez Herrero, Aug 27 2009
a(s) = 2^omega(s), if s > 1 is a squarefree number (A005117) and omega(s) is: A001221. - Enrique Pérez Herrero, Sep 08 2009
a(n) = A048691(n) - A055205(n). - Reinhard Zumkeller, Dec 08 2009
For n > 1, a(n) = 2 + Sum_{k=2..n-1} floor((cos(Pi*n/k))^2). And floor((cos(Pi*n/k))^2) = floor(1/4 * e^(-(2*i*Pi*n)/k) + 1/4 * e^((2*i*Pi*n)/k) + 1/2). - Eric Desbiaux, Mar 09 2010, corrected Apr 16 2011
a(n) = 1 + Sum_{k=1..n} (floor(2^n/(2^k-1)) mod 2) for every n. - Fabio Civolani (civox(AT)tiscali.it), Mar 12 2010
From Vladimir Shevelev, May 22 2010: (Start)
(Sum_{d|n} a(d))^2 = Sum_{d|n} a(d)^3 (J. Liouville).
Sum_{d|n} A008836(d)*a(d)^2 = A008836(n)*Sum_{d|n} a(d). (End)
a(n) = sigma_0(n) = 1 + Sum_{m>=2} Sum_{r>=1} (1/m^(r+1))*Sum_{j=1..m-1} Sum_{k=0..m^(r+1)-1} e^(2*k*Pi*i*(n+(m-j)*m^r)/m^(r+1)). - A. Neves, Oct 04 2010
a(n) = 2*A038548(n) - A010052(n). - Reinhard Zumkeller, Mar 08 2013
Sum_{n>=1} a(n)*q^n = (log(1-q) + psi_q(1)) / log(q), where psi_q(z) is the q-digamma function. - Vladimir Reshetnikov, Apr 23 2013
a(n) = Product_{k = 1..A001221(n)} (A124010(n,k) + 1). - Reinhard Zumkeller, Jul 12 2013
a(n) = Sum_{k=1..n} A238133(k)*A000041(n-k). - Mircea Merca, Feb 18 2013
G.f.: Sum_{k>=1} Sum_{j>=1} x^(j*k). - Mats Granvik, Jun 15 2013
The formula above is obtained by expanding the Lambert series Sum_{k>=1} x^k/(1-x^k). - Joerg Arndt, Mar 12 2014
G.f.: Sum_{n>=1} Sum_{d|n} ( -log(1 - x^(n/d)) )^d / d!. - Paul D. Hanna, Aug 21 2014
2*Pi*a(n) = Sum_{m=1..n} Integral_{x=0..2*Pi} r^(m-n)( cos((m-n)*x)-r^m cos(n*x) )/( 1+r^(2*m)-2r^m cos(m*x) )dx, 0 < r < 1 a free parameter. This formula is obtained as the sum of the residues of the Lambert series Sum_{k>=1} x^k/(1-x^k). - Seiichi Kirikami, Oct 22 2015
a(n) = A091220(A091202(n)) = A106737(A156552(n)). - Antti Karttunen, circa 2004 & Mar 06 2017
a(n) = A034296(n) - A237665(n+1) [Wang, Fokkink, Fokkink]. - George Beck, May 06 2017
G.f.: 2*x/(1-x) - Sum_{k>0} x^k*(1-2*x^k)/(1-x^k). - Mamuka Jibladze, Aug 29 2018
a(n) = Sum_{k=1..n} 1/phi(n / gcd(n, k)). - Daniel Suteu, Nov 05 2018
a(k*n) = a(n)*(f(k,n)+2)/(f(k,n)+1), where f(k,n) is the exponent of the highest power of k dividing n and k is prime. - Gary Detlefs, Feb 08 2019
a(n) = 2*log(p(n))/log(n), n > 1, where p(n)= the product of the factors of n = A007955(n). - Gary Detlefs, Feb 15 2019
a(n) = (1/n) * Sum_{k=1..n} sigma(gcd(n,k)), where sigma(n) = sum of divisors of n. - Orges Leka, May 09 2019
a(n) = A001227(n)*(A007814(n) + 1) = A001227(n)*A001511(n). - Ivan N. Ianakiev, Nov 14 2019
From Richard L. Ollerton, May 11 2021: (Start)
a(n) = A038040(n) / n = (1/n)*Sum_{d|n} phi(d)*sigma(n/d), where phi = A000010 and sigma = A000203.
a(n) = (1/n)*Sum_{k=1..n} phi(gcd(n,k))*sigma(n/gcd(n,k))/phi(n/gcd(n,k)). (End)
From Ridouane Oudra, Nov 12 2021: (Start)
a(n) = Sum_{j=1..n} Sum_{k=1..j} (1/j)*cos(2*k*n*Pi/j);
a(n) = Sum_{j=1..n} Sum_{k=1..j} (1/j)*e^(2*k*n*Pi*i/j), where i^2=-1. (End)

Extensions

Incorrect formula deleted by Ridouane Oudra, Oct 28 2021

A000010 Euler totient function phi(n): count numbers <= n and prime to n.

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8, 8, 16, 6, 18, 8, 12, 10, 22, 8, 20, 12, 18, 12, 28, 8, 30, 16, 20, 16, 24, 12, 36, 18, 24, 16, 40, 12, 42, 20, 24, 22, 46, 16, 42, 20, 32, 24, 52, 18, 40, 24, 36, 28, 58, 16, 60, 30, 36, 32, 48, 20, 66, 32, 44
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of elements in a reduced residue system modulo n.
Degree of the n-th cyclotomic polynomial (cf. A013595). - Benoit Cloitre, Oct 12 2002
Number of distinct generators of a cyclic group of order n. Number of primitive n-th roots of unity. (A primitive n-th root x is such that x^k is not equal to 1 for k = 1, 2, ..., n - 1, but x^n = 1.) - Lekraj Beedassy, Mar 31 2005
Also number of complex Dirichlet characters modulo n; Sum_{k=1..n} a(k) is asymptotic to (3/Pi^2)*n^2. - Steven Finch, Feb 16 2006
a(n) is the highest degree of irreducible polynomial dividing 1 + x + x^2 + ... + x^(n-1) = (x^n - 1)/(x - 1). - Alexander Adamchuk, Sep 02 2006, corrected Sep 27 2006
a(p) = p - 1 for prime p. a(n) is even for n > 2. For n > 2, a(n)/2 = A023022(n) = number of partitions of n into 2 ordered relatively prime parts. - Alexander Adamchuk, Jan 25 2007
Number of automorphisms of the cyclic group of order n. - Benoit Jubin, Aug 09 2008
a(n+2) equals the number of palindromic Sturmian words of length n which are "bispecial", prefix or suffix of two Sturmian words of length n + 1. - Fred Lunnon, Sep 05 2010
Suppose that a and n are coprime positive integers, then by Euler's totient theorem, any factor of n divides a^phi(n) - 1. - Lei Zhou, Feb 28 2012
If m has k prime factors, (p_1, p_2, ..., p_k), then phi(m*n) = (Product_{i=1..k} phi (p_i*n))/phi(n)^(k-1). For example, phi(42*n) = phi(2*n)*phi(3*n)*phi(7*n)/phi(n)^2. - Gary Detlefs, Apr 21 2012
Sum_{n>=1} a(n)/n! = 1.954085357876006213144... This sum is referenced in Plouffe's inverter. - Alexander R. Povolotsky, Feb 02 2013 (see A336334. - Hugo Pfoertner, Jul 22 2020)
The order of the multiplicative group of units modulo n. - Michael Somos, Aug 27 2013
A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all positive integers n and m. - Michael Somos, Dec 30 2016
From Eric Desbiaux, Jan 01 2017: (Start)
a(n) equals the Ramanujan sum c_n(n) (last term on n-th row of triangle A054533).
a(n) equals the Jordan function J_1(n) (cf. A007434, A059376, A059377, which are the Jordan functions J_2, J_3, J_4, respectively). (End)
For n > 1, a(n) appears to be equal to the number of semi-meander solutions for n with top arches containing exactly 2 mountain ranges and exactly 2 arches of length 1. - Roger Ford, Oct 11 2017
a(n) is the minimum dimension of a lattice able to generate, via cut-and-project, the quasilattice whose diffraction pattern features n-fold rotational symmetry. The case n=15 is the first n > 1 in which the following simpler definition fails: "a(n) is the minimum dimension of a lattice with n-fold rotational symmetry". - Felix Flicker, Nov 08 2017
Number of cyclic Latin squares of order n with the first row in ascending order. - Eduard I. Vatutin, Nov 01 2020
a(n) is the number of rational numbers p/q >= 0 (in lowest terms) such that p + q = n. - Rémy Sigrist, Jan 17 2021
From Richard L. Ollerton, May 08 2021: (Start)
Formulas for the numerous OEIS entries involving Dirichlet convolution of a(n) and some sequence h(n) can be derived using the following (n >= 1):
Sum_{d|n} phi(d)*h(n/d) = Sum_{k=1..n} h(gcd(n,k)) [see P. H. van der Kamp link] = Sum_{d|n} h(d)*phi(n/d) = Sum_{k=1..n} h(n/gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)). Similarly,
Sum_{d|n} phi(d)*h(d) = Sum_{k=1..n} h(n/gcd(n,k)) = Sum_{k=1..n} h(gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)).
More generally,
Sum_{d|n} h(d) = Sum_{k=1..n} h(gcd(n,k))/phi(n/gcd(n,k)) = Sum_{k=1..n} h(n/gcd(n,k))/phi(n/gcd(n,k)).
In particular, for sequences involving the Möbius transform:
Sum_{d|n} mu(d)*h(n/d) = Sum_{k=1..n} h(gcd(n,k))*mu(n/gcd(n,k))/phi(n/gcd(n,k)) = Sum_{k=1..n} h(n/gcd(n,k))*mu(gcd(n,k))/phi(n/gcd(n,k)), where mu = A008683.
Use of gcd(n,k)*lcm(n,k) = n*k and phi(gcd(n,k))*phi(lcm(n,k)) = phi(n)*phi(k) provide further variations. (End)
From Richard L. Ollerton, Nov 07 2021: (Start)
Formulas for products corresponding to the sums above may found using the substitution h(n) = log(f(n)) where f(n) > 0 (for example, cf. formulas for the sum A018804 and product A067911 of gcd(n,k)):
Product_{d|n} f(n/d)^phi(d) = Product_{k=1..n} f(gcd(n,k)) = Product_{d|n} f(d)^phi(n/d) = Product_{k=1..n} f(n/gcd(n,k))^(phi(gcd(n,k))/phi(n/gcd(n,k))),
Product_{d|n} f(d)^phi(d) = Product_{k=1..n} f(n/gcd(n,k)) = Product_{k=1..n} f(gcd(n,k))^(phi(gcd(n,k))/phi(n/gcd(n,k))),
Product_{d|n} f(d) = Product_{k=1..n} f(gcd(n,k))^(1/phi(n/gcd(n,k))) = Product_{k=1..n} f(n/gcd(n,k))^(1/phi(n/gcd(n,k))),
Product_{d|n} f(n/d)^mu(d) = Product_{k=1..n} f(gcd(n,k))^(mu(n/gcd(n,k))/phi(n/gcd(n,k))) = Product_{k=1..n} f(n/gcd(n,k))^(mu(gcd(n,k))/phi(n/gcd(n,k))), where mu = A008683. (End)
a(n+1) is the number of binary words with exactly n distinct subsequences (when n > 0). - Radoslaw Zak, Nov 29 2021

Examples

			G.f. = x + x^2 + 2*x^3 + 2*x^4 + 4*x^5 + 2*x^6 + 6*x^7 + 4*x^8 + 6*x^9 + 4*x^10 + ...
a(8) = 4 with {1, 3, 5, 7} units modulo 8. a(10) = 4 with {1, 3, 7, 9} units modulo 10. - _Michael Somos_, Aug 27 2013
From _Eduard I. Vatutin_, Nov 01 2020: (Start)
The a(5)=4 cyclic Latin squares with the first row in ascending order are:
  0 1 2 3 4   0 1 2 3 4   0 1 2 3 4   0 1 2 3 4
  1 2 3 4 0   2 3 4 0 1   3 4 0 1 2   4 0 1 2 3
  2 3 4 0 1   4 0 1 2 3   1 2 3 4 0   3 4 0 1 2
  3 4 0 1 2   1 2 3 4 0   4 0 1 2 3   2 3 4 0 1
  4 0 1 2 3   3 4 0 1 2   2 3 4 0 1   1 2 3 4 0
(End)

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 840.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 24.
  • M. Baake and U. Grimm, Aperiodic Order Vol. 1: A Mathematical Invitation, Encyclopedia of Mathematics and its Applications 149, Cambridge University Press, 2013: see Tables 3.1 and 3.2.
  • Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 409.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 193.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 154-156.
  • C. W. Curtis, Pioneers of Representation Theory ..., Amer. Math. Soc., 1999; see p. 3.
  • J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Ellipses, Paris, 2004, Problème 529, pp. 71-257.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, Chapter V.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 115-119.
  • Carl Friedrich Gauss, "Disquisitiones Arithmeticae", Yale University Press, 1965; see p. 21.
  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Math., 2n-d ed.; Addison-Wesley, 1994, p. 137.
  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section B36.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1979, th. 60, 62, 63, 288, 323, 328, 330.
  • Peter Hilton and Jean Pedersen, A Mathematical Tapestry, Demonstrating the Beautiful Unity of Mathematics, Cambridge University Press, pages 261-264, the Coach theorem.
  • Jean-Marie Monier, Analyse, Exercices corrigés, 2ème année MP, Dunod, 1997, Exercice 3.2.21 pp. 281-294.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, New York, Heidelberg, Berlin, 2 vols., 1976, Vol. II, problem 71, p. 126.
  • Paulo Ribenboim, The New Book of Prime Number Records.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 28-33.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 162-167.

Links

Crossrefs

Cf. A002088 (partial sums), A008683, A003434 (steps to reach 1), A007755, A049108, A002202 (values), A011755 (Sum k*phi(k)).
Cf. also A005277 (nontotient numbers). For inverse see A002181, A006511, A058277.
Jordan function J_k(n) is a generalization - see A059379 and A059380 (triangle of values of J_k(n)), this sequence (J_1), A007434 (J_2), A059376 (J_3), A059377 (J_4), A059378 (J_5).
Cf. A054521, A023022, A054525.
Row sums of triangles A134540, A127448, A143239, A143353 and A143276.
Equals right and left borders of triangle A159937. - Gary W. Adamson, Apr 26 2009
Values for prime powers p^e: A006093 (e=1), A036689 (e=2), A135177 (e=3), A138403 (e=4), A138407 (e=5), A138412 (e=6).
Values for perfect powers n^e: A002618 (e=2), A053191 (e=3), A189393 (e=4), A238533 (e=5), A306411 (e=6), A239442 (e=7), A306412 (e=8), A239443 (e=9).
Cf. A003558, A135303.
Cf. A152455, A080737.
Cf. A076479.
Cf. A023900 (Dirichlet inverse of phi), A306633 (Dgf at s=3).

Programs

  • Axiom
    [eulerPhi(n) for n in 1..100]
  • Haskell
    a n = length (filter (==1) (map (gcd n) [1..n])) -- Allan C. Wechsler, Dec 29 2014
  • Julia
    # Computes the first N terms of the sequence.
function A000010List(N)
    phi = [i for i in 1:N + 1]
    for i in 2:N + 1
        if phi[i] == i
            for j in i:i:N + 1
                phi[j] -= div(phi[j], i)
    end end end
return phi end
println(A000010List(68))  # Peter Luschny, Sep 03 2023
  • Magma
    [ EulerPhi(n) : n in [1..100] ]; // Sergei Haller (sergei(AT)sergei-haller.de), Dec 21 2006
  • Maple
    with(numtheory): A000010 := phi; [ seq(phi(n), n=1..100) ]; # version 1
with(numtheory): phi := proc(n) local i,t1,t2; t1 := ifactors(n)[2]; t2 := n*mul((1-1/t1[i][1]),i=1..nops(t1)); end; # version 2
# Alternative without library function:
A000010List := proc(N) local i, j, phi;
    phi := Array([seq(i, i = 1 .. N+1)]);
    for i from 2 to N + 1 do
        if phi[i] = i then
            for j from i by i to N + 1 do
                phi[j] := phi[j] - iquo(phi[j], i) od
        fi od;
return phi end:
A000010List(68);  # Peter Luschny, Sep 03 2023
  • Mathematica
    Array[EulerPhi, 70]
  • Maxima
    makelist(totient(n),n,0,1000); /* Emanuele Munarini, Mar 26 2011 */
  • PARI
    {a(n) = if( n==0, 0, eulerphi(n))}; /* Michael Somos, Feb 05 2011 */
  • Python
    from sympy.ntheory import totient
print([totient(i) for i in range(1, 70)])  # Indranil Ghosh, Mar 17 2017
  • Python
    # Note also the implementation in A365339.
  • Sage
    def A000010(n): return euler_phi(n) # Jaap Spies, Jan 07 2007
  • Sage
    [euler_phi(n) for n in range(1, 70)]  # Zerinvary Lajos, Jun 06 2009

Formula

phi(n) = n*Product_{distinct primes p dividing n} (1 - 1/p).
Sum_{d divides n} phi(d) = n.
phi(n) = Sum_{d divides n} mu(d)*n/d, i.e., the Moebius transform of the natural numbers; mu() = Moebius function A008683().
Dirichlet generating function Sum_{n>=1} phi(n)/n^s = zeta(s-1)/zeta(s). Also Sum_{n >= 1} phi(n)*x^n/(1 - x^n) = x/(1 - x)^2.
Multiplicative with a(p^e) = (p - 1)*p^(e-1). - David W. Wilson, Aug 01 2001
Sum_{n>=1} (phi(n)*log(1 - x^n)/n) = -x/(1 - x) for -1 < x < 1 (cf. A002088) - Henry Bottomley, Nov 16 2001
a(n) = binomial(n+1, 2) - Sum_{i=1..n-1} a(i)*floor(n/i) (see A000217 for inverse). - Jon Perry, Mar 02 2004
It is a classical result (certainly known to Landau, 1909) that lim inf n/phi(n) = 1 (taking n to be primes), lim sup n/(phi(n)*log(log(n))) = e^gamma, with gamma = Euler's constant (taking n to be products of consecutive primes starting from 2 and applying Mertens' theorem). See e.g. Ribenboim, pp. 319-320. - Pieter Moree, Sep 10 2004
a(n) = Sum_{i=1..n} |k(n, i)| where k(n, i) is the Kronecker symbol. Also a(n) = n - #{1 <= i <= n : k(n, i) = 0}. - Benoit Cloitre, Aug 06 2004 [Corrected by Jianing Song, Sep 25 2018]
Conjecture: Sum_{i>=2} (-1)^i/(i*phi(i)) exists and is approximately 0.558 (A335319). - Orges Leka (oleka(AT)students.uni-mainz.de), Dec 23 2004
From Enrique Pérez Herrero, Sep 07 2010: (Start)
a(n) = Sum_{i=1..n} floor(sigma_k(i*n)/sigma_k(i)*sigma_k(n)), where sigma_2 is A001157.
a(n) = Sum_{i=1..n} floor(tau_k(i*n)/tau_k(i)*tau_k(n)), where tau_3 is A007425.
a(n) = Sum_{i=1..n} floor(rad(i*n)/rad(i)*rad(n)), where rad is A007947. (End)
a(n) = A173557(n)*A003557(n). - R. J. Mathar, Mar 30 2011
a(n) = A096396(n) + A096397(n). - Reinhard Zumkeller, Mar 24 2012
phi(p*n) = phi(n)*(floor(((n + p - 1) mod p)/(p - 1)) + p - 1), for primes p. - Gary Detlefs, Apr 21 2012
For odd n, a(n) = 2*A135303((n-1)/2)*A003558((n-1)/2) or phi(n) = 2*c*k; the Coach theorem of Pedersen et al. Cf. A135303. - Gary W. Adamson, Aug 15 2012
G.f.: Sum_{n>=1} mu(n)*x^n/(1 - x^n)^2, where mu(n) = A008683(n). - Mamuka Jibladze, Apr 05 2015
a(n) = n - cototient(n) = n - A051953(n). - Omar E. Pol, May 14 2016
a(n) = lim_{s->1} n*zeta(s)*(Sum_{d divides n} A008683(d)/(e^(1/d))^(s-1)), for n > 1. - Mats Granvik, Jan 26 2017
Conjecture: a(n) = Sum_{a=1..n} Sum_{b=1..n} Sum_{c=1..n} 1 for n > 1. The sum is over a,b,c such that n*c - a*b = 1. - Benedict W. J. Irwin, Apr 03 2017
a(n) = Sum_{j=1..n} gcd(j, n) cos(2*Pi*j/n) = Sum_{j=1..n} gcd(j, n) exp(2*Pi*i*j/n) where i is the imaginary unit. Notice that the Ramanujan's sum c_n(k) := Sum_{j=1..n, gcd(j, n) = 1} exp(2*Pi*i*j*k/n) gives a(n) = Sum_{k|n} k*c_(n/k)(1) = Sum_{k|n} k*mu(n/k). - Michael Somos, May 13 2018
G.f.: x*d/dx(x*d/dx(log(Product_{k>=1} (1 - x^k)^(-mu(k)/k^2)))), where mu(n) = A008683(n). - Mamuka Jibladze, Sep 20 2018
a(n) = Sum_{d|n} A007431(d). - Steven Foster Clark, May 29 2019
G.f. A(x) satisfies: A(x) = x/(1 - x)^2 - Sum_{k>=2} A(x^k). - Ilya Gutkovskiy, Sep 06 2019
a(n) >= sqrt(n/2) (Nicolas). - Hugo Pfoertner, Jun 01 2020
a(n) > n/(exp(gamma)*log(log(n)) + 5/(2*log(log(n)))), except for n=223092870 (Rosser, Schoenfeld). - Hugo Pfoertner, Jun 02 2020
From Bernard Schott, Nov 28 2020: (Start)
Sum_{m=1..n} 1/a(m) = A028415(n)/A048049(n) -> oo when n->oo.
Sum_{n >= 1} 1/a(n)^2 = A109695.
Sum_{n >= 1} 1/a(n)^3 = A335818.
Sum_{n >= 1} 1/a(n)^k is convergent iff k > 1.
a(2n) = a(n) iff n is odd, and, a(2n) > a(n) iff n is even. (End) [Actually, a(2n) = 2*a(n) for even n. - Jianing Song, Sep 18 2022]
a(n) = 2*A023896(n)/n, n > 1. - Richard R. Forberg, Feb 03 2021
From Richard L. Ollerton, May 09 2021: (Start)
For n > 1, Sum_{k=1..n} phi^{(-1)}(n/gcd(n,k))*a(gcd(n,k))/a(n/gcd(n,k)) = 0, where phi^{(-1)} = A023900.
For n > 1, Sum_{k=1..n} a(gcd(n,k))*mu(rad(gcd(n,k)))*rad(gcd(n,k))/gcd(n,k) = 0.
For n > 1, Sum_{k=1..n} a(gcd(n,k))*mu(rad(n/gcd(n,k)))*rad(n/gcd(n,k))*gcd(n,k) = 0.
Sum_{k=1..n} a(gcd(n,k))/a(n/gcd(n,k)) = n. (End)
a(n) = Sum_{d|n, e|n} gcd(d, e)*mobius(n/d)*mobius(n/e) (the sum is a multiplicative function of n by Tóth, and takes the value p^e - p^(e-1) for n = p^e, a prime power). - Peter Bala, Jan 22 2024
Sum_{n >= 1} phi(n)*x^n/(1 + x^n) = x + 3*x^3 + 5*x^5 + 7*x^7 + ... = Sum_{n >= 1} phi(2*n-1)*x^(2*n-1)/(1 - x^(4*n-2)). For the first equality see Pólya and Szegő, problem 71, p. 126. - Peter Bala, Feb 29 2024
Conjecture: a(n) = lim_{k->oo} (n^(k + 1))/A000203(n^k). - Velin Yanev, Dec 04 2024 [A000010(p) = p-1, A000203(p^k) = (p^(k+1)-1)/(p-1), so the conjecture is true if n is prime. - Vaclav Kotesovec, Dec 19 2024]

A007318 Pascal's triangle read by rows: C(n,k) = binomial(n,k) = n!/(k!*(n-k)!), 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1
Offset: 0

Views

Author

N. J. A. Sloane and Mira Bernstein, Apr 28 1994

Keywords

Comments

A. W. F. Edwards writes: "It [the triangle] was first written down long before 1654, the year in which Blaise Pascal wrote his Traité du triangle arithmétique, but it was this work that brought together all the different aspects of the numbers for the first time. In it Pascal developed the properties of the number as a piece of pure mathematics ... and then, in a series of appendices, showed how these properties were relevant to the study of the figurate numbers, to the theory of combinations, to the expansion of binomial expressions, and to the solution of an important problem in the theory of probability." (A. W. F. Edwards, Pascal's Arithmetical Triangle, Johns Hopkins University Press (2002), p. xiii)
Edwards reports that the naming of the triangle after Pascal was done first by Montmort in 1708 as the "Table de M. Pascal pour les combinaisons" and then by De Moivre in 1730 as the "Triangulum Arithmeticum PASCALANIUM". (Edwards, p. xiv)
In China, Yang Hui in 1261 listed the coefficients of (a+b)^n up to n=6, crediting the expansion to Chia Hsein's Shih-so suan-shu circa 1100. Another prominent early use was in Chu Shih-Chieh's Precious Mirror of the Four Elements in 1303. (Edwards, p. 51)
In Persia, Al-Karaji discovered the binomial triangle "some time soon after 1007", and Al-Samawal published it in the Al-bahir some time before 1180. (Edwards, p. 52)
In India, Halayuda's commentary (circa 900) on Pingala's treatise on syllabic combinations (circa 200 B.C.E.) contains a clear description of the additive computation of the triangle. (Amulya Kumar Bag, Binomial Theorem in Ancient India, p. 72)
Also in India, the multiplicative formula for C(n,k) was known to Mahavira in 850 and restated by Bhaskara in 1150. (Edwards, p. 27)
In Italy, Tartaglia published the triangle in his General trattato (1556), and Cardano published it in his Opus novum (1570). (Edwards, p. 39, 44) - Russ Cox, Mar 29 2022
Also sometimes called Omar Khayyam's triangle.
Also sometimes called Yang Hui's triangle.
C(n,k) = number of k-element subsets of an n-element set.
Row n gives coefficients in expansion of (1+x)^n.
Binomial(n+k-1,n-1) is the number of ways of placing k indistinguishable balls into n boxes (the "bars and stars" argument - see Feller).
Binomial(n-1,k-1) is the number of compositions (ordered partitions) of n with k summands.
Binomial(n+k-1,k-1) is the number of weak compositions (ordered weak partitions) of n into exactly k summands. - Juergen Will, Jan 23 2016
Binomial(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0) and (1,1). - Joerg Arndt, Jul 01 2011
If thought of as an infinite lower triangular matrix, inverse begins:
+1
-1 +1
+1 -2 +1
-1 +3 -3 +1
+1 -4 +6 -4 +1
All 2^n palindromic binomial coefficients starting after the A006516(n)-th entry are odd. - Lekraj Beedassy, May 20 2003
Binomial(n+k-1,n-1) is the number of standard tableaux of shape (n,1^k). - Emeric Deutsch, May 13 2004
Can be viewed as an array, read by antidiagonals, where the entries in the first row and column are all 1's and A(i,j) = A(i-1,j) + A(i,j-1) for all other entries. The determinant of each of its n X n subarrays starting at (0,0) is 1. - Gerald McGarvey, Aug 17 2004
Also the lower triangular readout of the exponential of a matrix whose entry {j+1,j} equals j+1 (and all other entries are zero). - Joseph Biberstine (jrbibers(AT)indiana.edu), May 26 2006
Binomial(n-3,k-1) counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 132 and k descents. Binomial(n-3,k-1) also counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 213 and k descents. - David Hoek (david.hok(AT)telia.com), Feb 28 2007
Inverse of A130595 (as an infinite lower triangular matrix). - Philippe Deléham, Aug 21 2007
Consider integer lists LL of lists L of the form LL = [m#L] = [m#[k#2]] (where '#' means 'times') like LL(m=3,k=3) = [[2,2,2],[2,2,2],[2,2,2]]. The number of the integer list partitions of LL(m,k) is equal to binomial(m+k,k) if multiple partitions like [[1,1],[2],[2]] and [[2],[2],[1,1]] and [[2],[1,1],[2]] are counted only once. For the example, we find 4*5*6/3! = 20 = binomial(6,3). - Thomas Wieder, Oct 03 2007
The infinitesimal generator for Pascal's triangle and its inverse is A132440. - Tom Copeland, Nov 15 2007
Row n>=2 gives the number of k-digit (k>0) base n numbers with strictly decreasing digits; e.g., row 10 for A009995. Similarly, row n-1>=2 gives the number of k-digit (k>1) base n numbers with strictly increasing digits; see A009993 and compare A118629. - Rick L. Shepherd, Nov 25 2007
From Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008: (Start)
Binomial(n+k-1, k) is the number of ways a sequence of length k can be partitioned into n subsequences (see the Naish link).
Binomial(n+k-1, k) is also the number of n- (or fewer) digit numbers written in radix at least k whose digits sum to k. For example, in decimal, there are binomial(3+3-1,3)=10 3-digit numbers whose digits sum to 3 (see A052217) and also binomial(4+2-1,2)=10 4-digit numbers whose digits sum to 2 (see A052216). This relationship can be used to generate the numbers of sequences A052216 to A052224 (and further sequences using radix greater than 10). (End)
From Milan Janjic, May 07 2008: (Start)
Denote by sigma_k(x_1,x_2,...,x_n) the elementary symmetric polynomials. Then:
Binomial(2n+1,2k+1) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2(i*Pi/(2n+1)), (i=1,2,...,n).
Binomial(2n,2k+1) = 2n*sigma_{n-1-k}(x_1,x_2,...,x_{n-1}), where x_i = tan^2(i*Pi/(2n)), (i=1,2,...,n-1).
Binomial(2n,2k) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n)), (i=1,2,...,n).
Binomial(2n+1,2k) = (2n+1)sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n+2)), (i=1,2,...,n). (End)
Given matrices R and S with R(n,k) = binomial(n,k)*r(n-k) and S(n,k) = binomial(n,k)*s(n-k), then R*S = T where T(n,k) = binomial(n,k)*[r(.)+s(.)]^(n-k), umbrally. And, the e.g.f.s for the row polynomials of R, S and T are, respectively, exp(x*t)*exp[r(.)*x], exp(x*t)*exp[s(.)*x] and exp(x*t)*exp[r(.)*x]*exp[s(.)*x] = exp{[t+r(.)+s(.)]*x}. The row polynomials are essentially Appell polynomials. See A132382 for an example. - Tom Copeland, Aug 21 2008
As the rectangle R(m,n) = binomial(m+n-2,m-1), the weight array W (defined generally at A144112) of R is essentially R itself, in the sense that if row 1 and column 1 of W=A144225 are deleted, the remaining array is R. - Clark Kimberling, Sep 15 2008
If A007318 = M as an infinite lower triangular matrix, M^n gives A130595, A023531, A007318, A038207, A027465, A038231, A038243, A038255, A027466, A038279, A038291, A038303, A038315, A038327, A133371, A147716, A027467 for n=-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 respectively. - Philippe Deléham, Nov 11 2008
The coefficients of the polynomials with e.g.f. exp(x*t)*(cosh(t)+sinh(t)). - Peter Luschny, Jul 09 2009
The triangle or chess sums, see A180662 for their definitions, link Pascal's triangle with twenty different sequences, see the crossrefs. All sums come in pairs due to the symmetrical nature of this triangle. The knight sums Kn14 - Kn110 have been added. It is remarkable that all knight sums are related to the Fibonacci numbers, i.e., A000045, but none of the others. - Johannes W. Meijer, Sep 22 2010
Binomial(n,k) is also the number of ways to distribute n+1 balls into k+1 urns so that each urn gets at least one ball. See example in the example section below. - Dennis P. Walsh, Jan 29 2011
Binomial(n,k) is the number of increasing functions from {1,...,k} to {1,...,n} since there are binomial(n,k) ways to choose the k distinct, ordered elements of the range from the codomain {1,...,n}. See example in the example section below. - Dennis P. Walsh, Apr 07 2011
Central binomial coefficients: T(2*n,n) = A000984(n), T(n, floor(n/2)) = A001405(n). - Reinhard Zumkeller, Nov 09 2011
Binomial(n,k) is the number of subsets of {1,...,n+1} with k+1 as median element. To see this, note that Sum_{j=0..min(k,n-k)}binomial(k,j)*binomial(n-k,j) = binomial(n,k). See example in Example section below. - Dennis P. Walsh, Dec 15 2011
This is the coordinator triangle for the lattice Z^n, see Conway-Sloane, 1997. - N. J. A. Sloane, Jan 17 2012
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A046521 and A068555. For real r >= 0, C_r(n,k) := floor(r*n)!/(floor(r*k)!*floor(r*(n-k))!) is integral. See A211226 for the case r = 1/2. - Peter Bala, Apr 10 2012
Define a finite triangle T(m,k) with n rows such that T(m,0) = 1 is the left column, T(m,m) = binomial(n-1,m) is the right column, and the other entries are T(m,k) = T(m-1,k-1) + T(m-1,k) as in Pascal's triangle. The sum of all entries in T (there are A000217(n) elements) is 3^(n-1). - J. M. Bergot, Oct 01 2012
The lower triangular Pascal matrix serves as a representation of the operator exp(RLR) in a basis composed of a sequence of polynomials p_n(x) characterized by ladder operators defined by R p_n(x) = p_(n+1)(x) and L p_n(x) = n p_(n-1)(x). See A132440, A218272, A218234, A097805, and A038207. The transposed and padded Pascal matrices can be associated to the special linear group SL2. - Tom Copeland, Oct 25 2012
See A193242. - Alexander R. Povolotsky, Feb 05 2013
A permutation p_1...p_n of the set {1,...,n} has a descent at position i if p_i > p_(i+1). Let S(n) denote the subset of permutations p_1...p_n of {1,...,n} such that p_(i+1) - p_i <= 1 for i = 1,...,n-1. Then binomial(n,k) gives the number of permutations in S(n+1) with k descents. Alternatively, binomial(n,k) gives the number of permutations in S(n+1) with k+1 increasing runs. - Peter Bala, Mar 24 2013
Sum_{n=>0} binomial(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where binomial(n,k) = 0. Also Sum_{n>=0} binomial(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
The square n X n submatrix (first n rows and n columns) of the Pascal matrix P(x) defined in the formulas below when multiplying on the left the Vandermonde matrix V(x_1,...,x_n) (with ones in the first row) translates the matrix to V(x_1+x,...,x_n+x) while leaving the determinant invariant. - Tom Copeland, May 19 2014
For k>=2, n>=k, k/((k/(k-1) - Sum_{n=k..m} 1/binomial(n,k))) = m!/((m-k+1)!*(k-2)!). Note: k/(k-1) is the infinite sum. See A000217, A000292, A000332 for examples. - Richard R. Forberg, Aug 12 2014
Let G_(2n) be the subgroup of the symmetric group S_(2n) defined by G_(2n) = {p in S_(2n) | p(i) = i (mod n) for i = 1,2,...,2n}. G_(2n) has order 2^n. Binomial(n,k) gives the number of permutations in G_(2n) having n + k cycles. Cf. A130534 and A246117. - Peter Bala, Aug 15 2014
C(n,k) = the number of Dyck paths of semilength n+1, with k+1 "u"'s in odd numbered positions and k+1 returns to the x axis. Example: {U = u in odd position and = return to x axis} binomial(3,0)=1 (Uudududd); binomial(3,1)=3 [(Uududd_Ud_), (Ud_Uududd_), (Uudd_Uudd_)]; binomial(3,2)=3 [(Ud_Ud_Uudd_), (Uudd_Ud_Ud_), (Ud_Uudd_Ud_)]; binomial(3,3)=1 (Ud_Ud_Ud_Ud_). - Roger Ford, Nov 05 2014
From Daniel Forgues, Mar 12 2015: (Start)
The binomial coefficients binomial(n,k) give the number of individuals of the k-th generation after n population doublings. For each doubling of population, each individual's clone has its generation index incremented by 1, and thus goes to the next row. Just tally up each row from 0 to 2^n - 1 to get the binomial coefficients.
0 1 3 7 15
0: O | . | . . | . . . . | . . . . . . . . |
1: | O | O . | O . . . | O . . . . . . . |
2: | | O | O O . | O O . O . . . |
3: | | | O | O O O . |
4: | | | | O |
This is a fractal process: to get the pattern from 0 to 2^n - 1, append a shifted down (by one row) copy of the pattern from 0 to 2^(n-1) - 1 to the right of the pattern from 0 to 2^(n-1) - 1. (Inspired by the "binomial heap" data structure.)
Sequence of generation indices: 1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n) (see A000120):
{0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, ...}
Binary expansion of 0 to 15:
0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1111
(End)
A258993(n,k) = T(n+k,n-k), n > 0. - Reinhard Zumkeller, Jun 22 2015
T(n,k) is the number of set partitions w of [n+1] that avoid 1/2/3 with rb(w)=k. The same holds for ls(w)=k, where avoidance is in the sense of Klazar and ls,rb defined by Wachs and White.
Satisfies Benford's law [Diaconis, 1977] - N. J. A. Sloane, Feb 09 2017
Let {A(n)} be a set with exactly n identical elements, with {A(0)} being the empty set E. Let {A(n,k)} be the k-th iteration of {A(n)}, with {A(n,0)} = {A(n)}. {A(n,1)} = The set of all the subsets of A{(n)}, including {A(n)} and E. {A(n,k)} = The set of all subsets of {A(n,k-1)}, including all of the elements of {A(n,k-1)}. Let A(n,k) be the number of elements in {A(n,k)}. Then A(n,k) = C(n+k,k), with each successive iteration replicating the members of the k-th diagonal of Pascal's Triangle. See examples. - Gregory L. Simay, Aug 06 2018
Binomial(n-1,k) is also the number of permutations avoiding both 213 and 312 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n-1,k) is also the number of permutations avoiding both 132 and 213 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n,k) is the dimension of the k-th exterior power of a vector space of dimension n. - Stefano Spezia, Dec 22 2018
C(n,k-1) is the number of unoriented colorings of the facets (or vertices) of an n-dimensional simplex using exactly k colors. Each chiral pair is counted as one when enumerating unoriented arrangements. - Robert A. Russell, Oct 20 2020
From Dilcher and Stolarsky: "Two of the most ubiquitous objects in mathematics are the sequence of prime numbers and the binomial coefficients (and thus Pascal's triangle). A connection between the two is given by a well-known characterization of the prime numbers: Consider the entries in the k-th row of Pascal's triangle, without the initial and final entries. They are all divisible by k if and only if k is a prime." - Tom Copeland, May 17 2021
Named "Table de M. Pascal pour les combinaisons" by Pierre Remond de Montmort (1708) after the French mathematician, physicist and philosopher Blaise Pascal (1623-1662). - Amiram Eldar, Jun 11 2021
Consider the n-th diagonal of the triangle as a sequence b(n) with n starting at 0. From it form a new sequence by leaving the 0th term as is, and thereafter considering all compositions of n, taking the product of b(i) over the respective numbers i in each composition, adding terms corresponding to compositions with an even number of parts subtracting terms corresponding to compositions with an odd number of parts. Then the n-th row of the triangle is obtained, with every second term multiplied by -1, followed by infinitely many zeros. For sequences starting with 1, this operation is a special case of a self-inverse operation, and therefore the converse is true. - Thomas Anton, Jul 05 2021
C(n,k) is the number of vertices in an n-dimensional unit hypercube, at an L1 distance of k (or: with a shortest path of k 1d-edges) from a given vertex. - Eitan Y. Levine, May 01 2023
C(n+k-1,k-1) is the number of vertices at an L1 distance from a given vertex in an infinite-dimensional box, which has k sides of length 2^m for each m >= 0. Equivalently, given a set of tokens containing k distinguishable tokens with value 2^m for each m >= 0, C(n+k-1,k-1) is the number of subsets of tokens with a total value of n. - Eitan Y. Levine, Jun 11 2023
Numbers in the k-th column, i.e., numbers of the form C(n,k) for n >= k, are known as k-simplex numbers. - Pontus von Brömssen, Jun 26 2023
Let r(k) be the k-th row and c(k) the k-th column. Denote convolution by * and repeated convolution by ^. Then r(k)*r(m)=r(k+m) and c(k)*c(m)=c(k+m+1). This is because r(k) = r(1) ^ k and c(k) = c(0) ^ k+1. - Eitan Y. Levine, Jul 23 2023
Number of permutations of length n avoiding simultaneously the patterns 231 and 312(resp., 213 and 231; 213 and 312) with k descents (equivalently, with k ascents). An ascent (resp., descent) in a permutation a(1)a(2)...a(n) is position i such that a(i)a(i+1)). - Tian Han, Nov 25 2023
C(n,k) are generalized binomial coefficients of order m=0. Calculated by the formula C(n,k) = Sum_{i=0..n-k} binomial(n+1, n-k-i)*Stirling2(i+ m+ 1, i+1) *(-1)^i, where m = 0 for n>= 0, 0 <= k <= n. - Igor Victorovich Statsenko, Feb 26 2023
The Akiyama-Tanigawa algorithm applied to the diagonals, binomial(n+k,k), yields the powers of n. - Shel Kaphan, May 03 2024

Examples

			Triangle T(n,k) begins:
   n\k 0   1   2   3   4   5   6   7   8   9  10  11 ...
   0   1
   1   1   1
   2   1   2   1
   3   1   3   3   1
   4   1   4   6   4   1
   5   1   5  10  10   5   1
   6   1   6  15  20  15   6   1
   7   1   7  21  35  35  21   7   1
   8   1   8  28  56  70  56  28   8   1
   9   1   9  36  84 126 126  84  36   9   1
  10   1  10  45 120 210 252 210 120  45  10   1
  11   1  11  55 165 330 462 462 330 165  55  11   1
  ...
There are C(4,2)=6 ways to distribute 5 balls BBBBB, among 3 different urns, < > ( ) [ ], so that each urn gets at least one ball, namely, <BBB>(B)[B], <B>(BBB)[B], <B>(B)[BBB], <BB>(BB)[B], <BB>(B)[BB], and <B>(BB)[BB].
There are C(4,2)=6 increasing functions from {1,2} to {1,2,3,4}, namely, {(1,1),(2,2)},{(1,1),(2,3)}, {(1,1),(2,4)}, {(1,2),(2,3)}, {(1,2),(2,4)}, and {(1,3),(2,4)}. - _Dennis P. Walsh_, Apr 07 2011
There are C(4,2)=6 subsets of {1,2,3,4,5} with median element 3, namely, {3}, {1,3,4}, {1,3,5}, {2,3,4}, {2,3,5}, and {1,2,3,4,5}. - _Dennis P. Walsh_, Dec 15 2011
The successive k-iterations of {A(0)} = E are E;E;E;...; the corresponding number of elements are 1,1,1,... The successive k-iterations of {A(1)} = {a} are (omitting brackets) a;a,E; a,E,E;...; the corresponding number of elements are 1,2,3,... The successive k-iterations of {A(2)} = {a,a} are aa; aa,a,E; aa, a, E and a,E and E;...; the corresponding number of elements are 1,3,6,... - _Gregory L. Simay_, Aug 06 2018
Boas-Buck type recurrence for column k = 4: T(8, 4) = (5/4)*(1 + 5 + 15 + 35) = 70. See the Boas-Buck comment above. - _Wolfdieter Lang_, Nov 12 2018

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • Amulya Kumar Bag, Binomial theorem in ancient India, Indian Journal of History of Science, vol. 1 (1966), pp. 68-74.
  • Arthur T. Benjamin and Jennifer Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 63ff.
  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 306.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 68-74.
  • Paul Curtz, Intégration numérique des systèmes différentiels à conditions initiales, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969.
  • A. W. F. Edwards, Pascal's Arithmetical Triangle, 2002.
  • William Feller, An Introduction to Probability Theory and Its Application, Vol. 1, 2nd ed. New York: Wiley, p. 36, 1968.
  • Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, p. 155.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §4.4 Powers and Roots, pp. 140-141.
  • David Hök, Parvisa mönster i permutationer [Swedish], 2007.
  • Donald E. Knuth, The Art of Computer Programming, Vol. 1, 2nd ed., p. 52.
  • Sergei K. Lando, Lecture on Generating Functions, Amer. Math. Soc., Providence, R.I., 2003, pp. 60-61.
  • Blaise Pascal, Traité du triangle arithmétique, avec quelques autres petits traitez sur la mesme matière, Desprez, Paris, 1665.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 71.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 271-275.
  • A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 6.
  • John Riordan, Combinatorial Identities, Wiley, 1968, p. 2.
  • Robert Sedgewick and Philippe Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996, p. 143.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, pages 43-52.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 13, 30-33.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 115-118.
  • Douglas B. West, Combinatorial Mathematics, Cambridge, 2021, p. 25.

Links

Crossrefs

Equals differences between consecutive terms of A102363. - David G. Williams (davidwilliams(AT)Paxway.com), Jan 23 2006
Row sums give A000079 (powers of 2).
Cf. A083093 (triangle read mod 3), A214292 (first differences of rows).
Partial sums of rows give triangle A008949.
The triangle of the antidiagonals is A011973.
Infinite matrix squared: A038207, cubed: A027465.
Cf. A101164. If rows are sorted we get A061554 or A107430.
Another version: A108044.
Cf. A008277, A132311, A132312, A052216, A052217, A052218, A052219, A052220, A052221, A052222, A052223, A144225, A202750, A211226, A047999, A026729, A052553, A051920, A193242.
Triangle sums (see the comments): A000079 (Row1); A000007 (Row2); A000045 (Kn11 & Kn21); A000071 (Kn12 & Kn22); A001924 (Kn13 & Kn23); A014162 (Kn14 & Kn24); A014166 (Kn15 & Kn25); A053739 (Kn16 & Kn26); A053295 (Kn17 & Kn27); A053296 (Kn18 & Kn28); A053308 (Kn19 & Kn29); A053309 (Kn110 & Kn210); A001519 (Kn3 & Kn4); A011782 (Fi1 & Fi2); A000930 (Ca1 & Ca2); A052544 (Ca3 & Ca4); A003269 (Gi1 & Gi2); A055988 (Gi3 & Gi4); A034943 (Ze1 & Ze2); A005251 (Ze3 & Ze4). - Johannes W. Meijer, Sep 22 2010
Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074, A228196, A228576.
Cf. A137948, A245334.
Cf. A085478, A258993.
Cf. A115940 (pandigital binomial coefficients C(m,k) with k>1).
Cf. (simplex colorings) A325002 (oriented), [k==n+1] (chiral), A325003 (achiral), A325000 (k or fewer colors), A325009 (orthotope facets, orthoplex vertices), A325017 (orthoplex facets, orthotope vertices).
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 2..12: A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • Axiom
    -- (start)
)set expose add constructor OutputForm
pascal(0,n) == 1
pascal(n,n) == 1
pascal(i,j | 0 < i and i < j) == pascal(i-1,j-1) + pascal(i,j-1)
pascalRow(n) == [pascal(i,n) for i in 0..n]
displayRow(n) == output center blankSeparate pascalRow(n)
for i in 0..20 repeat displayRow i -- (end)
  • GAP
    Flat(List([0..12],n->List([0..n],k->Binomial(n,k)))); # Stefano Spezia, Dec 22 2018
  • Haskell
    a007318 n k = a007318_tabl !! n !! k
a007318_row n = a007318_tabl !! n
a007318_list = concat a007318_tabl
a007318_tabl = iterate (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1]
-- Cf. http://www.haskell.org/haskellwiki/Blow_your_mind#Mathematical_sequences
-- Reinhard Zumkeller, Nov 09 2011, Oct 22 2010
  • Magma
    /* As triangle: */ [[Binomial(n, k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jul 29 2015
  • Maple
    A007318 := (n,k)->binomial(n,k);
  • Mathematica
    Flatten[Table[Binomial[n, k], {n, 0, 11}, {k, 0, n}]] (* Robert G. Wilson v, Jan 19 2004 *)
Flatten[CoefficientList[CoefficientList[Series[1/(1 - x - x*y), {x, 0, 12}], x], y]] (* Mats Granvik, Jul 08 2014 *)
  • Maxima
    create_list(binomial(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
  • PARI
    C(n,k)=binomial(n,k) \\ Charles R Greathouse IV, Jun 08 2011
  • Python
    # See Hobson link. Further programs:
from math import prod,factorial
def C(n,k): return prod(range(n,n-k,-1))//factorial(k) # M. F. Hasler, Dec 13 2019, updated Apr 29 2022, Feb 17 2023
  • Python
    from math import comb, isqrt
def A007318(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Nov 11 2024
  • Sage
    def C(n,k): return Subsets(range(n), k).cardinality() # Ralf Stephan, Jan 21 2014

Formula

a(n, k) = C(n,k) = binomial(n, k).
C(n, k) = C(n-1, k) + C(n-1, k-1).
The triangle is symmetric: C(n,k) = C(n,n-k).
a(n+1, m) = a(n, m) + a(n, m-1), a(n, -1) := 0, a(n, m) := 0, n
C(n, k) = n!/(k!(n-k)!) if 0<=k<=n, otherwise 0.
C(n, k) = ((n-k+1)/k) * C(n, k-1) with C(n, 0) = 1. - Michael B. Porter, Mar 23 2025
G.f.: 1/(1-y-x*y) = Sum_(C(n, k)*x^k*y^n, n, k>=0)
G.f.: 1/(1-x-y) = Sum_(C(n+k, k)*x^k*y^n, n, k>=0).
G.f. for row n: (1+x)^n = Sum_{k=0..n} C(n, k)*x^k.
G.f. for column k: x^k/(1-x)^(k+1); [corrected by Werner Schulte, Jun 15 2022].
E.g.f.: A(x, y) = exp(x+x*y).
E.g.f. for column n: x^n*exp(x)/n!.
In general the m-th power of A007318 is given by: T(0, 0) = 1, T(n, k) = T(n-1, k-1) + m*T(n-1, k), where n is the row-index and k is the column; also T(n, k) = m^(n-k)*C(n, k).
Triangle T(n, k) read by rows; given by A000007 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
Let P(n+1) = the number of integer partitions of (n+1); let p(i) = the number of parts of the i-th partition of (n+1); let d(i) = the number of different parts of the i-th partition of (n+1); let m(i, j) = multiplicity of the j-th part of the i-th partition of (n+1). Define the operator Sum_{i=1..P(n+1), p(i)=k+1} as the sum running from i=1 to i=P(n+1) but taking only partitions with p(i)=(k+1) parts into account. Define the operator Product_{j=1..d(i)} = product running from j=1 to j=d(i). Then C(n, k) = Sum_{p(i)=(k+1), i=1..P(n+1)} p(i)! / [Product_{j=1..d(i)} m(i, j)!]. E.g., C(5, 3) = 10 because n=6 has the following partitions with m=3 parts: (114), (123), (222). For their multiplicities one has: (114): 3!/(2!*1!) = 3; (123): 3!/(1!*1!*1!) = 6; (222): 3!/3! = 1. The sum is 3 + 6 + 1 = 10 = C(5, 3). - Thomas Wieder, Jun 03 2005
C(n, k) = Sum_{j=0..k} (-1)^j*C(n+1+j, k-j)*A000108(j). - Philippe Deléham, Oct 10 2005
G.f.: 1 + x*(1 + x) + x^3*(1 + x)^2 + x^6*(1 + x)^3 + ... . - Michael Somos, Sep 16 2006
Sum_{k=0..floor(n/2)} x^(n-k)*T(n-k,k) = A000007(n), A000045(n+1), A002605(n), A030195(n+1), A057087(n), A057088(n), A057089(n), A057090(n), A057091(n), A057092(n), A057093(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, respectively. Sum_{k=0..floor(n/2)} (-1)^k*x^(n-k)*T(n-k,k) = A000007(n), A010892(n), A009545(n+1), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n), A084329(n+1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, respectively. - Philippe Deléham, Sep 16 2006
C(n,k) <= A062758(n) for n > 1. - Reinhard Zumkeller, Mar 04 2008
C(t+p-1, t) = Sum_{i=0..t} C(i+p-2, i) = Sum_{i=1..p} C(i+t-2, t-1). A binomial number is the sum of its left parent and all its right ancestors, which equals the sum of its right parent and all its left ancestors. - Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008
From Paul D. Hanna, Mar 24 2011: (Start)
Let A(x) = Sum_{n>=0} x^(n*(n+1)/2)*(1+x)^n be the g.f. of the flattened triangle:
A(x) = 1 + (x + x^2) + (x^3 + 2*x^4 + x^5) + (x^6 + 3*x^7 + 3*x^8 + x^9) + ...
then A(x) equals the series Sum_{n>=0} (1+x)^n*x^n*Product_{k=1..n} (1-(1+x)*x^(2*k-1))/(1-(1+x)*x^(2*k));
also, A(x) equals the continued fraction 1/(1- x*(1+x)/(1+ x*(1-x)*(1+x)/(1- x^3*(1+x)/(1+ x^2*(1-x^2)*(1+x)/(1- x^5*(1+x)/(1+ x^3*(1-x^3)*(1+x)/(1- x^7*(1+x)/(1+ x^4*(1-x^4)*(1+x)/(1- ...))))))))).
These formulas are due to (1) a q-series identity and (2) a partial elliptic theta function expression. (End)
For n > 0: T(n,k) = A029600(n,k) - A029635(n,k), 0 <= k <= n. - Reinhard Zumkeller, Apr 16 2012
Row n of the triangle is the result of applying the ConvOffs transform to the first n terms of the natural numbers (1, 2, 3, ..., n). See A001263 or A214281 for a definition of this transformation. - Gary W. Adamson, Jul 12 2012
From L. Edson Jeffery, Aug 02 2012: (Start)
Row n (n >= 0) of the triangle is given by the n-th antidiagonal of the infinite matrix P^n, where P = (p_{i,j}), i,j >= 0, is the production matrix
0, 1,
1, 0, 1,
0, 1, 0, 1,
0, 0, 1, 0, 1,
0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 0, 1, 0, 1,
... (End)
Row n of the triangle is also given by the n+1 coefficients of the polynomial P_n(x) defined by the recurrence P_0(x) = 1, P_1(x) = x + 1, P_n(x) = x*P_{n-1}(x) + P_{n-2}(x), n > 1. - L. Edson Jeffery, Aug 12 2013
For a closed-form formula for arbitrary left and right borders of Pascal-like triangles see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013
(1+x)^n = Sum_{k=0..n} (-1)^(n-k)*binomial(n,k)*Sum_{i=0..k} k^(n-i)*binomial(k,i)*x^(n-i)/(n-i)!. - Vladimir Kruchinin, Oct 21 2013
E.g.f.: A(x,y) = exp(x+x*y) = 1 + (x+y*x)/( E(0)-(x+y*x)), where E(k) = 1 + (x+y*x)/(1 + (k+1)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 08 2013
E.g.f.: E(0) -1, where E(k) = 2 + x*(1+y)/(2*k+1 - x*(1+y)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
G.f.: 1 + x*(1+x)*(1+x^2*(1+x)/(W(0)-x^2-x^3)), where W(k) = 1 + (1+x)*x^(k+2) - (1+x)*x^(k+3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
Sum_{n>=0} C(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where C(n,k) = 0. Also Sum_{n>=0} C(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
Sum_{n>=k} 1/C(n,k) = k/(k-1) for k>=1. - Richard R. Forberg, Feb 10 2014
From Tom Copeland, Apr 26 2014: (Start)
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result by A007318(x) = P(x). Then with :xD:^n = x^n*(d/dx)^n and B(n,x), the Bell polynomials (A008277),
A) P(x)= exp(x*dP) = exp[x*(e^M-I)] = exp[M*B(.,x)] = (I+dP)^B(.,x)
with dP = A132440, M = A238385-I, and I = identity matrix, and
B) P(:xD:) = exp(dP:xD:) = exp[(e^M-I):xD:] = exp[M*B(.,:xD:)] = exp[M*xD] = (I+dP)^(xD) with action P(:xD:)g(x) = exp(dP:xD:)g(x) = g[(I+dP)*x] (cf. also A238363).
C) P(x)^y = P(y*x). P(2x) = A038207(x) = exp[M*B(.,2x)], the face vectors of the n-dim hypercubes.
D) P(x) = [St2]*exp(x*M)*[St1] = [St2]*(I+dP)^x*[St1]
E) = [St1]^(-1)*(I+dP)^x*[St1] = [St2]*(I+dP)^x*[St2]^(-1)
where [St1]=padded A008275 just as [St2]=A048993=padded A008277 and exp(x*M) = (I+dP)^x = Sum_{k>=0} C(x,k) dP^k. (End)
T(n,k) = A245334(n,k) / A137948(n,k), 0 <= k <= n. - Reinhard Zumkeller, Aug 31 2014
From Peter Bala, Dec 21 2014: (Start)
Recurrence equation: T(n,k) = T(n-1,k)*(n + k)/(n - k) - T(n-1,k-1) for n >= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1. Note, changing the minus sign in the recurrence to a plus sign gives a recurrence for the square of the binomial coefficients - see A008459.
There is a relation between the e.g.f.'s of the rows and the diagonals of the triangle, namely, exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + 3*x + 3*x^2/2! + x^3/3!) = 1 + 4*x + 10*x^2/2! + 20*x^3/3! + 35*x^4/4! + .... This property holds more generally for the Riordan arrays of the form ( f(x), x/(1 - x) ), where f(x) is an o.g.f. of the form 1 + f_1*x + f_2*x^2 + .... See, for example, A055248 and A106516.
Let P denote the present triangle. For k = 0,1,2,... define P(k) to be the lower unit triangular block array
/I_k 0\
\ 0 P/ having the k X k identity matrix I_k as the upper left block; in particular, P(0) = P. The infinite product P(0)*P(1)*P(2)*..., which is clearly well-defined, is equal to the triangle of Stirling numbers of the second kind A008277. The infinite product in the reverse order, that is, ...*P(2)*P(1)*P(0), is equal to the triangle of Stirling cycle numbers A130534. (End)
C(a+b,c) = Sum_{k=0..a} C(a,k)*C(b,b-c+k). This is a generalization of equation 1 from section 4.2.5 of the Prudnikov et al. reference, for a=b=c=n: C(2*n,n) = Sum_{k=0..n} C(n,k)^2. See Links section for animation of new formula. - Hermann Stamm-Wilbrandt, Aug 26 2015
The row polynomials of the Pascal matrix P(n,x) = (1+x)^n are related to the Bernoulli polynomials Br(n,x) and their umbral compositional inverses Bv(n,x) by the umbral relation P(n,x) = (-Br(.,-Bv(.,x)))^n = (-1)^n Br(n,-Bv(.,x)), which translates into the matrix relation P = M * Br * M * Bv, where P is the Pascal matrix, M is the diagonal matrix diag(1,-1,1,-1,...), Br is the matrix for the coefficients of the Bernoulli polynomials, and Bv that for the umbral inverse polynomials defined umbrally by Br(n,Bv(.,x)) = x^n = Bv(n,Br(.,x)). Note M = M^(-1). - Tom Copeland, Sep 05 2015
1/(1-x)^k = (r(x) * r(x^2) * r(x^4) * ...) where r(x) = (1+x)^k. - Gary W. Adamson, Oct 17 2016
Boas-Buck type recurrence for column k for Riordan arrays (see the Aug 10 2017 remark in A046521, also for the reference) with the Boas-Buck sequence b(n) = {repeat(1)}. T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} T(j, k), for n >= 1, with T(n, n) = 1. This reduces, with T(n, k) = binomial(n, k), to a known binomial identity (e.g, Graham et al. p. 161). - Wolfdieter Lang, Nov 12 2018
C((p-1)/a, b) == (-1)^b * fact_a(a*b-a+1)/fact_a(a*b) (mod p), where fact_n denotes the n-th multifactorial, a divides p-1, and the denominator of the fraction on the right side of the equation represents the modular inverse. - Isaac Saffold, Jan 07 2019
C(n,k-1) = A325002(n,k) - [k==n+1] = (A325002(n,k) + A325003(n,k)) / 2 = [k==n+1] + A325003(n,k). - Robert A. Russell, Oct 20 2020
From Hermann Stamm-Wilbrandt, May 13 2021: (Start)
Binomial sums are Fibonacci numbers A000045:
Sum_{k=0..n} C(n + k, 2*k + 1) = F(2*n).
Sum_{k=0..n} C(n + k, 2*k) = F(2*n + 1). (End)
C(n,k) = Sum_{i=0..k} A000108(i) * C(n-2i-1, k-i), for 0 <= k <= floor(n/2)-1. - Tushar Bansal, May 17 2025

Extensions

Checked all links, deleted 8 that seemed lost forever and were probably not of great importance. - N. J. A. Sloane, May 08 2018

A000120 1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n).

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

The binary weight of n is also called Hamming weight of n. [The term "Hamming weight" was named after the American mathematician Richard Wesley Hamming (1915-1998). - Amiram Eldar, Jun 16 2021]
a(n) is also the largest integer such that 2^a(n) divides binomial(2n, n) = A000984(n). - Benoit Cloitre, Mar 27 2002
To construct the sequence, start with 0 and use the rule: If k >= 0 and a(0), a(1), ..., a(2^k-1) are the first 2^k terms, then the next 2^k terms are a(0) + 1, a(1) + 1, ..., a(2^k-1) + 1. - Benoit Cloitre, Jan 30 2003
An example of a fractal sequence. That is, if you omit every other number in the sequence, you get the original sequence. And of course this can be repeated. So if you form the sequence a(0 * 2^n), a(1 * 2^n), a(2 * 2^n), a(3 * 2^n), ... (for any integer n > 0), you get the original sequence. - Christopher.Hills(AT)sepura.co.uk, May 14 2003
The n-th row of Pascal's triangle has 2^k distinct odd binomial coefficients where k = a(n) - 1. - Lekraj Beedassy, May 15 2003
Fixed point of the morphism 0 -> 01, 1 -> 12, 2 -> 23, 3 -> 34, 4 -> 45, etc., starting from a(0) = 0. - Robert G. Wilson v, Jan 24 2006
a(n) is the number of times n appears among the mystery calculator sequences: A005408, A042964, A047566, A115419, A115420, A115421. - Jeremy Gardiner, Jan 25 2006
a(n) is the number of solutions of the Diophantine equation 2^m*k + 2^(m-1) + i = n, where m >= 1, k >= 0, 0 <= i < 2^(m-1); a(5) = 2 because only (m, k, i) = (1, 2, 0) [2^1*2 + 2^0 + 0 = 5] and (m, k, i) = (3, 0, 1) [2^3*0 + 2^2 + 1 = 5] are solutions. - Hieronymus Fischer, Jan 31 2006
The first appearance of k, k >= 0, is at a(2^k-1). - Robert G. Wilson v, Jul 27 2006
Sequence is given by T^(infinity)(0) where T is the operator transforming any word w = w(1)w(2)...w(m) into T(w) = w(1)(w(1)+1)w(2)(w(2)+1)...w(m)(w(m)+1). I.e., T(0) = 01, T(01) = 0112, T(0112) = 01121223. - Benoit Cloitre, Mar 04 2009
For n >= 2, the minimal k for which a(k(2^n-1)) is not multiple of n is 2^n + 3. - Vladimir Shevelev, Jun 05 2009
Triangle inequality: a(k+m) <= a(k) + a(m). Equality holds if and only if C(k+m, m) is odd. - Vladimir Shevelev, Jul 19 2009
a(k*m) <= a(k) * a(m). - Robert Israel, Sep 03 2023
The number of occurrences of value k in the first 2^n terms of the sequence is equal to binomial(n, k), and also equal to the sum of the first n - k + 1 terms of column k in the array A071919. Example with k = 2, n = 7: there are 21 = binomial(7,2) = 1 + 2 + 3 + 4 + 5 + 6 2's in a(0) to a(2^7-1). - Brent Spillner (spillner(AT)acm.org), Sep 01 2010, simplified by R. J. Mathar, Jan 13 2017
Let m be the number of parts in the listing of the compositions of n as lists of parts in lexicographic order, a(k) = n - length(composition(k)) for all k < 2^n and all n (see example); A007895 gives the equivalent for compositions into odd parts. - Joerg Arndt, Nov 09 2012
From Daniel Forgues, Mar 13 2015: (Start)
Just tally up row k (binary weight equal k) from 0 to 2^n - 1 to get the binomial coefficient C(n,k). (See A007318.)
0 1 3 7 15
0: O | . | . . | . . . . | . . . . . . . . |
1: | O | O . | O . . . | O . . . . . . . |
2: | | O | O O . | O O . O . . . |
3: | | | O | O O O . |
4: | | | | O |
Due to its fractal nature, the sequence is quite interesting to listen to.
(End)
The binary weight of n is a particular case of the digit sum (base b) of n. - Daniel Forgues, Mar 13 2015
The mean of the first n terms is 1 less than the mean of [a(n+1),...,a(2n)], which is also the mean of [a(n+2),...,a(2n+1)]. - Christian Perfect, Apr 02 2015
a(n) is also the largest part of the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2, 2, 2, 1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 20 2017
a(n) is also known as the population count of the binary representation of n. - Chai Wah Wu, May 19 2020

Examples

			Using the formula a(n) = a(floor(n / floor_pow4(n))) + a(n mod floor_pow4(n)):
  a(4) = a(1) + a(0) = 1,
  a(8) = a(2) + a(0) = 1,
  a(13) = a(3) + a(1) = 2 + 1 = 3,
  a(23) = a(1) + a(7) = 1 + a(1) + a(3) = 1 + 1 + 2 = 4.
_Gary W. Adamson_ points out (Jun 03 2009) that this can be written as a triangle:
  0,
  1,
  1,2,
  1,2,2,3,
  1,2,2,3,2,3,3,4,
  1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,
  1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,
  1,2,2,3,2,3,...
where the rows converge to A063787.
From _Joerg Arndt_, Nov 09 2012: (Start)
Connection to the compositions of n as lists of parts (see comment):
[ #]:   a(n)  composition
[ 0]:   [0]   1 1 1 1 1
[ 1]:   [1]   1 1 1 2
[ 2]:   [1]   1 1 2 1
[ 3]:   [2]   1 1 3
[ 4]:   [1]   1 2 1 1
[ 5]:   [2]   1 2 2
[ 6]:   [2]   1 3 1
[ 7]:   [3]   1 4
[ 8]:   [1]   2 1 1 1
[ 9]:   [2]   2 1 2
[10]:   [2]   2 2 1
[11]:   [3]   2 3
[12]:   [2]   3 1 1
[13]:   [3]   3 2
[14]:   [3]   4 1
[15]:   [4]   5
(End)

References

  • Jean-Paul Allouche and Jeffrey Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 119.
  • Donald E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.1.3, Problem 41, p. 589. - N. J. A. Sloane, Aug 03 2012
  • Manfred R. Schroeder, Fractals, Chaos, Power Laws. W.H. Freeman, 1991, p. 383.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

The basic sequences concerning the binary expansion of n are this one, A000788, A000069, A001969, A023416, A059015, A007088.
Partial sums see A000788. For run lengths see A131534. See also A001792, A010062.
Number of 0's in n: A023416 and A080791.
a(n) = n - A011371(n).
Sum of digits of n written in bases 2-16: this sequence, A053735, A053737, A053824, A053827, A053828, A053829, A053830, A007953, A053831, A053832, A053833, A053834, A053835, A053836.
Cf. A001620, A344716, A007814, A063787.
This is Guy Steele's sequence GS(3, 4) (see A135416).
Cf. A055640, A055641, A102669-A102685, A117804, A122840, A122841, A160093, A160094, A196563, A196564 (for base 10).
Cf. A230952 (boustrophedon transform).
Cf. A070939 (length of binary representation of n).

Programs

  • Fortran
    c See link in A139351
  • Haskell
    import Data.Bits (Bits, popCount)
a000120 :: (Integral t, Bits t) => t -> Int
a000120 = popCount
a000120_list = 0 : c [1] where c (x:xs) = x : c (xs ++ [x,x+1])
-- Reinhard Zumkeller, Aug 26 2013, Feb 19 2012, Jun 16 2011, Mar 07 2011
  • Haskell
    a000120 = concat r
    where r = [0] : (map.map) (+1) (scanl1 (++) r)
-- Luke Palmer, Feb 16 2014
  • Magma
    [Multiplicity(Intseq(n, 2), 1): n in [0..104]]; // Marius A. Burtea, Jan 22 2020
  • Magma
    [&+Intseq(n, 2):n in [0..104]]; // Marius A. Burtea, Jan 22 2020
  • Maple
    A000120 := proc(n) local w,m,i; w := 0; m := n; while m > 0 do i := m mod 2; w := w+i; m := (m-i)/2; od; w; end: wt := A000120;
A000120 := n -> add(i, i=convert(n,base,2)): # Peter Luschny, Feb 03 2011
with(Bits): p:=n->ilog2(n-And(n,n-1)): seq(p(binomial(2*n,n)),n=0..200) # Gary Detlefs, Jan 27 2019
  • Mathematica
    Table[DigitCount[n, 2, 1], {n, 0, 105}]
Nest[Flatten[# /. # -> {#, # + 1}] &, {0}, 7] (* Robert G. Wilson v, Sep 27 2011 *)
Table[Plus @@ IntegerDigits[n, 2], {n, 0, 104}]
Nest[Join[#, # + 1] &, {0}, 7] (* IWABUCHI Yu(u)ki, Jul 19 2012 *)
Log[2, Nest[Join[#, 2#] &, {1}, 14]] (* gives 2^14 term, Carlos Alves, Mar 30 2014 *)
  • PARI
    {a(n) = if( n<0, 0, 2*n - valuation((2*n)!, 2))};
  • PARI
    {a(n) = if( n<0, 0, subst(Pol(binary(n)), x ,1))};
  • PARI
    {a(n) = if( n<1, 0, a(n\2) + n%2)}; /* Michael Somos, Mar 06 2004 */
  • PARI
    a(n)=my(v=binary(n));sum(i=1,#v,v[i]) \\ Charles R Greathouse IV, Jun 24 2011
  • PARI
    a(n)=norml2(binary(n)) \\ better use {A000120=hammingweight}. - M. F. Hasler, Oct 09 2012, edited Feb 27 2020
  • PARI
    a(n)=hammingweight(n) \\ Michel Marcus, Oct 19 2013
(Common Lisp) (defun floor-to-power (n pow) (declare (fixnum pow)) (expt pow (floor (log n pow)))) (defun enabled-bits (n) (if (< n 4) (n-th n (list 0 1 1 2)) (+ (enabled-bits (floor (/ n (floor-to-power n 4)))) (enabled-bits (mod n (floor-to-power n 4)))))) ; Stephen K. Touset (stephen(AT)touset.org), Apr 04 2007
  • Python
    def A000120(n): return bin(n).count('1') # Chai Wah Wu, Sep 03 2014
  • Python
    import numpy as np
A000120 = np.array([0], dtype="uint8")
for bitrange in range(25): A000120 = np.append(A000120, np.add(A000120, 1))
print([A000120[n] for n in range(0, 105)]) # Karl-Heinz Hofmann, Nov 07 2022
  • Python
    def A000120(n): return n.bit_count() # Requires Python 3.10 or higher. - Pontus von Brömssen, Nov 08 2022
  • Python
    # Also see links.
  • SageMath
    def A000120(n):
    if n <= 1: return Integer(n)
    return A000120(n//2) + n%2
[A000120(n) for n in range(105)]  # Peter Luschny, Nov 19 2012
  • SageMath
    def A000120(n) : return sum(n.digits(2)) # Eric M. Schmidt, Apr 26 2013
  • Scala
    (0 to 127).map(Integer.bitCount()) // _Alonso del Arte, Mar 05 2019

Formula

a(0) = 0, a(2*n) = a(n), a(2*n+1) = a(n) + 1.
a(0) = 0, a(2^i) = 1; otherwise if n = 2^i + j with 0 < j < 2^i, a(n) = a(j) + 1.
G.f.: Product_{k >= 0} (1 + y*x^(2^k)) = Sum_{n >= 0} y^a(n)*x^n. - N. J. A. Sloane, Jun 04 2009
a(n) = a(n-1) + 1 - A007814(n) = log_2(A001316(n)) = 2n - A005187(n) = A070939(n) - A023416(n). - Henry Bottomley, Apr 04 2001; corrected by Ralf Stephan, Apr 15 2002
a(n) = log_2(A000984(n)/A001790(n)). - Benoit Cloitre, Oct 02 2002
For n > 0, a(n) = n - Sum_{k=1..n} A007814(k). - Benoit Cloitre, Oct 19 2002
a(n) = n - Sum_{k>=1} floor(n/2^k) = n - A011371(n). - Benoit Cloitre, Dec 19 2002
G.f.: (1/(1-x)) * Sum_{k>=0} x^(2^k)/(1+x^(2^k)). - Ralf Stephan, Apr 19 2003
a(0) = 0, a(n) = a(n - 2^floor(log_2(n))) + 1. Examples: a(6) = a(6 - 2^2) + 1 = a(2) + 1 = a(2 - 2^1) + 1 + 1 = a(0) + 2 = 2; a(101) = a(101 - 2^6) + 1 = a(37) + 1 = a(37 - 2^5) + 2 = a(5 - 2^2) + 3 = a(1 - 2^0) + 4 = a(0) + 4 = 4; a(6275) = a(6275 - 2^12) + 1 = a(2179 - 2^11) + 2 = a(131 - 2^7) + 3 = a(3 - 2^1) + 4 = a(1 - 2^0) + 5 = 5; a(4129) = a(4129 - 2^12) + 1 = a(33 - 2^5) + 2 = a(1 - 2^0) + 3 = 3. - Hieronymus Fischer, Jan 22 2006
A fixed point of the mapping 0 -> 01, 1 -> 12, 2 -> 23, 3 -> 34, 4 -> 45, ... With f(i) = floor(n/2^i), a(n) is the number of odd numbers in the sequence f(0), f(1), f(2), f(3), f(4), f(5), ... - Philippe Deléham, Jan 04 2004
When read mod 2 gives the Morse-Thue sequence A010060.
Let floor_pow4(n) denote n rounded down to the next power of four, floor_pow4(n) = 4 ^ floor(log4 n). Then a(0) = 0, a(1) = 1, a(2) = 1, a(3) = 2, a(n) = a(floor(n / floor_pow4(n))) + a(n % floor_pow4(n)). - Stephen K. Touset (stephen(AT)touset.org), Apr 04 2007
a(n) = n - Sum_{k=2..n} Sum_{j|n, j >= 2} (floor(log_2(j)) - floor(log_2(j-1))). - Hieronymus Fischer, Jun 18 2007
a(n) = A138530(n, 2) for n > 1. - Reinhard Zumkeller, Mar 26 2008
a(A077436(n)) = A159918(A077436(n)); a(A000290(n)) = A159918(n). - Reinhard Zumkeller, Apr 25 2009
a(n) = A063787(n) - A007814(n). - Gary W. Adamson, Jun 04 2009
a(n) = A007814(C(2n, n)) = 1 + A007814(C(2n-1, n)). - Vladimir Shevelev, Jul 20 2009
For odd m >= 1, a((4^m-1)/3) = a((2^m+1)/3) + (m-1)/2 (mod 2). - Vladimir Shevelev, Sep 03 2010
a(n) - a(n-1) = { 1 - a(n-1) if and only if A007814(n) = a(n-1), 1 if and only if A007814(n) = 0, -1 for all other A007814(n) }. - Brent Spillner (spillner(AT)acm.org), Sep 01 2010
a(A001317(n)) = 2^a(n). - Vladimir Shevelev, Oct 25 2010
a(n) = A139351(n) + A139352(n) = Sum_k {A030308(n, k)}. - Philippe Deléham, Oct 14 2011
From Hieronymus Fischer, Jun 10 2012: (Start)
a(n) = Sum_{j = 1..m+1} (floor(n/2^j + 1/2) - floor(n/2^j)), where m = floor(log_2(n)).
General formulas for the number of digits >= d in the base p representation of n, where 1 <= d < p: a(n) = Sum_{j = 1..m+1} (floor(n/p^j + (p-d)/p) - floor(n/p^j)), where m=floor(log_p(n)); g.f.: g(x) = (1/(1-x))*Sum_{j>=0} (x^(d*p^j) - x^(p*p^j))/(1-x^(p*p^j)). (End)
a(n) = A213629(n, 1) for n > 0. - Reinhard Zumkeller, Jul 04 2012
a(n) = A240857(n,n). - Reinhard Zumkeller, Apr 14 2014
a(n) = log_2(C(2*n,n) - (C(2*n,n) AND C(2*n,n)-1)). - Gary Detlefs, Jul 10 2014
Sum_{n >= 1} a(n)/2n(2n+1) = (gamma + log(4/Pi))/2 = A344716, where gamma is Euler's constant A001620; see Sondow 2005, 2010 and Allouche, Shallit, Sondow 2007. - Jonathan Sondow, Mar 21 2015
For any integer base b >= 2, the sum of digits s_b(n) of expansion base b of n is the solution of this recurrence relation: s_b(n) = 0 if n = 0 and s_b(n) = s_b(floor(n/b)) + (n mod b). Thus, a(n) satisfies: a(n) = 0 if n = 0 and a(n) = a(floor(n/2)) + (n mod 2). This easily yields a(n) = Sum_{i = 0..floor(log_2(n))} (floor(n/2^i) mod 2). From that one can compute a(n) = n - Sum_{i = 1..floor(log_2(n))} floor(n/2^i). - Marek A. Suchenek, Mar 31 2016
Sum_{k>=1} a(k)/2^k = 2 * Sum_{k >= 0} 1/(2^(2^k)+1) = 2 * A051158. - Amiram Eldar, May 15 2020
Sum_{k>=1} a(k)/(k*(k+1)) = A016627 = log(4). - Bernard Schott, Sep 16 2020
a(m*(2^n-1)) >= n. Equality holds when 2^n-1 >= A000265(m), but also in some other cases, e.g., a(11*(2^2-1)) = 2 and a(19*(2^3-1)) = 3. - Pontus von Brömssen, Dec 13 2020
G.f.: A(x) satisfies A(x) = (1+x)*A(x^2) + x/(1-x^2). - Akshat Kumar, Nov 04 2023

A056239 If n = Product_{k >= 1} (p_k)^(c_k) where p_k is k-th prime and c_k >= 0 then a(n) = Sum_{k >= 1} k*c_k.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 4, 6, 5, 5, 4, 7, 5, 8, 5, 6, 6, 9, 5, 6, 7, 6, 6, 10, 6, 11, 5, 7, 8, 7, 6, 12, 9, 8, 6, 13, 7, 14, 7, 7, 10, 15, 6, 8, 7, 9, 8, 16, 7, 8, 7, 10, 11, 17, 7, 18, 12, 8, 6, 9, 8, 19, 9, 11, 8, 20, 7, 21, 13, 8, 10, 9, 9, 22, 7, 8, 14, 23, 8, 10, 15, 12, 8, 24, 8, 10
Offset: 1

Views

Author

Leroy Quet, Aug 19 2000

Keywords

Comments

A pseudo-logarithmic function in the sense that a(b*c) = a(b)+a(c) and so a(b^c) = c*a(b) and f(n) = k^a(n) is a multiplicative function. [Cf. A248692 for example.] Essentially a function from the positive integers onto the partitions of the nonnegative integers (1->0, 2->1, 3->2, 4->1+1, 5->3, 6->1+2, etc.) so each value a(n) appears A000041(a(n)) times, first with the a(n)-th prime and last with the a(n)-th power of 2. Produces triangular numbers from primorials. - Henry Bottomley, Nov 22 2001
Michael Nyvang writes (May 08 2006) that the Danish composer Karl Aage Rasmussen discovered this sequence in the 1990's: it has excellent musical properties.
All A000041(a(n)) different n's with the same value a(n) are listed in row a(n) of triangle A215366. - Alois P. Heinz, Aug 09 2012
a(n) is the sum of the parts of the partition having Heinz number n. We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product_{j=1..r} (p_j-th prime) (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] we get 2*2*3*7*29 = 2436. Example: a(33) = 7 because the partition with Heinz number 33 = 3 * 11 is [2,5]. - Emeric Deutsch, May 19 2015

Examples

			a(12) = 1*2 + 2*1 = 4, since 12 = 2^2 *3^1 = (p_1)^2 *(p_2)^1.

Links

Crossrefs

Row sums of A112798.
Cf. A003963 (gives the corresponding products).
Cf. also A000041, A000217, A000720, A001221, A001222, A002110, A005940, A008475, A027748, A049084, A060437, A081401, A082090, A088314, A088318, A088850, A088880, A088881, A088887, A088902, A108951, A122111, A124010, A127668, A141128, A153734, A154351, A156552, A163517, A178503, A215366, A215369, A215501, A242422, A242423, A242424, A243070, A243353, A243354, A243503, A248692, A249336, A249337, A329605, A329902.

Programs

  • Haskell
    a056239 n = sum $ zipWith (*) (map a049084 $ a027748_row n) (a124010_row n)
-- Reinhard Zumkeller, Apr 27 2013
  • Maple
    # To get 10000 terms. First make prime table: M:=10000; pl:=array(1..M); for i from 1 to M do pl[i]:=0; od: for i from 1 to M do if ithprime(i) > M then break; fi; pl[ithprime(i)]:=i; od:
# Decode Maple's amazing syntax for factoring integers: g:=proc(n) local e,p,t1,t2,t3,i,j,k; global pl; t1:=ifactor(n); t2:=nops(t1); if t2 = 2 and whattype(t1) <> `*` then p:=op(1,op(1,t1)); e:=op(2,t1); t3:=pl[p]*e; else
t3:=0; for i from 1 to t2 do j:=op(i,t1); if nops(j) = 1 then e:=1; p:=op(1,j); else e:=op(2,j); p:=op(1,op(1,j)); fi; t3:=t3+pl[p]*e; od: fi; t3; end; # N. J. A. Sloane, May 10 2006
A056239 := proc(n) add( numtheory[pi](op(1,p))*op(2,p), p = ifactors(n)[2]) ; end proc: # R. J. Mathar, Apr 20 2010
# alternative:
with(numtheory): a := proc (n) local B: B := proc (n) local nn, j, m: nn := op(2, ifactors(n)): for j to nops(nn) do m[j] := op(j, nn) end do: [seq(seq(pi(op(1, m[i])), q = 1 .. op(2, m[i])), i = 1 .. nops(nn))] end proc: add(B(n)[i], i = 1 .. nops(B(n))) end proc: seq(a(n), n = 1 .. 130); # Emeric Deutsch, May 19 2015
  • Mathematica
    a[1] = 0; a[2] = 1; a[p_?PrimeQ] := a[p] = PrimePi[p];
a[n_] := a[n] = Total[#[[2]]*a[#[[1]]] & /@ FactorInteger[n]]; a /@ Range[91] (* Jean-François Alcover, May 19 2011 *)
Table[Total[FactorInteger[n] /. {p_, c_} /; p > 0 :> PrimePi[p] c], {n, 91}] (* Michael De Vlieger, Jul 12 2017 *)
  • PARI
    A056239(n) = if(1==n,0,my(f=factor(n)); sum(i=1, #f~, f[i,2] * primepi(f[i,1]))); \\ Antti Karttunen, Oct 26 2014, edited Jan 13 2020
  • Python
    from sympy import primepi, factorint
def A056239(n): return sum(primepi(p)*e for p, e in factorint(n).items()) # Chai Wah Wu, Jan 01 2023
  • Scheme
    (require 'factor) ;; Uses the function factor available in Aubrey Jaffer's SLIB Scheme library.
(define (A056239 n) (apply + (map A049084 (factor n))))
;; Antti Karttunen, Oct 26 2014

Formula

Totally additive with a(p) = PrimePi(p), where PrimePi(n) = A000720(n).
a(n) = Sum_{k=1..A001221(n)} A049084(A027748(k))*A124010(k). - Reinhard Zumkeller, Apr 27 2013
From Antti Karttunen, Oct 11 2014: (Start)
a(n) = n - A178503(n).
a(n) = A161511(A156552(n)).
a(n) = A227183(A243354(n)).
For all n >= 0:
a(A002110(n)) = A000217(n). [Cf. Henry Bottomley's comment above.]
a(A005940(n+1)) = A161511(n).
a(A243353(n)) = A227183(n).
Also, for all n >= 1:
a(A241909(n)) = A243503(n).
a(A122111(n)) = a(n).
a(A242424(n)) = a(n).
A248692(n) = 2^a(n). (End)
a(n) < A329605(n), a(n) = A001222(A108951(n)), a(A329902(n)) = A112778(n). - Antti Karttunen, Jan 14 2020

A003415 a(n) = n' = arithmetic derivative of n: a(0) = a(1) = 0, a(prime) = 1, a(m*n) = m*a(n) + n*a(m).

Original entry on oeis.org

0, 0, 1, 1, 4, 1, 5, 1, 12, 6, 7, 1, 16, 1, 9, 8, 32, 1, 21, 1, 24, 10, 13, 1, 44, 10, 15, 27, 32, 1, 31, 1, 80, 14, 19, 12, 60, 1, 21, 16, 68, 1, 41, 1, 48, 39, 25, 1, 112, 14, 45, 20, 56, 1, 81, 16, 92, 22, 31, 1, 92, 1, 33, 51, 192, 18, 61, 1, 72, 26, 59, 1, 156, 1, 39, 55, 80, 18, 71
Offset: 0

Views

Author

N. J. A. Sloane, R. K. Guy

Keywords

Comments

Can be extended to negative numbers by defining a(-n) = -a(n).
Based on the product rule for differentiation of functions: for functions f(x) and g(x), (fg)' = f'g + fg'. So with numbers, (ab)' = a'b + ab'. This implies 1' = 0. - Kerry Mitchell, Mar 18 2004
The derivative of a number x with respect to a prime number p as being the number "dx/dp" = (x-x^p)/p, which is an integer due to Fermat's little theorem. - Alexandru Buium, Mar 18 2004
The relation (ab)' = a'b + ab' implies 1' = 0, but it does not imply p' = 1 for p a prime. In fact, any function f defined on the primes can be extended uniquely to a function on the integers satisfying this relation: f(Product_i p_i^e_i) = (Product_i p_i^e_i) * (Sum_i e_i*f(p_i)/p_i). - Franklin T. Adams-Watters, Nov 07 2006
See A131116 and A131117 for record values and where they occur. - Reinhard Zumkeller, Jun 17 2007
Let n be the product of a multiset P of k primes. Consider the k-dimensional box whose edges are the elements of P. Then the (k-1)-dimensional surface of this box is 2*a(n). For example, 2*a(25) = 20, the perimeter of a 5 X 5 square. Similarly, 2*a(18) = 42, the surface area of a 2 X 3 X 3 box. - David W. Wilson, Mar 11 2011
The arithmetic derivative n' was introduced, probably for the first time, by the Spanish mathematician José Mingot Shelly in June 1911 with "Una cuestión de la teoría de los números", work presented at the "Tercer Congreso Nacional para el Progreso de las Ciencias, Granada", cf. link to the abstract on Zentralblatt MATH, and L. E. Dickson, History of the Theory of Numbers. - Giorgio Balzarotti, Oct 19 2013
a(A235991(n)) odd; a(A235992(n)) even. - Reinhard Zumkeller, Mar 11 2014
Sequence A157037 lists numbers with prime arithmetic derivative, i.e., indices of primes in this sequence. - M. F. Hasler, Apr 07 2015
Maybe the simplest "natural extension" of the arithmetic derivative, in the spirit of the above remark by Franklin T. Adams-Watters (2006), is the "pi based" version where f(p) = primepi(p), see sequence A258851. When f is chosen to be the identity map (on primes), one gets A066959. - M. F. Hasler, Jul 13 2015
When n is composite, it appears that a(n) has lower bound 2*sqrt(n), with equality when n is the square of a prime, and a(n) has upper bound (n/2)*log_2(n), with equality when n is a power of 2. - Daniel Forgues, Jun 22 2016
If n = p1*p2*p3*... where p1, p2, p3, ... are all the prime factors of n (not necessarily distinct), and h is a real number (we assume h nonnegative and < 1), the arithmetic derivative of n is equivalent to n' = lim_{h->0} ((p1+h)*(p2+h)*(p3+h)*... - (p1*p2*p3*...))/h. It also follows that the arithmetic derivative of a prime is 1. We could assume h = 1/N, where N is an integer; then the limit becomes {N -> oo}. Note that n = 1 is not a prime and plays the role of constant. - Giorgio Balzarotti, May 01 2023

Examples

			6' = (2*3)' = 2'*3 + 2*3' = 1*3 + 2*1 = 5.
Note that, for example, 2' + 3' = 1 + 1 = 2, (2+3)' = 5' = 1. So ' is not linear.
G.f. = x^2 + x^3 + 4*x^4 + x^5 + 5*x^6 + x^7 + 12*x^8 + 6*x^9 + 7*x^10 + ...

References

  • G. Balzarotti, P. P. Lava, La derivata aritmetica, Editore U. Hoepli, Milano, 2013.
  • E. J. Barbeau, Problem, Canad. Math. Congress Notes, 5 (No. 8, April 1973), 6-7.
  • L. E. Dickson, History of the Theory of Numbers, Vol. 1, Chapter XIX, p. 451, Dover Edition, 2005. (Work originally published in 1919.)
  • A. M. Gleason et al., The William Lowell Putnam Mathematical Competition: Problems and Solutions 1938-1964, Math. Assoc. America, 1980, p. 295.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A086134 (least prime factor of n').
Cf. A086131 (greatest prime factor of n').
Cf. A068719 (derivative of 2n).
Cf. A068720 (derivative of n^2).
Cf. A068721 (derivative of n^3).
Cf. A001787 (derivative of 2^n).
Cf. A027471 (derivative of 3^(n-1)).
Cf. A085708 (derivative of 10^n).
Cf. A068327 (derivative of n^n).
Cf. A024451 (derivative of p#).
Cf. A068237 (numerator of derivative of 1/n).
Cf. A068238 (denominator of derivative of 1/n).
Cf. A068328 (derivative of squarefree numbers).
Cf. A068311 (derivative of n!).
Cf. A168386 (derivative of n!!).
Cf. A260619 (derivative of hyperfactorial(n)).
Cf. A260620 (derivative of superfactorial(n)).
Cf. A068312 (derivative of triangular numbers).
Cf. A068329 (derivative of Fibonacci(n)).
Cf. A096371 (derivative of partition number).
Cf. A099301 (derivative of d(n)).
Cf. A099310 (derivative of phi(n)).
Cf. A342925 (derivative of sigma(n)).
Cf. A349905 (derivative of prime shift).
Cf. A327860 (derivative of primorial base exp-function).
Cf. A369252 (derivative of products of three odd primes), A369251 (same sorted).
Cf. A068346 (second derivative of n).
Cf. A099306 (third derivative of n).
Cf. A258644 (fourth derivative of n).
Cf. A258645 (fifth derivative of n).
Cf. A258646 (sixth derivative of n).
Cf. A258647 (seventh derivative of n).
Cf. A258648 (eighth derivative of n).
Cf. A258649 (ninth derivative of n).
Cf. A258650 (tenth derivative of n).
Cf. A185232 (n-th derivative of n).
Cf. A258651 (A(n,k) = k-th arithmetic derivative of n).
Cf. A085731 (gcd(n,n')), A083345 (n'/gcd(n,n')), A057521 (gcd(n, (n')^k) for k>1).
Cf. A342014 (n' mod n), A369049 (n mod n').
Cf. A341998 (A003557(n')), A342001 (n'/A003557(n)).
Cf. A098699 (least x such that x' = n, antiderivative of n).
Cf. A098700 (n such that x' = n has no integer solution).
Cf. A099302 (number of solutions to x' = n).
Cf. A099303 (greatest x such that x' = n).
Cf. A051674 (n such that n' = n).
Cf. A083347 (n such that n' < n).
Cf. A083348 (n such that n' > n).
Cf. A099304 (least k such that (n+k)' = n' + k').
Cf. A099305 (number of solutions to (n+k)' = n' + k').
Cf. A328235 (least k > 0 such that (n+k)' = u * n' for some natural number u).
Cf. A328236 (least m > 1 such that (m*n)' = u * n' for some natural number u).
Cf. A099307 (least k such that the k-th arithmetic derivative of n is zero).
Cf. A099308 (k-th arithmetic derivative of n is zero for some k).
Cf. A099309 (k-th arithmetic derivative of n is nonzero for all k).
Cf. A129150 (n-th derivative of 2^3).
Cf. A129151 (n-th derivative of 3^4).
Cf. A129152 (n-th derivative of 5^6).
Cf. A189481 (x' = n has a unique solution).
Cf. A190121 (partial sums).
Cf. A258057 (first differences).
Cf. A229501 (n divides the n-th partial sum).
Cf. A165560 (parity).
Cf. A235991 (n' is odd), A235992 (n' is even).
Cf. A327863, A327864, A327865 (n' is a multiple of 3, 4, 5).
Cf. A157037 (n' is prime), A192192 (n'' is prime), A328239 (n''' is prime).
Cf. A328393 (n' is squarefree), A328234 (squarefree and > 1).
Cf. A328244 (n'' is squarefree), A328246 (n''' is squarefree).
Cf. A328303 (n' is not squarefree), A328252 (n' is squarefree, but n is not).
Cf. A328248 (least k such that the (k-1)-th derivative of n is squarefree).
Cf. A328251 (k-th arithmetic derivative is never squarefree for any k >= 0).
Cf. A256750 (least k such that the k-th derivative is either 0 or has a factor p^p).
Cf. A327928 (number of distinct primes p such that p^p divides n').
Cf. A342003 (max. exponent k for any prime power p^k that divides n').
Cf. A327929 (n' has at least one divisor of the form p^p).
Cf. A327978 (n' is primorial number > 1).
Cf. A328243 (n' is a partial sum of primorial numbers and larger than one).
Cf. A328310 (maximal prime exponent of n' minus maximal prime exponent of n).
Cf. A328320 (max. prime exponent of n' is less than that of n).
Cf. A328321 (max. prime exponent of n' is >= that of n).
Cf. A328383 (least k such that the k-th derivative of n is either a multiple or a divisor of n, but not both).
Cf. A263111 (the ordinal transform of a).
Cf. A300251, A319684 (Möbius and inverse Möbius transform).
Cf. A305809 (Dirichlet convolution square).
Cf. A349133, A349173, A349394, A349380, A349618, A349619, A349620, A349621 (for miscellaneous Dirichlet convolutions).
Cf. A069359 (similar formula which agrees on squarefree numbers).
Cf. A258851 (the pi-based arithmetic derivative of n).
Cf. A328768, A328769 (primorial-based arithmetic derivatives of n).
Cf. A328845, A328846 (Fibonacci-based arithmetic derivatives of n).
Cf. A302055, A327963, A327965, A328099 (for other variants and modifications).
Cf. A038554 (another sequence using "derivative" in its name, but involving binary expansion of n).
Cf. A322582, A348507 (lower and upper bounds), also A002620.

Programs

  • GAP
    A003415:= Concatenation([0,0],List(List([2..10^3],Factors),
i->Product(i)*Sum(i,j->1/j))); # Muniru A Asiru, Aug 31 2017
(APL, Dyalog dialect) A003415 ← { ⍺←(0 1 2) ⋄ ⍵≤1:⊃⍺ ⋄ 0=(3⊃⍺)|⍵:((⊃⍺+(2⊃⍺)×(⍵÷3⊃⍺)) ((2⊃⍺)×(3⊃⍺)) (3⊃⍺)) ∇ ⍵÷3⊃⍺ ⋄ ((⊃⍺) (2⊃⍺) (1+(3⊃⍺))) ∇ ⍵} ⍝ Antti Karttunen, Feb 18 2024
  • Haskell
    a003415 0 = 0
a003415 n = ad n a000040_list where
  ad 1 _             = 0
  ad n ps'@(p:ps)
     | n < p * p     = 1
     | r > 0         = ad n ps
     | otherwise     = n' + p * ad n' ps' where
       (n',r) = divMod n p
-- Reinhard Zumkeller, May 09 2011
  • Magma
    Ad:=func; [n le 1 select 0 else Ad(n): n in [0..80]]; // Bruno Berselli, Oct 22 2013
  • Maple
    A003415 := proc(n) local B,m,i,t1,t2,t3; B := 1000000000039; if n<=1 then RETURN(0); fi; if isprime(n) then RETURN(1); fi; t1 := ifactor(B*n); m := nops(t1); t2 := 0; for i from 1 to m do t3 := op(i,t1); if nops(t3) = 1 then t2 := t2+1/op(t3); else t2 := t2+op(2,t3)/op(op(1,t3)); fi od: t2 := t2-1/B; n*t2; end;
A003415 := proc(n)
        local a,f;
        a := 0 ;
        for f in ifactors(n)[2] do
                a := a+ op(2,f)/op(1,f);
        end do;
        n*a ;
end proc: # R. J. Mathar, Apr 05 2012
  • Mathematica
    a[ n_] := If[ Abs @ n < 2, 0, n Total[ #2 / #1 & @@@ FactorInteger[ Abs @ n]]]; (* Michael Somos, Apr 12 2011 *)
dn[0] = 0; dn[1] = 0; dn[n_?Negative] := -dn[-n]; dn[n_] := Module[{f = Transpose[FactorInteger[n]]}, If[PrimeQ[n], 1, Total[n*f[[2]]/f[[1]]]]]; Table[dn[n], {n, 0, 100}] (* T. D. Noe, Sep 28 2012 *)
  • PARI
    A003415(n) = {local(fac);if(n<1,0,fac=factor(n);sum(i=1,matsize(fac)[1],n*fac[i,2]/fac[i,1]))} /* Michael B. Porter, Nov 25 2009 */
  • PARI
    apply( A003415(n)=vecsum([n/f[1]*f[2]|f<-factor(n+!n)~]), [0..99]) \\ M. F. Hasler, Sep 25 2013, updated Nov 27 2019
  • PARI
    A003415(n) = { my(s=0, m=1, spf); while(n>1, spf = A020639(n); n /= spf; s += m*n; m *= spf); (s); }; \\ Antti Karttunen, Mar 10 2021
  • PARI
    a(n) = my(f=factor(n), r=[1/(e+!e)|e<-f[,1]], c=f[,2]); n*r*c; \\ Ruud H.G. van Tol, Sep 03 2023
  • Python
    from sympy import factorint
def A003415(n):
    return sum([int(n*e/p) for p,e in factorint(n).items()]) if n > 1 else 0
# Chai Wah Wu, Aug 21 2014
  • Sage
    def A003415(n):
    F = [] if n == 0 else factor(n)
    return n * sum(g / f for f, g in F)
[A003415(n) for n in range(79)] # Peter Luschny, Aug 23 2014

Formula

If n = Product p_i^e_i, a(n) = n * Sum (e_i/p_i).
a(m*p^p) = (m + a(m))*p^p, p prime: a(m*A051674(k))=A129283(m)*A051674(k). - Reinhard Zumkeller, Apr 07 2007
For n > 1: a(n) = a(A032742(n)) * A020639(n) + A032742(n). - Reinhard Zumkeller, May 09 2011
a(n) = n * Sum_{p|n} v_p(n)/p, where v_p(n) is the largest power of the prime p dividing n. - Wesley Ivan Hurt, Jul 12 2015
For n >= 2, Sum_{k=2..n} floor(1/a(k)) = pi(n) = A000720(n) (see K. T. Atanassov article). - Ivan N. Ianakiev, Mar 22 2019
From A.H.M. Smeets, Jan 17 2020: (Start)
Limit_{n -> oo} (1/n^2)*Sum_{i=1..n} a(i) = A136141/2.
Limit_{n -> oo} (1/n)*Sum_{i=1..n} a(i)/i = A136141.
a(n) = n if and only if n = p^p, where p is a prime number. (End)
Dirichlet g.f.: zeta(s-1)*Sum_{p prime} 1/(p^s-p), see A136141 (s=2), A369632 (s=3) [Haukkanen, Merikoski and Tossavainen]. - Sebastian Karlsson, Nov 25 2021
From Antti Karttunen, Nov 25 2021: (Start)
a(n) = Sum_{d|n} d * A349394(n/d).
For all n >= 1, A322582(n) <= a(n) <= A348507(n).
If n is not a prime, then a(n) >= 2*sqrt(n), or in other words, for all k >= 1 for which A002620(n)+k is not a prime, we have a(A002620(n)+k) > n. [See Ufnarovski and Åhlander, Theorem 9, point (3).]
(End)

Extensions

More terms from Michel ten Voorde, Apr 11 2001

A001223 Prime gaps: differences between consecutive primes.

Original entry on oeis.org

1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 2, 4, 14, 4, 6, 2, 10, 2, 6, 6, 4, 6, 6, 2, 10, 2, 4, 2, 12, 12, 4, 2, 4, 6, 2, 10, 6, 6, 6, 2, 6, 4, 2, 10, 14, 4, 2, 4, 14, 6, 10, 2, 4, 6, 8, 6, 6, 4, 6, 8, 4, 8, 10, 2, 10, 2, 6, 4, 6, 8, 4, 2, 4, 12, 8, 4, 8, 4, 6, 12
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

There is a unique decomposition of the primes: provided the weight A117078(n) is > 0, we have prime(n) = weight * level + gap, or A000040(n) = A117078(n) * A117563(n) + a(n). - Rémi Eismann, Feb 14 2008
Let rho(m) = A179196(m), for any n, let m be an integer such that p_(rho(m)) <= p_n and p_(n+1) <= p_(rho(m+1)), then rho(m) <= n < n + 1 <= rho(m + 1), therefore a(n) = p_(n+1) - p_n <= p_rho(m+1) - p_rho(m) = A182873(m). For all rho(m) = A179196(m), a(rho(m)) < A165959(m). - John W. Nicholson, Dec 14 2011
A solution (modular square root) of x^2 == A001248(n) (mod A000040(n+1)). - L. Edson Jeffery, Oct 01 2014
There exists a constant C such that for n -> infinity, Cramer conjecture a(n) < C log^2 prime(n) is equivalent to (log prime(n+1)/log prime(n))^n < e^C. - Thomas Ordowski, Oct 11 2014
a(n) = A008347(n+1) - A008347(n-1). - Reinhard Zumkeller, Feb 09 2015
Yitang Zhang proved lim inf_{n -> infinity} a(n) is finite. - Robert Israel, Feb 12 2015
lim sup_{n -> infinity} a(n)/log^2 prime(n) = C <==> lim sup_{n -> infinity}(log prime(n+1)/log prime(n))^n = e^C. - Thomas Ordowski, Mar 09 2015
a(A038664(n)) = 2*n and a(m) != 2*n for m < A038664(n). - Reinhard Zumkeller, Aug 23 2015
If j and k are positive integers then there are no two consecutive primes gaps of the form 2+6j and 2+6k (A016933) or 4+6j and 4+6k (A016957). - Andres Cicuttin, Jul 14 2016
Conjecture: For any positive numbers x and y, there is an index k such that x/y = a(k)/a(k+1). - Andres Cicuttin, Sep 23 2018
Conjecture: For any three positive numbers x, y and j, there is an index k such that x/y = a(k)/a(k+j). - Andres Cicuttin, Sep 29 2018
Conjecture: For any three positive numbers x, y and j, there are infinitely many indices k such that x/y = a(k)/a(k+j). - Andres Cicuttin, Sep 29 2018
Row m of A174349 lists all indices n for which a(n) = 2m. - M. F. Hasler, Oct 26 2018
Since (6a, 6b) is an admissible pattern of gaps for any integers a, b > 0 (and also if other multiples of 6 are inserted in between), the above conjecture follows from the prime k-tuple conjecture which states that any admissible pattern occurs infinitely often (see, e.g., the Caldwell link). This also means that any subsequence a(n .. n+m) with n > 2 (as to exclude the untypical primes 2 and 3) should occur infinitely many times at other starting points n'. - M. F. Hasler, Oct 26 2018
Conjecture: Defining b(n,j,k) as the number of pairs of prime gaps {a(i),a(i+j)} such that i < n, j > 0, and a(i)/a(i+j) = k with k > 0, then
lim_{n -> oo} b(n,j,k)/b(n,j,1/k) = 1, for any j > 0 and k > 0, and
lim_{n -> oo} b(n,j,k1)/b(n,j,k2) = C with C = C(j,k1,k2) > 0. - Andres Cicuttin, Sep 01 2019

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 870.
  • GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See page 92.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 186-192.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A000040 (primes), A001248 (primes squared), A000720, A037201, A007921, A030173, A036263-A036274, A167770, A008347.
Second difference is A036263, first occurrence is A000230.
For records see A005250, A005669.
Cf. A038664, A031131, A031165, A031166, A031167, A031168, A031169, A031170, A031171, A031172.
Cf. A174349, A029707, A029709, A320701, ..., A320720.
Sequences related to the differences between successive primes: A001223 (Delta(p)), A028334, A080378, A104120, A330556-A330561.

Programs

  • Haskell
    a001223 n = a001223_list !! (n-1)
a001223_list = zipWith (-) (tail a000040_list) a000040_list
-- Reinhard Zumkeller, Oct 29 2011
  • Magma
    [(NthPrime(n+1) - NthPrime(n)): n in [1..100]]; // Vincenzo Librandi, Apr 02 2011
  • Maple
    with(numtheory): for n from 1 to 500 do printf(`%d,`,ithprime(n+1) - ithprime(n)) od:
  • Mathematica
    Differences[Prime[Range[100]]] (* Harvey P. Dale, May 15 2011 *)
  • PARI
    diff(v)=vector(#v-1,i,v[i+1]-v[i]);
diff(primes(100)) \\ Charles R Greathouse IV, Feb 11 2011
  • PARI
    forprime(p=1, 1e3, print1(nextprime(p+1)-p, ", ")) \\ Felix Fröhlich, Sep 06 2014
  • Python
    from sympy import prime
def A001223(n): return prime(n+1)-prime(n) # Chai Wah Wu, Jul 07 2022
  • Sage
    differences(prime_range(1000)) # Joerg Arndt, May 15 2011

Formula

G.f.: b(x)*(1-x), where b(x) is the g.f. for the primes. - Franklin T. Adams-Watters, Jun 15 2006
a(n) = prime(n+1) - prime(n). - Franklin T. Adams-Watters, Mar 31 2010
Conjectures: (i) a(n) = ceiling(prime(n)*log(prime(n+1)/prime(n))). (ii) a(n) = floor(prime(n+1)*log(prime(n+1)/prime(n))). (iii) a(n) = floor((prime(n)+prime(n+1))*log(prime(n+1)/prime(n))/2). - Thomas Ordowski, Mar 21 2013
A167770(n) == a(n)^2 (mod A000040(n+1)). - L. Edson Jeffery, Oct 01 2014
a(n) = Sum_{k=1..2^(n+1)-1} (floor(cos^2(Pi*(n+1)^(1/(n+1))/(1+primepi(k))^(1/(n+1))))). - Anthony Browne, May 11 2016
G.f.: (Sum_{k>=1} x^pi(k)) - 1, where pi(k) is the prime counting function. - Benedict W. J. Irwin, Jun 13 2016
Conjecture: Limit_{N->oo} (Sum_{n=2..N} log(a(n))) / (Sum_{n=2..N} log(log(prime(n)))) = 1. - Alain Rocchelli, Dec 16 2022
Conjecture: The asymptotic limit of the average of log(a(n)) ~ log(log(prime(n))) - gamma (where gamma is Euler's constant). Also, for n tending to infinity, the geometric mean of a(n) is equivalent to log(prime(n)) / e^gamma. - Alain Rocchelli, Jan 23 2023
It has been conjectured that primes are distributed around their average spacing in a Poisson distribution (cf. D. A. Goldston in above links). This is the basis of the last two conjectures above. - Alain Rocchelli, Feb 10 2023

Extensions

More terms from James Sellers, Feb 19 2001

A001511 The ruler function: exponent of the highest power of 2 dividing 2n. Equivalently, the 2-adic valuation of 2n.

Original entry on oeis.org

1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 7, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of 2's dividing 2*n.
a(n) is equivalently the exponent of the smallest power of 2 which does not divide n. - David James Sycamore, Oct 02 2023
a(n) - 1 is the number of trailing zeros in the binary expansion of n.
If you are counting in binary and the least significant bit is numbered 1, the next bit is 2, etc., a(n) is the bit that is incremented when increasing from n-1 to n. - Jud McCranie, Apr 26 2004
Number of steps to reach an integer starting with (n+1)/2 and using the map x -> x*ceiling(x) (cf. A073524).
a(n) is the number of the disk to be moved at the n-th step of the optimal solution to Towers of Hanoi problem (comment from Andreas M. Hinz).
Shows which bit to flip when creating the binary reflected Gray code (bits are numbered from the right, offset is 1). This is essentially equivalent to Hinz's comment. - Adam Kertesz, Jul 28 2001
a(n) is the Hamming distance between n and n-1 (in binary). This is equivalent to Kertesz's comments above. - Tak-Shing Chan (chan12(AT)alumni.usc.edu), Feb 25 2003
Let S(0) = {1}, S(n) = {S(n-1), S(n-1)-{x}, x+1} where x = last term of S(n-1); sequence gives S(infinity). - Benoit Cloitre, Jun 14 2003
The sum of all terms up to and including the first occurrence of m is 2^m-1. - Donald Sampson (marsquo(AT)hotmail.com), Dec 01 2003
m appears every 2^m terms starting with the 2^(m-1)th term. - Donald Sampson (marsquo(AT)hotmail.com), Dec 08 2003
Sequence read mod 4 gives A092412. - Philippe Deléham, Mar 28 2004
If q = 2n/2^A001511(n) and if b(m) is defined by b(0)=q-1 and b(m)=2*b(m-1)+1, then 2n = b(A001511(n)) + 1. - Gerald McGarvey, Dec 18 2004
Repeating pattern ABACABADABACABAE ... - Jeremy Gardiner, Jan 16 2005
Relation to C(n) = Collatz function iteration using only odd steps: a(n) is the number of right bits set in binary representation of A004767(n) (numbers of the form 4*m+3). So for m=A004767(n) it follows that there are exactly a(n) recursive steps where m
Between every two instances of any positive integer m there are exactly m distinct values (1 through m-1 and one value greater than m). - Franklin T. Adams-Watters, Sep 18 2006
Number of divisors of n of the form 2^k. - Giovanni Teofilatto, Jul 25 2007
Every prefix up to (but not including) the first occurrence of some k >= 2 is a palindrome. - Gary W. Adamson, Sep 24 2008
1 interleaved with (2 interleaved with (3 interleaved with ( ... ))). - Eric D. Burgess (ericdb(AT)gmail.com), Oct 17 2009
A054525 (Möbius transform) * A001511 = A036987 = A047999^(-1) * A001511. - Gary W. Adamson, Oct 26 2009
Equals A051731 * A036987, (inverse Möbius transform of the Fredholm-Rueppel sequence) = A047999 * A036987. - Gary W. Adamson, Oct 26 2009
Cf. A173238, showing links between generalized ruler functions and A000041. - Gary W. Adamson, Feb 14 2010
Given A000041, P(x) = A(x)/A(x^2) with P(x) = (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...), A(x) = (1 + x + 3x^2 + 4x^3 + 10x^4 + 13x^5 + ...), A(x^2) = (1 + x^2 + 3x^4 + 4x^6 + 10x^8 + ...), where A092119 = (1, 1, 3, 4, 10, ...) = Euler transform of the ruler sequence, A001511. - Gary W. Adamson, Feb 11 2010
Subtracting 1 from every term and deleting any 0's yields the same sequence, A001511. - Ben Branman, Dec 28 2011
In the listing of the compositions of n as lists in lexicographic order, a(k) is the last part of composition(k) for all k <= 2^(n-1) and all n, see example. - Joerg Arndt, Nov 12 2012
According to Hinz, et al. (see links), this sequence was studied by Louis Gros in his 1872 pamphlet "Théorie du Baguenodier" and has therefore been called the Gros sequence.
First n terms comprise least squarefree word of length n using positive integers, where "squarefree" means that the word contains no consecutive identical subwords; e.g., 1 contains no square; 11 contains a square but 12 does not; 121 contains no square; both 1211 and 1212 have squares but 1213 does not; etc. - Clark Kimberling, Sep 05 2013
Length of 0-run starting from 2 (10, 100, 110, 1000, 1010, ...), or length of 1-run starting from 1 (1, 11, 101, 111, 1001, 1011, ...) of every second number, from right to left in binary representation. - Armands Strazds, Apr 13 2017
a(n) is also the frequency of the largest part in the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2,2,2,1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 24 2017
As A000005(n) equals the number of even divisors of 2n and A001227(n) = A001227(2n), the formula A001511(n) = A000005(n)/A001227(n) might be read as "The number of even divisors of 2n is always divisible by the number of odd divisors of 2n" (where number of divisors means sum of zeroth powers of divisors). Conjecture: For any nonnegative integer k, the sum of the k-th powers of even divisors of n is always divisible by the sum of the k-th powers of odd divisors of n. - Ivan N. Ianakiev, Jul 06 2019
From Benoit Cloitre, Jul 14 2022: (Start)
To construct the sequence, start from 1's separated by a place 1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,...
Then put the 2's in every other remaining place
1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,...
Then the 3's in every other remaining place
1,2,1,3,1,2,1,,1,2,1,3,1,2,1,,1,2,1,3,1,2,1,,1,2,1,...
Then the 4's in every other remaining place
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,,1,2,1,3,1,2,1,4,1,2,1,...
By iterating this process, we get the ruler function 1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,... (End)
a(n) is the least positive integer k for which there does not exist i+j=n and a(i)=a(j)=k (cf. A322523). - Rémy Sigrist and Jianing Song, Aug 23 2022
a(n) is the smallest positive integer that does not occur in the coincidences of the sequence so far a(1..n-1) and its reverse. - Neal Gersh Tolunsky, Jan 18 2023
The geometric mean of this sequence approaches the Somos constant (A112302). - Jwalin Bhatt, Jan 31 2025

Examples

			For example, 2^1|2, 2^2|4, 2^1|6, 2^3|8, 2^1|10, 2^2|12, ... giving the initial terms 1, 2, 1, 3, 1, 2, ...
From _Omar E. Pol_, Jun 12 2009: (Start)
Triangle begins:
1;
2,1;
3,1,2,1;
4,1,2,1,3,1,2,1;
5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1;
6,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1;
7,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6,1,2,1,3,...
(End)
S(0) = {} S(1) = 1 S(2) = 1, 2, 1 S(3) = 1, 2, 1, 3, 1, 2, 1 S(4) = 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1. - Yann David (yann_david(AT)hotmail.com), Mar 21 2010
From _Joerg Arndt_, Nov 12 2012: (Start)
The 16 compositions of 5 as lists in lexicographic order:
[ n]  a(n)  composition
[ 1]  [ 1]  [ 1 1 1 1 1 ]
[ 2]  [ 2]  [ 1 1 1 2 ]
[ 3]  [ 1]  [ 1 1 2 1 ]
[ 4]  [ 3]  [ 1 1 3 ]
[ 5]  [ 1]  [ 1 2 1 1 ]
[ 6]  [ 2]  [ 1 2 2 ]
[ 7]  [ 1]  [ 1 3 1 ]
[ 8]  [ 4]  [ 1 4 ]
[ 9]  [ 1]  [ 2 1 1 1 ]
[10]  [ 2]  [ 2 1 2 ]
[11]  [ 1]  [ 2 2 1 ]
[12]  [ 3]  [ 2 3 ]
[13]  [ 1]  [ 3 1 1 ]
[14]  [ 2]  [ 3 2 ]
[15]  [ 1]  [ 4 1 ]
[16]  [ 5]  [ 5 ]
a(n) is the last part in each list.
(End)
From _Omar E. Pol_, Aug 20 2013: (Start)
Also written as a triangle in which the right border gives A000027 and row lengths give A011782 and row sums give A000079 the sequence begins:
1;
2;
1,3;
1,2,1,4;
1,2,1,3,1,2,1,5;
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6;
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,7;
(End)
G.f. = x + 2*x^2 + x^3 + 3*x^4 + x^5 + 2*x^6 + x^7 + 4*x^8 + x^9 + 2*x^10 + ...

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
  • E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 2nd ed., 2001-2003; see Dim- and Dim+ on p. 98; Dividing Rulers, on pp. 436-437; The Ruler Game, pp. 469-470; Ruler Fours, Fives, ... Fifteens on p. 470.
  • L. Gros, Théorie du Baguenodier, Aimé Vingtrinier, Lyon, 1872.
  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E22.
  • A. M. Hinz, The Tower of Hanoi, in Algebras and combinatorics (Hong Kong, 1997), 277-289, Springer, Singapore, 1999.
  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.1.3, Problem 41, p. 589.
  • Andrew Schloss, "Towers of Hanoi" composition, in The Digital Domain. Elektra/Asylum Records 9 60303-2, 1983. Works by Jaffe (Finale to "Silicon Valley Breakdown"), McNabb ("Love in the Asylum"), Schloss ("Towers of Hanoi"), Mattox ("Shaman"), Rush, Moorer ("Lions are Growing") and others.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Column 1 of table A050600.
Sequence read mod 2 gives A035263.
Sequence is bisection of A007814, A050603, A050604, A067029, A089309.
This is Guy Steele's sequence GS(4, 2) (see A135416).
Cf. A005187 (partial sums), A085058 (bisection), A112302 (geometric mean), A171977 (2^a(n)).
Cf. A000005, A000041, A000079, A003188, A003278, A003602, A007949, A018238, A047999, A051731, A054525, A054852, A065176, A082850, A089080, A092119, A117303, A129360, A130093, A173238, A181988, A220466.
Cf. A287896, A002487, A209229 (Mobius trans.), A092673 (Dirichlet inv.).
Cf. generalized ruler functions for k=3,4,5: A051064, A115362, A055457.

Programs

  • Haskell
    a001511 n = length $ takeWhile ((== 0) . (mod n)) a000079_list
-- Reinhard Zumkeller, Sep 27 2011
  • Haskell
    a001511 n | odd n = 1 | otherwise = 1 + a001511 (n `div` 2)
-- Walt Rorie-Baety, Mar 22 2013
  • MATLAB
    nmax=5;r=1;for n=2:nmax;r=[r n r];end % Adriano Caroli, Feb 26 2016
  • Magma
    [Valuation(2*n,2): n in [1..105]]; // Bruno Berselli, Nov 23 2015
  • Maple
    A001511 := n->2-wt(n)+wt(n-1); # where wt is defined in A000120
# This is the binary logarithm of the denominator of (256^n-1)B_{8n}/n, in Maple parlance a := n -> log[2](denom((256^n-1)*bernoulli(8*n)/n)). - Peter Luschny, May 31 2009
A001511 := n -> padic[ordp](2*n,2): seq(A001511(n), n=1..105);  # Peter Luschny, Nov 26 2010
a:= n-> ilog2((Bits[Xor](2*n, 2*n-1)+1)/2): seq(a(n), n=1..50);  # Gary Detlefs, Dec 13 2018
  • Mathematica
    Array[ If[ Mod[ #, 2] == 0, FactorInteger[ # ][[1, 2]], 0] &, 105] + 1 (* or *)
Nest[ Flatten[ # /. a_Integer -> {1, a + 1}] &, {1}, 7] (* Robert G. Wilson v, Mar 04 2005 *)
IntegerExponent[2*n, 2] (* Alexander R. Povolotsky, Aug 19 2011 *)
myHammingDistance[n_, m_] := Module[{g = Max[m, n], h = Min[m, n]}, b1 = IntegerDigits[g, 2]; b2 = IntegerDigits[h, 2, Length[b1]]; HammingDistance[b1, b2]] (* Vladimir Shevelev A206853 *) Table[ myHammingDistance[n, n - 1], {n, 111}] (* Robert G. Wilson v, Apr 05 2012 *)
Table[Position[Reverse[IntegerDigits[n,2]],1,1,1],{n,110}]//Flatten (* Harvey P. Dale, Aug 18 2017 *)
  • PARI
    a(n) = sum(k=0,floor(log(n)/log(2)),floor(n/2^k)-floor((n-1)/2^k)) /* Ralf Stephan */
  • PARI
    a(n)=if(n%2,1,factor(n)[1,2]+1) /* Jon Perry, Jun 06 2004 */
  • PARI
    {a(n) = if( n, valuation(n, 2) + 1, 0)}; /* Michael Somos, Sep 30 2006 */
  • PARI
    {a(n)=if(n==1,1,polcoeff(x-sum(k=1, n-1, a(k)*x^k*(1-x^k)*(1-x+x*O(x^n))), n))} /* Paul D. Hanna, Jun 22 2007 */
  • Python
    def a(n): return bin(n)[2:][::-1].index("1") + 1 # Indranil Ghosh, May 11 2017
  • Python
    A001511 = lambda n: (n&-n).bit_length() # M. F. Hasler, Apr 09 2020
  • Python
    def A001511(n): return (~n & n-1).bit_length()+1 # Chai Wah Wu, Jul 01 2022
  • Sage
    [valuation(2*n,2) for n in (1..105)]  # Bruno Berselli, Nov 23 2015
  • Scheme
    (define (A001511 n) (let loop ((n n) (e 1)) (if (odd? n) e (loop (/ n 2) (+ 1 e))))) ;; Antti Karttunen, Oct 06 2017

Formula

a(n) = A007814(n) + 1 = A007814(2*n).
a(2*n+1) = 1; a(2*n) = 1 + a(n). - Philippe Deléham, Dec 08 2003
a(n) = 2 - A000120(n) + A000120(n-1), n >= 1. - Daniele Parisse
a(n) = 1 + log_2(abs(A003188(n) - A003188(n-1))).
Multiplicative with a(p^e) = e+1 if p = 2; 1 if p > 2. - David W. Wilson, Aug 01 2001
For any real x > 1/2: lim_{N->infinity} (1/N)*Sum_{n=1..N} x^(-a(n)) = 1/(2*x-1); also lim_{N->infinity} (1/N)*Sum_{n=1..N} 1/a(n) = log(2). - Benoit Cloitre, Nov 16 2001
s(n) = Sum_{k=1..n} a(k) is asymptotic to 2*n since s(n) = 2*n - A000120(n). - Benoit Cloitre, Aug 31 2002
For any n >= 0, for any m >= 1, a(2^m*n + 2^(m-1)) = m. - Benoit Cloitre, Nov 24 2002
a(n) = Sum_{d divides n and d is odd} mu(d)*tau(n/d). - Vladeta Jovovic, Dec 04 2002
G.f.: A(x) = Sum_{k>=0} x^(2^k)/(1-x^(2^k)). - Ralf Stephan, Dec 24 2002
a(1) = 1; for n > 1, a(n) = a(n-1) + (-1)^n*a(floor(n/2)). - Vladeta Jovovic, Apr 25 2003
A fixed point of the mapping 1->12; 2->13; 3->14; 4->15; 5->16; ... . - Philippe Deléham, Dec 13 2003
Product_{k>0} (1+x^k)^a(k) is g.f. for A000041(). - Vladeta Jovovic, Mar 26 2004
G.f. A(x) satisfies A(x) = A(x^2) + x/(1-x). - Franklin T. Adams-Watters, Feb 09 2006
a(A118413(n,k)) = A002260(n,k); = a(A118416(n,k)) = A002024(n,k); a(A014480(n)) = A003602(A014480(n)). - Reinhard Zumkeller, Apr 27 2006
Ordinal transform of A003602. - Franklin T. Adams-Watters, Aug 28 2006 (The ordinal transform of a sequence b_0, b_1, b_2, ... is the sequence a_0, a_1, a_2, ... where a_n is the number of times b_n has occurred in {b_0 ... b_n}.)
Could be extended to n <= 0 using a(-n) = a(n), a(0) = 0, a(2*n) = a(n)+1 unless n=0. - Michael Somos, Sep 30 2006
A094267(2*n) = A050603(2*n) = A050603(2*n + 1) = a(n). - Michael Somos, Sep 30 2006
Sequence = A129360 * A000005 = M*V, where M = an infinite lower triangular matrix and V = d(n) as a vector: [1, 2, 2, 3, 2, 4, ...]. - Gary W. Adamson, Apr 15 2007
Row sums of triangle A130093. - Gary W. Adamson, May 13 2007
Dirichlet g.f.: zeta(s)*2^s/(2^s-1). - Ralf Stephan, Jun 17 2007
a(n) = -Sum_{d divides n} mu(2*d)*tau(n/d). - Benoit Cloitre, Jun 21 2007
G.f.: x/(1-x) = Sum_{n>=1} a(n)*x^n*( 1 - x^n ). - Paul D. Hanna, Jun 22 2007
2*n = 2^a(n)* A000265(n). - Eric Desbiaux, May 14 2009 [corrected by Alejandro Erickson, Apr 17 2012]
Multiplicative with a(2^k) = k + 1, a(p^k) = 1 for any odd prime p. - Franklin T. Adams-Watters, Jun 09 2009
With S(n): 2^n - 1 first elements of the sequence then S(0) = {} (empty list) and if n > 0, S(n) = S(n-1), n, S(n-1). - Yann David (yann_david(AT)hotmail.com), Mar 21 2010
a(n) = log_2(A046161(n)/A046161(n-1)). - Johannes W. Meijer, Nov 04 2012
a((2*n-1)*2^p) = p+1, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 05 2013
a(n+1) = 1 + Sum_{j=0..ceiling(log_2(n+1))} (j * (1 - abs(sign((n mod 2^(j + 1)) - 2^j + 1)))). - Enrico Borba, Oct 01 2015
Conjecture: a(n) = A181988(n)/A003602(n). - L. Edson Jeffery, Nov 21 2015
a(n) = log_2(A006519(n)) + 1. - Doug Bell, Jun 02 2017
Inverse Moebius transform of A209229. - Andrew Howroyd, Aug 04 2018
a(n) = 1 + (A183063(n)/A001227(n)). - Omar E. Pol, Nov 06 2018 (after Franklin T. Adams-Watters)
a(n) = log_2((Xor(2*n,2*n-1)+1)/2). - Gary Detlefs, Dec 13 2018
(2^(a(n)-1)-1)*(n mod 4) = 2*floor(((n+1) mod 4)/3). - Gary Detlefs, Dec 14 2018
a(n) = A000005(n)/A001227(n). - Ivan N. Ianakiev, Jul 05 2019
a(n) = Sum_{j=1..r} (j/2^j)*(Product_{k=1..j} (1 - (-1)^floor( (n+2^(j-1))/2^(k-1) ))), for n < a predefined 2^r. - Adriano Caroli, Sep 30 2019

Extensions

Name edited following suggestion by David James Sycamore, Oct 05 2023

A064413 EKG sequence (or ECG sequence): a(1) = 1; a(2) = 2; for n > 2, a(n) = smallest number not already used which shares a factor with a(n-1).

Original entry on oeis.org

1, 2, 4, 6, 3, 9, 12, 8, 10, 5, 15, 18, 14, 7, 21, 24, 16, 20, 22, 11, 33, 27, 30, 25, 35, 28, 26, 13, 39, 36, 32, 34, 17, 51, 42, 38, 19, 57, 45, 40, 44, 46, 23, 69, 48, 50, 52, 54, 56, 49, 63, 60, 55, 65, 70, 58, 29, 87, 66, 62, 31, 93, 72, 64, 68, 74, 37, 111, 75, 78, 76, 80, 82
Offset: 1

Views

Author

Jonathan Ayres (Jonathan.ayres(AT)btinternet.com), Sep 30 2001

Keywords

Comments

Locally, the graph looks like an EKG (American English) or ECG (British English).
Calculating the square of A064413 and plotting the results shows the EKG behavior even more dramatically - see A104125. - Parthasarathy Nambi, Jan 27 2005
Theorem: (1) Every number appears exactly once: this is a permutation of the positive numbers. - J. C. Lagarias, E. M. Rains, N. J. A. Sloane, Oct 03 2001
The permutation has cycles (1) (2) (3, 4, 6, 9, 10, 5) (..., 20, 18, 12, 7, 14, 13, 28, 26, ...) (8) ...
Theorem: (2) The primes appear in increasing order. - J. C. Lagarias, E. M. Rains, N. J. A. Sloane, Oct 03 2001
Theorem: (3) When an odd prime p appears it is immediately preceded by 2p and followed by 3p. - Conjectured by Lagarias-Rains-Sloane, proved by Hofman-Pilipczuk.
Theorem: (4) Let a'(n) be the same sequence but with all terms p and 3p (p prime) changed to 2p (see A256417). Then lim a'(n)/n = 1, i.e., a(n) ~ n except for the values p and 3p for p prime. - Conjectured by Lagarias-Rains-Sloane, proved by Hofman-Pilipczuk.
Conjecture: If a(n) != p, then almost everywhere a(n) > n. - Thomas Ordowski, Jan 23 2009
Conjecture: lim #(a_n > n) / n = 1, i.e., #(a_n > n) ~ n. - Thomas Ordowski, Jan 23 2009
Conjecture: A term p^2, p a prime, is immediately preceded by p*(p+1) and followed by p*(p+2). - Vladimir Baltic, Oct 03 2001. This is false, for example the sequence contains the 3 terms p*(p+2), p^2, p*(p+3) for p = 157. - Eric Rains
Theorem: If a(k) = 3p, then |{a(m) : a(m>k) < 3p}| = 3p - k. Proof: If a(k) = 3p, then all a(mk) > p and |{a(m) : a(m>k) < 3p}| = 3p - k. - Thomas Ordowski, Jan 22 2009
Let ...,a_i,...,2p,p,3p,...,a_j,... There does not exist a_i > 3p. There does not exist a_j < p. - Thomas Ordowski, Jan 20 2009
Let...,a,...,2p,p,3p,...,b,... All a<3p and b>p. #(a>2p) <= #(b<2p). - Thomas Ordowski, Jan 21 2009
If a(k)=3p then |{a(m):a(m>k)<3p}|=3p-k. - Thomas Ordowski, Jan 22 2009
GCD(a(n),n) = A247379(n). - Reinhard Zumkeller, Sep 16 2014
If the definition is changed to require that the GCD of successive terms be a prime power > 1, the sequence stays the same until a(578)=620, at which point a(579)=610 has GCD = 10 with the previous term. - N. J. A. Sloane, Mar 30 2015
From Michael De Vlieger, Dec 06 2021: (Start)
For prime p > 2, we have the chain {j : 2|j} -> 2p -> p -> 3p -> {k : 3|k}. The term j introducing 2p must be even, since 2p is an even squarefree semiprime proved by Hofman-Pilipczuk to introduce p itself. Hence no term a(i) such that p | a(i) exists in the sequence for i < n-1, where a(n) = p, leaving 2|j. Similarly, the term k following 3p must be divisible by 3 since the terms mp that are not coprime to p (thus implying p | mp) have m >= 4, thereby large compared to numbers k such that 3|k that belong to the cototient of 3p. For the chain {4, 6, 3, 9, 12}, the term 12 following 3p indeed is 4p, but p = 3; this is the only case of 4p following 3p in the sequence. As a consequence, for i > 1, A073734(A064955(i)-1) = 2 and A073734(A064955(i)+2) = 3.
For Fermat primes p, we have the chain {j : 2|j} -> 2^e-> {2p = 2^e + 2} -> {p = 2^(e-1) + 1} -> 3p -> {k : 3|k}.
a(3) = 4 = 2^2, a(5) = 3 = 2^1 + 1;
a(8) = 8 = 2^3, a(10) = 5 = 2^2 + 1;
a(31) = 32 = 2^5, a(33) = 17 = 2^4 + 1;
a(485) = 512 = 2^9, a(487) = 257 = 2^8 + 1;
a(127354) = 131072 = 2^17, a(127356) = 65537 = 2^16 + 1.
(End)

Examples

			a(2) = 2, a(3) = 4 (gcd(2,4) = 2), a(4) = 6 (gcd(4,6) = 2), a(5) = 3 (gcd(6,3) = 3), a(6) = 9 (6 already used so next number which shares a factor is 9 since gcd(3,9) = 3).

References

  • N. J. A. Sloane, Seven Staggering Sequences, in Homage to a Pied Puzzler, E. Pegg Jr., A. H. Schoen and T. Rodgers (editors), A. K. Peters, Wellesley, MA, 2009, pp. 93-110.

Links

Crossrefs

A073734 gives GCD's of successive terms.
See A064664 for the inverse permutation. See A064665-A064668 for the first two infinite cycles of this permutation. A064669 gives cycle representatives.
See A064421 for sequence giving term at which n appears.
See A064424, A074177 for records.
Cf. A064955 & A352194 (prime positions), A195376 (parity), A064957 (positions of odd terms), A064953 (positions of even terms), A064426 (first differences).
See A169857 and A119415 for the effect of changing the start.
Cf. A240024 (nonprime version).
Cf. A152458 (fixed points), A247379, A247383.
For other initial terms, see A169841, A169837, A169843, A169855, A169849.
A256417 is a smoothed version.
See also A255582, A256466, A257218, A257311-A257315, A257405, A253279 (two-dimensional analog).
See also A276127.

Programs

  • Haskell
    import Data.List (delete, genericIndex)
a064413 n = genericIndex a064413_list (n - 1)
a064413_list = 1 : f 2 [2..] where
   ekg x zs = f zs where
       f (y:ys) = if gcd x y > 1 then y : ekg y (delete y zs) else f ys
-- Reinhard Zumkeller, May 01 2014, Sep 17 2011
  • Maple
    h := array(1..20000); a := array(1..10000); maxa := 300; maxn := 2*maxa; for n from 1 to maxn do h[n] := -1; od: a[1] := 2; h[2] := 1; c := 2; for n from 2 to maxa do for m from 2 to maxn do t1 := gcd(m,c); if t1 > 1 and h[m] = -1 then c := m; a[n] := c; h[c] := n; break; fi; od: od: ap := []: for n from 1 to maxa do ap := [op(ap),a[n]]; od: hp := []: for n from 2 to maxa do hp := [op(hp),h[n]]; od: convert(ap,list); convert(hp,list); # this is very crude!
N:= 1000: # to get terms before the first term > N
V:= Vector(N):
A[1]:= 1:
A[2]:= 2: V[2]:= 1:
for n from 3 do
  S:= {seq(seq(k*p,k=1..N/p),p=numtheory:-factorset(A[n-1]))};
  for s in sort(convert(S,list)) do
    if V[s] = 0 then
      A[n]:= s;
      break
    fi
  od;
  if V[s] = 1 then break fi;
  V[s]:= 1;
od:
seq(A[i],i=1..n-1); # Robert Israel, Jan 18 2016
  • Mathematica
    maxN = 100; ekg = {1, 2}; unused = Range[3, maxN]; found = True; While[found, found = False; i = 0; While[ !found && i < Length[unused], i++; If[GCD[ekg[[-1]], unused[[i]]] > 1, found = True; AppendTo[ekg, unused[[i]]]; unused = Delete[unused, i]]]]; ekg (* Ayres *)
ekGrapher[s_List] := Block[{m = s[[-1]], k = 3}, While[MemberQ[s, k] || GCD[m, k] == 1, k++ ]; Append[s, k]]; Nest[ekGrapher, {1, 2}, 71] (* Robert G. Wilson v, May 20 2009 *)
  • PARI
    a1=1; a2=2; v=[1,2];
for(n=3,100,a3=if(n<0,0,t=1;while(vecmin(vector(length(v),i,abs(v[i]-t)))*(gcd(a2,t)-1)==0,t++);t);a2=a3;v=concat(v,a3););
a(n)=v[n];
/* Benoit Cloitre, Sep 23 2012 */
  • Python
    from math import gcd
A064413_list, l, s, b = [1,2], 2, 3, {}
for _ in range(10**5):
    i = s
    while True:
        if not i in b and gcd(i, l) > 1:
            A064413_list.append(i)
            l, b[i] = i, True
            while s in b:
                b.pop(s)
                s += 1
            break
        i += 1 # Chai Wah Wu, Dec 08 2014

Formula

a(n) = smallest number not already used such that gcd(a(n), a(n-1)) > 1.
In Lagarias-Rains-Sloane (2002), it is conjectured that almost all a(n) satisfy the asymptotic formula a(n) = n (1+ 1/(3 log n)) + o(n/log n) as n -> oo and that the exceptional terms when the sequence is a prime or 3 times a prime p produce the spikes in the sequence. See the paper for a more precise statement of the conjecture. - N. J. A. Sloane, Mar 07 2015

Extensions

More terms from Naohiro Nomoto, Sep 30 2001
Entry extensively revised by N. J. A. Sloane, Oct 10 2001
Showing 1-10 of 172 results. Next

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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