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

A000204 Lucas numbers (beginning with 1): L(n) = L(n-1) + L(n-2) with L(1) = 1, L(2) = 3.

Original entry on oeis.org

1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761, 271443, 439204, 710647, 1149851, 1860498, 3010349, 4870847, 7881196, 12752043, 20633239, 33385282, 54018521, 87403803, 141422324
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

See A000032 for the version beginning 2, 1, 3, 4, 7, ...
Also called Schoute's accessory series (see Jean, 1984). - N. J. A. Sloane, Jun 08 2011
L(n) is the number of matchings in a cycle on n vertices: L(4)=7 because the matchings in a square with edges a, b, c, d (labeled consecutively) are the empty set, a, b, c, d, ac and bd. - Emeric Deutsch, Jun 18 2001
This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 1..m - 1, a(m) = m + 1. The generating function is (x + m*x^m)/(1 - x - x^m). Also a(n) = 1 + n*Sum_{i=1..n/m} binomial(n - 1 - (m - 1)*i, i - 1)/i. This gives the number of ways to cover (without overlapping) a ring lattice (or necklace) of n sites with molecules that are m sites wide. Special cases: m = 2: A000204, m = 3: A001609, m = 4: A014097, m = 5: A058368, m = 6: A058367, m = 7: A058366, m = 8: A058365, m = 9: A058364.
L(n) is the number of points of period n in the golden mean shift. The number of orbits of length n in the golden mean shift is given by the n-th term of the sequence A006206. - Thomas Ward, Mar 13 2001
Row sums of A029635 are 1, 1, 3, 4, 7, ... - Paul Barry, Jan 30 2005
a(n) counts circular n-bit strings with no repeated 1's. E.g., for a(5): 00000 00001 00010 00100 00101 01000 01001 01010 10000 10010 10100. Note #{0...} = fib(n+1), #{1...} = fib(n-1), #{000..., 001..., 100...} = a(n-1), #{010..., 101...} = a(n-2). - Len Smiley, Oct 14 2001
Row sums of the triangle in A182579. - Reinhard Zumkeller, May 07 2012
If p is prime then L(p) == 1 (mod p). L(2^k) == -1 (mod 2^(k+1)) for k = 0,1,2,... - Thomas Ordowski, Sep 25 2013
Satisfies Benford's law [Brown-Duncan, 1970; Berger-Hill, 2017]. - N. J. A. Sloane, Feb 08 2017

Examples

			G.f. = x + 3*x^2 + 4*x^3 + 7*x^4 + 11*x^5 + 18*x^6 + 29*x^7 + 47*x^8 + ...

References

  • P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 69.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 46.
  • Leonhard Euler, Introductio in analysin infinitorum (1748), sections 216 and 229.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 148.
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers. Houghton, Boston, MA, 1969.
  • R. V. Jean, Mathematical Approach to Pattern and Form in Plant Growth, Wiley, 1984. See p. 5. - N. J. A. Sloane, Jun 08 2011
  • Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.
  • 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).
  • S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Links

Crossrefs

Cf. A000032, A000045, A061084, A027960, A001609, A014097, A000079, A003269, A003520, A005708, A005709, A005710, A006206, A101033, A101032, A100492, A099731, A094216, A094638, A000108, A090946 (complement).

Programs

  • Haskell
    a000204 n = a000204_list !! n
a000204_list = 1 : 3 : zipWith (+) a000204_list (tail a000204_list)
-- Reinhard Zumkeller, Dec 18 2011
  • Magma
    [Lucas(n): n in [1..30]]; // G. C. Greubel, Dec 17 2017
  • Maple
    A000204 := proc(n) option remember; if n <=2 then 2*n-1; else procname(n-1)+procname(n-2); fi; end;
with(combinat): A000204 := n->fibonacci(n+1)+fibonacci(n-1);
# alternative Maple program:
L:= n-> (<<1|1>, <1|0>>^n. <<2, -1>>)[1, 1]:
seq(L(n), n=1..50);  # Alois P. Heinz, Jul 25 2008
# Alternative:
a := n -> `if`(n=1, 1, `if`(n=2, 3, hypergeom([(1-n)/2, -n/2], [1-n], -4))):
seq(simplify(a(n)), n=1..39); # Peter Luschny, Sep 03 2019
  • Mathematica
    c = (1 + Sqrt[5])/2; Table[Expand[c^n + (1 - c)^n], {n, 30}] (* Artur Jasinski, Oct 05 2008 *)
Table[LucasL[n, 1], {n, 36}] (* Zerinvary Lajos, Jul 09 2009 *)
LinearRecurrence[{1, 1},{1, 3}, 50] (* Sture Sjöstedt, Nov 28 2011 *)
lukeNum[n_] := If[n < 1, 0, LucasL[n]]; (* Michael Somos, May 18 2015 *)
lukeNum[n_] := SeriesCoefficient[x D[Log[1 / (1 - x - x^2)], x], {x, 0, n}]; (* Michael Somos, May 18 2015 *)
  • PARI
    A000204(n)=fibonacci(n+1)+fibonacci(n-1) \\ Michael B. Porter, Nov 05 2009
  • Python
    from functools import cache
@cache
def a(n): return [1, 3][n-1] if n < 3 else a(n-1) + a(n-2)
print([a(n) for n in range(1, 41)]) # Michael S. Branicky, Nov 13 2022
  • Python
    [(i:=-1)+(j:=2)] + [(j:=i+j)+(i:=j-i) for  in range(100)] # _Jwalin Bhatt, Apr 02 2025
  • Sage
    def A000204():
    x, y = 1, 2
    while true:
       yield x
       x, y = x + y, x
a = A000204(); print([next(a) for i in range(39)])  # Peter Luschny, Dec 17 2015
  • Scala
    def lucas(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, 2, 1)
}
(1 to 50).map(lucas()) // _Alonso del Arte, Oct 20 2019

Formula

Expansion of x(1 + 2x)/(1 - x - x^2). - Simon Plouffe, dissertation 1992; multiplied by x. - R. J. Mathar, Nov 14 2007
a(n) = A000045(2n)/A000045(n). - Benoit Cloitre, Jan 05 2003
For n > 1, L(n) = F(n + 2) - F(n - 2), where F(n) is the n-th Fibonacci number (A000045). - Gerald McGarvey, Jul 10 2004
a(n+1) = 4*A054886(n+3) - A022388(n) - 2*A022120(n+1) (a conjecture; note that the above sequences have different offsets). - Creighton Dement, Nov 27 2004
a(n) = Sum_{k=0..floor((n+1)/2)} (n+1)*binomial(n - k + 1, k)/(n - k + 1). - Paul Barry, Jan 30 2005
L(n) = A000045(n+3) - 2*A000045(n). - Creighton Dement, Oct 07 2005
L(n) = A000045(n+1) + A000045(n-1). - John Blythe Dobson, Sep 29 2007
a(n) = 2*Fibonacci(n-1) + Fibonacci(n), n >= 1. - Zerinvary Lajos, Oct 05 2007
L(n) is the term (1, 1) in the 1 X 2 matrix [2, -1].[1, 1; 1, 0]^n. - Alois P. Heinz, Jul 25 2008
a(n) = phi^n + (1 - phi)^n = phi^n + (-phi)^(-n) = ((1 + sqrt(5))^n + (1 - sqrt(5))^n)/(2^n) where phi is the golden ratio (A001622). - Artur Jasinski, Oct 05 2008
a(n) = A014217(n+1) - A014217(n-1). See A153263. - Paul Curtz, Dec 22 2008
a(n) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5)) + ((1 + sqrt(5))^(n - 1) - (1 - sqrt(5))^(n - 1))/(2^(n - 2)*sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009, Jan 14 2009
From Hieronymus Fischer, Oct 20 2010 (Start)
Continued fraction for n odd: [L(n); L(n), L(n), ...] = L(n) + fract(Fib(n) * phi).
Continued fraction for n even: [L(n); -L(n), L(n), -L(n), L(n), ...] = L(n) - 1 + fract(Fib(n)*phi). Also: [L(n) - 2; 1, L(n) - 2, 1, L(n) - 2, ...] = L(n) - 2 + fract(Fib(n)*phi). (End)
INVERT transform of (1, 2, -1, -2, 1, 2, ...). - Gary W. Adamson, Mar 07 2012
L(2n - 1) = floor(phi^(2n - 1)); L(2n) = ceiling(phi^(2n)). - Thomas Ordowski, Jun 15 2012
a(n) = hypergeom([(1 - n)/2, -n/2], [1 - n], -4) for n >= 3. - Peter Luschny, Sep 03 2019
E.g.f.: 2*(exp(x/2)*cosh(sqrt(5)*x/2) - 1). - Stefano Spezia, Jul 26 2022

Extensions

Additional comments from Yong Kong (ykong(AT)curagen.com), Dec 16 2000
Plouffe Maple line edited by N. J. A. Sloane, May 13 2008

A000931 Padovan sequence (or Padovan numbers): a(n) = a(n-2) + a(n-3) with a(0) = 1, a(1) = a(2) = 0.

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351, 465, 616, 816, 1081, 1432, 1897, 2513, 3329, 4410, 5842, 7739, 10252, 13581, 17991, 23833, 31572, 41824, 55405, 73396, 97229, 128801, 170625
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of compositions of n into parts congruent to 2 mod 3 (offset -1). - Vladeta Jovovic, Feb 09 2005
a(n) is the number of compositions of n into parts that are odd and >= 3. Example: a(10)=3 counts 3+7, 5+5, 7+3. - David Callan, Jul 14 2006
Referred to as N0102 in R. K. Guy's "Anyone for Twopins?" - Rainer Rosenthal, Dec 05 2006
Zagier conjectures that a(n+3) is the maximum number of multiple zeta values of weight n > 1 which are linearly independent over the rationals. - Jonathan Sondow and Sergey Zlobin (sirg_zlobin(AT)mail.ru), Dec 20 2006
Starting with offset 6: (1, 1, 2, 2, 3, 4, 5, ...) = INVERT transform of A106510: (1, 1, -1, 0, 1, -1, 0, 1, -1, ...). - Gary W. Adamson, Oct 10 2008
Starting with offset 7, the sequence 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, ... is called the Fibonacci quilt sequence by Catral et al., in Fib. Q. 2017. - N. J. A. Sloane, Dec 24 2021
Triangle A145462: right border = A000931 starting with offset 6. Row sums = Padovan sequence starting with offset 7. - Gary W. Adamson, Oct 10 2008
Starting with offset 3 = row sums of triangle A146973 and INVERT transform of [1, -1, 2, -2, 3, -3, ...]. - Gary W. Adamson, Nov 03 2008
a(n+5) corresponds to the diagonal sums of "triangle": 1; 1; 1,1; 1,1; 1,2,1; 1,2,1; 1,3,3,1; 1,3,3,1; 1,4,6,4,1; ..., rows of Pascal's triangle (A007318) repeated. - Philippe Deléham, Dec 12 2008
With offset 3: (1, 0, 1, 1, 1, 2, 2, ...) convolved with the tribonacci numbers prefaced with a "1": (1, 1, 1, 2, 4, 7, 13, ...) = the tribonacci numbers, A000073. (Cf. triangle A153462.) - Gary W. Adamson, Dec 27 2008
a(n) is also the number of strings of length (n-8) from an alphabet {A, B} with no more than one A or 2 B's consecutively. (E.g., n = 4: {ABAB,ABBA,BABA,BABB,BBAB} and a(4+8) = 5.) - Toby Gottfried, Mar 02 2010
p(n):=A000931(n+3), n >= 1, is the number of partitions of the numbers {1,2,3,...,n} into lists of length two or three containing neighboring numbers. The 'or' is inclusive. For n=0 one takes p(0)=1. For details see the W. Lang link. There the explicit formula for p(n) (analog of the Binet-de Moivre formula for Fibonacci numbers) is also given. Padovan sequences with different inputs are also considered there. - Wolfdieter Lang, Jun 15 2010
Equals the INVERTi transform of Fibonacci numbers prefaced with three 1's, i.e., (1 + x + x^2 + x^3 + x^4 + 2x^5 + 3x^6 + 5x^7 + 8x^8 + 13x^9 + ...). - Gary W. Adamson, Apr 01 2011
When run backwards gives (-1)^n*A050935(n).
a(n) is the top left entry of the n-th power of the 3 X 3 matrix [0, 0, 1; 1, 0, 1; 0, 1, 0] or of the 3 X 3 matrix [0, 1, 0; 0, 0, 1; 1, 1, 0]. - R. J. Mathar, Feb 03 2014
Figure 4 of Brauchart et al., 2014, shows a way to "visualize the Padovan sequence as cuboid spirals, where the dimensions of each cuboid made up by the previous ones are given by three consecutive numbers in the sequence". - N. J. A. Sloane, Mar 26 2014
a(n) is the number of closed walks from a vertex of a unidirectional triangle containing an opposing directed edge (arc) between the second and third vertices. Equivalently the (1,1) entry of A^n where the adjacency matrix of digraph is A=(0,1,0;0,0,1;1,1,0). - David Neil McGrath, Dec 19 2014
Number of compositions of n-3 (n >= 4) into 2's and 3's. Example: a(12)=5 because we have 333, 3222, 2322, 2232, and 2223. - Emeric Deutsch, Dec 28 2014
The Hoffman (2015) paper "offers significant evidence that the number of quantities needed to generate the weight-n multiple harmonic sums mod p is" a(n). - N. J. A. Sloane, Jun 24 2016
a(n) gives the number of compositions of n-5 into odd parts where the order of the 1's does not matter. For example, a(11)=4 counts the following compositions of 6: (5,1)=(1,5), (3,3), (3,1,1,1)=(1,3,1,1)=(1,1,3,1)=(1,1,1,3), (1,1,1,1,1,1). - Gregory L. Simay, Aug 04 2016
For n > 6, a(n) is the number of maximal matchings in the (n-5)-path graph, maximal independent vertex sets and minimal vertex covers in the (n-6)-path graph, and minimal edge covers in the (n-5)-pan graph and (n-3)-path graphs. - Eric W. Weisstein, Mar 30, Aug 03, and Aug 07 2017
From James Mitchell and Wilf A. Wilson, Jul 21 2017: (Start)
a(2n + 5) + 2n - 4, n > 2, is the number of maximal subsemigroups of the monoid of order-preserving mappings on a set with n elements.
a(n + 6) + n - 3, n > 3, is the number of maximal subsemigroups of the monoid of order-preserving or reversing mappings on a set with n elements.
(End)
Has the property that the largest of any four consecutive terms equals the sum of the two smallest. - N. J. A. Sloane, Aug 29 2017 [David Nacin points out that there are many sequences with this property, such as 1,1,1,2,1,1,1,2,1,1,1,2,... or 2,3,4,5,2,3,4,5,2,3,4,5,... or 2,2,1,3,3, 4,1,4, 5,5,1,6,6, 7,1,7, 8,8,1,9,9, 10,1,10, ... (spaces added for clarity), and a conjecture I made here in 2017 was simply wrong. I have deleted it. - N. J. A. Sloane, Oct 23 2018]
a(n) is also the number of maximal cliques in the (n+6)-path complement graph. - Eric W. Weisstein, Apr 12 2018
a(n+8) is the number of solus bitstrings of length n with no runs of 3 zeros. - Steven Finch, Mar 25 2020
Named after the architect Richard Padovan (b. 1935). - Amiram Eldar, Jun 08 2021
Shannon et al. (2006) credit a French architecture student Gérard Cordonnier with the discovery of these numbers.
For n >= 3, a(n) is the number of sequences of 0s and 1s of length (n-2) that begin with a 0, end with a 0, contain no two consecutive 0s, and contain no three consecutive 1s. - Yifan Xie, Oct 20 2022
For n >= 2, a(n+5) is the number of ways to tile the 1xn board with dominoes and squares (ie. size 1x1) such that are either none or one squares between dominoes, none or one squares at both ends of the board, and there is at least one domino. For example, for n=6, a(11)=4 since the tilings are |2|2, |22|, 2|2| and 222 (where 2 represents a domino and | a square). - Enrique Navarrete, Aug 31 2024

Examples

			G.f. = 1 + x^3 + x^5 + x^6 + x^7 + 2*x^8 + 2*x^9 + 3*x^10 + 4*x^11 + ...

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 47, ex. 4.
  • Minerva Catral, Pari L. Ford, Pamela E. Harris, Steven J. Miller, Dawn Nelson, Zhao Pan, and Huanzhong Xu, Legal Decompositions Arising from Non-positive Linear Recurrences, Fib. Quart., 55:3 (2017), 252-275. [Note that there is an earlier version of this paper, with only five authors, on the arXiv in 2016. Note to editors: do not merge these two citations. - N. J. A. Sloane, Dec 24 2021]
  • Richard K. Guy, "Anyone for Twopins?" in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 10-11.
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • A. G. Shannon, P. G. Anderson and A. F. Horadam, Properties of Cordonnier, Perrin and Van der Laan numbers, International Journal of Mathematical Education in Science and Technology, Volume 37:7 (2006), 825-831. See P_n.
  • 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).
  • Ian Stewart, L'univers des nombres, "La sculpture et les nombres", pp. 19-20, Belin-Pour La Science, Paris, 2000.
  • Hans van der Laan, Het plastische getal. XV lessen over de grondslagen van de architectonische ordonnantie. Leiden, E.J. Brill, 1967.
  • Don Zagier, Values of zeta functions and their applications, in First European Congress of Mathematics (Paris, 1992), Vol. II, A. Joseph et al. (eds.), Birkhäuser, Basel, 1994, pp. 497-512.

Links

Crossrefs

The following are basically all variants of the same sequence: A000931, A078027, A096231, A124745, A133034, A134816, A164001, A182097, A228361 and probably A020720. However, each one has its own special features and deserves its own entry.
Closely related to A001608.
Cf. A000073, A005682-A005691, A103372-A103380, A106510, A145462, A146973, A153462.
Doubling every term gives A291289.

Programs

  • GAP
    a:=[1,0,0];; for n in [4..50] do a[n]:=a[n-2]+a[n-3]; od; a; # G. C. Greubel, Dec 30 2019
  • Haskell
    a000931 n = a000931_list !! n
a000931_list = 1 : 0 : 0 : zipWith (+) a000931_list (tail a000931_list)
-- Reinhard Zumkeller, Feb 10 2011
  • Magma
    I:=[1,0,0]; [n le 3 select I[n] else Self(n-2) + Self(n-3): n in [1..60]]; // Vincenzo Librandi, Jul 21 2015
  • Maple
    A000931 := proc(n) option remember; if n = 0 then 1 elif n <= 2 then 0 else procname(n-2)+procname(n-3); fi; end;
A000931:=-(1+z)/(-1+z^2+z^3); # Simon Plouffe in his 1992 dissertation; gives sequence without five leading terms
a[0]:=1; a[1]:=0; a[2]:=0; for n from 3 to 50 do a[n]:=a[n-2]+a[n-3]; end do; # Francesco Daddi, Aug 04 2011
  • Mathematica
    CoefficientList[Series[(1-x^2)/(1-x^2-x^3), {x, 0, 50}], x]
a[0]=1; a[1]=a[2]=0; a[n_]:= a[n]= a[n-2] + a[n-3]; Table[a[n], {n, 0, 50}] (* Robert G. Wilson v, May 04 2006 *)
LinearRecurrence[{0,1,1}, {1,0,0}, 50] (* Harvey P. Dale, Jan 10 2012 *)
Table[RootSum[-1 -# +#^3 &, 5#^n -6#^(n+1) +4#^(n+2) &]/23, {n,0,50}] (* Eric W. Weisstein, Nov 09 2017 *)
  • PARI
    Vec((1-x^2)/(1-x^2-x^3) + O(x^50)) \\ Charles R Greathouse IV, Feb 11 2011
  • PARI
    {a(n) = if( n<0, polcoeff(1/(1+x-x^3) + x * O(x^-n), -n), polcoeff( (1 - x^2)/(1-x^2-x^3) + x * O(x^n), n))}; /* Michael Somos, Sep 18 2012 */
  • Python
    def aupton(nn):
    alst = [1, 0, 0]
    for n in range(3, nn+1): alst.append(alst[n-2]+alst[n-3])
    return alst
print(aupton(49)) # Michael S. Branicky, Mar 28 2022
  • Sage
    def A000931_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1-x^2)/(1-x^2-x^3) ).list()
A000931_list(50) # G. C. Greubel, Dec 30 2019

Formula

G.f.: (1-x^2)/(1-x^2-x^3).
a(n) is asymptotic to r^n / (2*r+3) where r = 1.3247179572447... = A060006, the real root of x^3 = x + 1. - Philippe Deléham, Jan 13 2004
a(n)^2 + a(n+2)^2 + a(n+6)^2 = a(n+1)^2 + a(n+3)^2 + a(n+4)^2 + a(n+5)^2 (Barniville, Question 16884, Ed. Times 1911).
a(n+5) = a(0) + a(1) + ... + a(n).
a(n) = central and lower right terms in the (n-3)-th power of the 3 X 3 matrix M = [0 1 0 / 0 0 1 / 1 1 0]. E.g., a(13) = 7. M^10 = [3 5 4 / 4 7 5 / 5 9 7]. - Gary W. Adamson, Feb 01 2004
G.f.: 1/(1 - x^3 - x^5 - x^7 - x^9 - ...). - Jon Perry, Jul 04 2004
a(n+4) = Sum_{k=0..floor((n-1)/2)} binomial(floor((n+k-2)/3), k). - Paul Barry, Jul 06 2004
a(n+3) = Sum_{k=0..floor(n/2)} binomial(k, n-2k). - Paul Barry, Sep 17 2004, corrected by Greg Dresden and Zi Ye, Jul 06 2021
a(n+3) is diagonal sum of A026729 (as a number triangle), with formula a(n+3) = Sum_{k=0..floor(n/2)} Sum_{i=0..n-k} (-1)^(n-k+i)*binomial(n-k, i)*binomial(i+k, i-k). - Paul Barry, Sep 23 2004
a(n) = a(n-1) + a(n-5) = A003520(n-4) + A003520(n-13) = A003520(n-3) - A003520(n-9). - Henry Bottomley, Jan 30 2005
a(n+3) = Sum_{k=0..floor(n/2)} binomial((n-k)/2, k)(1+(-1)^(n-k))/2. - Paul Barry, Sep 09 2005
The sequence 1/(1-x^2-x^3) (a(n+3)) is given by the diagonal sums of the Riordan array (1/(1-x^3), x/(1-x^3)). The row sums are A000930. - Paul Barry, Feb 25 2005
a(n) = A023434(n-7) + 1 for n >= 7. - David Callan, Jul 14 2006
a(n+5) corresponds to the diagonal sums of A030528. The binomial transform of a(n+5) is A052921. a(n+5) = Sum_{k=0..floor(n/2)} Sum_{k=0..n} (-1)^(n-k+i)*binomial(n-k, i)binomial(i+k+1, 2k+1). - Paul Barry, Jun 21 2004
r^(n-1) = (1/r)*a(n) + r*a(n+1) + a(n+2), where r = 1.32471... is the real root of x^3 - x - 1 = 0. Example: r^8 = (1/r)*a(9) + r*a(10) + a(11) = (1/r)*2 + r*3 + 4 = 9.483909... - Gary W. Adamson, Oct 22 2006
a(n) = (r^n)/(2r+3) + (s^n)/(2s+3) + (t^n)/(2t+3) where r, s, t are the three roots of x^3-x-1. - Keith Schneider (schneidk(AT)email.unc.edu), Sep 07 2007
a(n) = -k*a(n-1) + a(n-2) + (k+1)a(n-2) + k*a(n-4), n > 3, for any value of k. - Gary Detlefs, Sep 13 2010
From Francesco Daddi, Aug 04 2011: (Start)
a(0) + a(2) + a(4) + a(6) + ... + a(2*n) = a(2*n+3).
a(0) + a(3) + a(6) + a(9) + ... + a(3*n) = a(3*n+2)+1.
a(0) + a(5) + a(10) + a(15) + ... + a(5*n) = a(5*n+1)+1.
a(0) + a(7) + a(14) + a(21) + ... + a(7*n) = (a(7*n) + a(7*n+1) + 1)/2. (End)
a(n+3) = Sum_{k=0..floor((n+1)/2)} binomial((n+k)/3,k), where binomial((n+k)/3,k)=0 for noninteger (n+k)/3. - Nikita Gogin, Dec 07 2012
a(n) = A182097(n-3) for n > 2. - Jonathan Sondow, Mar 14 2014
a(n) = the k-th difference of a(n+5k) - a(n+5k-1), k>=1. For example, a(10)=3 => a(15)-a(14) => 2nd difference of a(20)-a(19) => 3rd difference of a(25)-a(24)... - Bob Selcoe, Mar 18 2014
Construct the power matrix T(n,j) = [A^*j]*[S^*(j-1)] where A=(0,0,1,0,1,0,1,...) and S=(0,1,0,0,...) or A063524. [* is convolution operation] Define S^*0=I with I=(1,0,0,...). Then a(n) = Sum_{j=1...n} T(n,j). - David Neil McGrath, Dec 19 2014
If x=a(n), y=a(n+1), z=a(n+2), then x^3 + 2*y*x^2 - z^2*x - 3*y*z*x + y^2*x + y^3 - y^2*z + z^3 = 1. - Alexander Samokrutov, Jul 20 2015
For the sequence shifted by 6 terms, a(n) = Sum_{k=ceiling(n/3)..ceiling(n/2)} binomial(k+1,3*k-n) [Doslic-Zubac]. - N. J. A. Sloane, Apr 23 2017
From Joseph M. Shunia, Jan 21 2020: (Start)
a(2n) = 2*a(n-1)*a(n) + a(n)^2 + a(n+1)^2, for n > 8.
a(2n-1) = 2*a(n)*a(n+1) + a(n-1)^2, for n > 8.
a(2n+1) = 2*a(n+1)*a(n+2) + a(n)^2, for n > 7. (End)
0*a(0) + 1*a(1) + 2*a(2) + ... + n*a(n) = n*a(n+5) - a(n+9) + 2. - Greg Dresden and Zi Ye, Jul 02 2021
From Greg Dresden and Zi Ye, Jul 06 2021: (Start)
2*a(n) = a(n+2) + a(n-5) for n >= 5.
3*a(n) = a(n+4) - a(n-9) for n >= 9.
4*a(n) = a(n+5) - a(n-9) for n >= 9. (End)

Extensions

Edited by Charles R Greathouse IV, Mar 17 2010
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A011973 Irregular triangle read by rows: T(n,k) = binomial(n-k, k), n >= 0, 0 <= k <= floor(n/2); or, coefficients of (one version of) Fibonacci polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 4, 3, 1, 5, 6, 1, 1, 6, 10, 4, 1, 7, 15, 10, 1, 1, 8, 21, 20, 5, 1, 9, 28, 35, 15, 1, 1, 10, 36, 56, 35, 6, 1, 11, 45, 84, 70, 21, 1, 1, 12, 55, 120, 126, 56, 7, 1, 13, 66, 165, 210, 126, 28, 1, 1, 14, 78, 220, 330, 252, 84, 8, 1, 15, 91, 286, 495, 462
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

T(n,k) is the number of subsets of {1,2,...,n-1} of size k and containing no consecutive integers. Example: T(6,2)=6 because the subsets of size 2 of {1,2,3,4,5} with no consecutive integers are {1,3},{1,4},{1,5},{2,4},{2,5} and {3,5}. Equivalently, T(n,k) is the number of k-matchings of the path graph P_n. - Emeric Deutsch, Dec 10 2003
T(n,k) = number of compositions of n+2 into k+1 parts, all >= 2. Example: T(6,2)=6 because we have (2,2,4),(2,4,2),(4,2,2),(2,3,3),(3,2,3) and (3,3,2). - Emeric Deutsch, Apr 09 2005
Given any recurrence sequence S(k) = x*a(k-1) + a(k-2), starting (1, x, x^2+1, ...); the (k+1)-th term of the series = f(x) in the k-th degree polynomial: (1, (x), (x^2 + 1), (x^3 + 2x), (x^4 + 3x^2 + 1), (x^5 + 4x^3 + 3x), (x^6 + 5x^4 + 6x^2 + 1), ...). Example: let x = 2, then S(k) = 1, 2, 5, 12, 29, 70, 169, ... such that A000129(7) = 169 = f(x), x^6 + 5x^4 + 6x^2 + 1 = (64 + 80 + 24 + 1). - Gary W. Adamson, Apr 16 2008
Row k gives the nonzero coefficients of U(k,x/2) where U is the Chebyshev polynomial of the second kind. For example, row 6 is 1,5,6,1 and U(6,x/2) = x^6 - 5x^4 + 6x^2 - 1. - David Callan, Jul 22 2008
T(n,k) is the number of nodes at level k in the Fibonacci tree f(k-1). The Fibonacci trees f(k) of order k are defined as follows: 1. f(-1) and f(0) each consist of a single node. 2. For k >= 1, to the root of f(k-1), taken as the root of f(k), we attach with a rightmost edge the tree f(k-2). See the Iyer and Reddy references. These trees are not the same as the Fibonacci trees in A180566. Example: T(3,0)=1 and T(3,1)=2 because in f(2) = /\ we have 1 node at level 0 and 2 nodes at level 1. - Emeric Deutsch, Jun 21 2011
Triangle, with zeros omitted, given by (1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2011
Riordan array (1/(1-x),x^2/(1-x). - Philippe Deléham, Dec 12 2011
This sequence is the elements on the rising diagonals of the Pascal triangle, where the sum of the elements in each rising diagonal represents a Fibonacci number. - Mohammad K. Azarian, Mar 08 2012
If we set F(0;x) = 0, F(1;x) = 1, F(n+1;x) = x*F(n;x) + F(n-1;x), then we obtain the sequence of Vieta-Fibonacci polynomials discussed by Gary W. Adamson above. We note that F(n;x) = (-i)^n * U(n;i*x/2), where U denotes the respective Chebyshev polynomial of the second kind (see David Callan's remark above). Let us fix a,b,f(0),f(1) in C, b is not the zero, and set f(n) = a*f(n-1) + b*f(n-2). Then we deduce the relation: f(n) = b^((n-1)/2) * F(n;a/sqrt(b))*f(1) + b^(n/2) * F(n-1;a/sqrt(b))*f(0), where for a given value of the complex root sqrt(b) we set b^(n/2) = (sqrt(b))^n. Moreover, if b=1 then we get f(n+k) + (-1)^k * f(n-k) = L(k;a)*f(n), for every k=0,1,...,n, and where L(0;a)=2, L(1;a)=a, L(n+1;a)=a*L(n;a) + L(n-1;a) are the Vieta-Lucas polynomials. Let us observe that L(n+2;a) = F(n+2;a) + F(n;a), L(m+n;a) = L(m;a)*F(n;a) + L(m-1;a)*F(n-1;a), which implies also L(n+1;a) = a*F(n;a) + 2*F(n-1;a). Further we have L(n;a) = 2*(-i)^n * T(n;i*x/2), where T(n;x) denotes the n-th Chebyshev polynomial of the first kind. For the proofs, other relations and facts - see Witula-Slota's papers. - Roman Witula, Oct 12 2012
The diagonal sums of this triangle are A000930. - John Molokach, Jul 04 2013
Aside from signs and index shift, the coefficients of the characteristic polynomial of the Coxeter adjacency matrix for the Coxeter group A_n related to the Chebyshev polynomial of the second kind (cf. Damianou link p. 19). - Tom Copeland, Oct 11 2014
For a mirrored, shifted version showing the relation of these coefficients to the Pascal triangle, Fibonacci, and other number triangles, see A030528. See also A053122 for a relation to Cartan matrices. - Tom Copeland, Nov 04 2014
For a relation to a formulation for a universal Lie Weyl algebra for su(1,1), see page 16 of Durov et al. - Tom Copeland, Nov 29 2014
A reversed, signed and aerated version is given by A049310, related to Chebyshev polynomials. - Tom Copeland, Dec 06 2015
For n >= 3, the n-th row gives the coefficients of the independence polynomial of the (n-2)-path graph P_{n-2}. - Eric W. Weisstein, Apr 07 2017
For n >= 2, the n-th row gives the coefficients of the matching-generating polynomial of the (n-1)-path graph P_{n-1}. - Eric W. Weisstein, Apr 10 2017
Antidiagonals of the Pascal matrix A007318 read bottom to top. These are also the antidiagonals read from top to bottom of the numerical coefficients of the Maurer-Cartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper), which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Reverse is A102426. - Tom Copeland, Jul 02 2018
T(n,k) is the number of Markov equivalence classes with skeleton the path on n+1 nodes having exactly k immoralities. See Theorem 2.1 in the article by A. Radhakrishnan et al. below. - Liam Solus, Aug 23 2018
T(n, k) = number of compositions of n+1 into n+1-2*k odd parts. For example, T(6,2) = 6 because 7 = 5+1+1 = 3+3+1 = 3+1+3 = 1+1+5 = 1+3+3 = 1+1+5. - Michael Somos, Sep 19 2019
From Gary W. Adamson, Apr 25 2022: (Start)
Alternate rows can be parsed into those with odd integer coefficients to the right of the leftmost 1, and those with even integer coefficients to the right of the leftmost 1. The first set is shown in A054142 and are characteristic polynomials of submatrices of an infinite tridiagonal matrix (A332602) with all -1's in the super and subdiagonals and (1,2,2,2,...) as the main diagonal. For example, the characteristic equation of the 3 X 3 submatrix (1,-1,0; -1,2,-1; 0,-1,2) is x^3 - 5x^2 + 6x - 1. The roots are the Beraha constants B(7,1) = 3.24697...; B(7,2) = 1.55495...; and B(7,3) = 0.198062.... For n X n matrices of this form, the largest eigenvalue is B(2n+1, 1). The 3 X 3 matrix has an eigenvalue of 3.24697... = B(7,1).
Polynomials with even integer coefficients to the right of the leftmost 1 are in A053123 with roots being the even-indexed Beraha constants. The generating Cartan matrices are those with (2,2,2,...) as the main diagonal and -1's as the sub- and superdiagonals. The largest eigenvalue of n X n matrices of this form are B(2n+2,1). For example, the largest eigenvalue of (2,-1,0; -1,2,-1; 0,-1,2) is 3.414... = B(8,1) = a root to x^3 - 6x^2 + 10x - 4. (End)
T(n,k) is the number of edge covers of P_(n+2) with (n-k) edges. For example, T(6,2)=6 because among edges 1, 2, ..., 7 of P_8, we can eliminate any two non-consecutive edges among 2-6. These numbers can be found using the recurrence relation for the edge cover polynomial of P_n, which is E(P_n,x) = xE(P_(n-1),x)+xE(P_(n-2),x) and E(P_1,x)=0, E(P_2,x)=x (ref. Akbari and Oboudi). - Feryal Alayont, Jun 03 2022
T(n,k) is the number of ways to tile an n-board (an n X 1 array of 1 X 1 cells) using k dominoes and n-2*k squares. - Michael A. Allen, Dec 28 2022
T(n,k) is the number of positive integer sequences (s(1),s(2),...,s(n-2k)) such that s(i) < s(i+1), s(1) is odd, s(n-2k) <= n, and s(i) and s(i+1) have opposite parity (ref. Donnelly, Dunkum, and McCoy). Example: T(6,0)=1 corresponds to 123456; T(6,1)=5 corresponds to 1234, 1236, 1256, 1456, 3456; T(6,2)=6 corresponds to 12, 14, 16, 34, 36; and T(6,3)=1 corresponds to the empty sequence () with length 0. - Molly W. Dunkum, Jun 27 2023

Examples

			The first few Fibonacci polynomials (defined here by F(0,x) = 0, F(1,x) = 1; F(n+1, x) = F(n, x) + x*F(n-1, x)) are:
0: 0
1: 1
2: 1
3: 1 + x
4: 1 + 2*x
5: 1 + 3*x + x^2
6: (1 + x)*(1 + 3*x)
7: 1 + 5*x + 6*x^2 + x^3
8: (1 + 2*x)*(1 + 4*x + 2*x^2)
9: (1 + x)*(1 + 6*x + 9*x^2 + x^3)
10: (1 + 3*x + x^2 )*(1 + 5*x + 5*x^2)
11: 1 + 9*x + 28*x^2 + 35*x^3 + 15*x^4 + x^5
From _Roger L. Bagula_, Feb 20 2009: (Start)
  1
  1
  1   1
  1   2
  1   3   1
  1   4   3
  1   5   6   1
  1   6  10   4
  1   7  15  10   1
  1   8  21  20   5
  1   9  28  35  15   1
  1  10  36  56  35   6
  1  11  45  84  70  21   1
  1  12  55 120 126  56   7 (End)
For n=9 and k=4, T(9,4) = C(5,4) = 5 since there are exactly five size-4 subsets of {1,2,...,8} that contain no consecutive integers, namely, {1,3,5,7}, {1,3,5,8}, {1,3,6,8}, {1,4,6,8}, and {2,4,6,8}. - _Dennis P. Walsh_, Mar 31 2011
When the rows of the triangle are displayed as centered text, the falling diagonal sums are A005314. The first few terms are row1 = 1 = 1; row2 = 1+1 = 2; row3 = 2+1 = 3; row4 = 1+3+1 = 5; row5 = 1+3+4+1 = 9; row6 = 4+6+5+1 = 16; row7 = 1+10+10+6+1 = 28; row8 = 1+5+20+15+7+1 = 49; row9 = 6+15+35+21+8+1 = 86; row10 = 1+21+35+56+28+9+1 = 151. - _John Molokach_, Jul 08 2013
In the example, you can see that the n-th row of Pascal's triangle is given by T(n, 0), T(n+1, 1), ..., T(2n-1, n-1), T(2n, n). - _Daniel Forgues_, Jul 07 2018

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 141ff.
  • C. D. Godsil, Algebraic Combinatorics, Chapman and Hall, New York, 1993.
  • I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. See p. 117.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 182-183.

Links

Crossrefs

Row sums = A000045(n+1) (Fibonacci numbers). - Michael Somos, Apr 02 1999
Cf. A000129, A054123, A052553, A000930, A030528, A053122.
All of A011973, A092865, A098925, A102426, A169803 describe essentially the same triangle in different ways.
Cf. A007318, A025581, A055087.
Cf. A000108, A049310, A055151, A098474, A124644, A134264.
Cf. A054142, A053123, A332602.

Programs

  • Haskell
    a011973 n k = a011973_tabf !! n !! k
a011973_row n = a011973_tabf !! n
a011973_tabf = zipWith (zipWith a007318) a025581_tabl a055087_tabf
-- Reinhard Zumkeller, Jul 14 2015
  • Maple
    a := proc(n) local k; [ seq(binomial(n-k,k),k=0..floor(n/2)) ]; end;
T := proc(n, k): if k<0 or k>floor(n/2) then return(0) fi: binomial(n-k, k) end: seq(seq(T(n,k), k=0..floor(n/2)), n=0..15); # Johannes W. Meijer, Aug 26 2013
  • Mathematica
    (* first: sum method *) Table[CoefficientList[Sum[Binomial[n - m + 1, m]*x^m, {m, 0, Floor[(n + 1)/2]}], x], {n, 0, 12}] (* Roger L. Bagula, Feb 20 2009 *)
(* second: polynomial recursion method *) Clear[L, p, x, n, m]; L[x, 0] = 1; L[x, 1] = 1 + x; L[x_, n_] := L[x, n - 1] + x*L[x, n - 2]; Table[ExpandAll[L[x, n]], {n, 0, 10}]; Table[CoefficientList[ExpandAll[L[x, n]], x], {n, 0, 12}]; Flatten[%] (* Roger L. Bagula, Feb 20 2009 *)
(* Center option shows falling diagonals are A224838 *) Column[Table[Binomial[n - m, m], {n, 0, 25}, {m, 0, Floor[n/2]}], Center] (* John Molokach, Jul 26 2013 *)
Table[ Select[ CoefficientList[ Fibonacci[n, x], x], Positive] // Reverse, {n, 1, 18} ] // Flatten (* Jean-François Alcover, Oct 21 2013 *)
CoefficientList[LinearRecurrence[{1, x}, {1 + x, 1 + 2 x}, {-1, 10}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)
CoefficientList[Table[x^((n - 1)/2) Fibonacci[n, 1/Sqrt[x]], {n, 15}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)
  • PARI
    {T(n, k) = if( k<0 || 2*k>n, 0, binomial(n-k, k))};
  • Sage
    # Prints the table; cf. A145574.
for n in (2..20): [Compositions(n, length=m, min_part=2).cardinality() for m in (1..n//2)]  # Peter Luschny, Oct 18 2012

Formula

Let F(n, x) be the n-th Fibonacci polynomial in x; the g.f. for F(n, x) is Sum_{n>=0} F(n, x)*y^n = (1 + x*y)/(1 - y - x*y^2). - Paul D. Hanna
T(m, n) = 0 for n != 0 and m <= 1 T(0, 0) = T(1, 0) = 1 T(m, n) = T(m - 1, n) + T(m-2, n-1) for m >= 2 (i.e., like the recurrence for Pascal's triangle A007318, but going up one row as well as left one column for the second summand). E.g., T(7, 2) = 10 = T(6, 2) + T(5, 1) = 6 + 4. - Rob Arthan, Sep 22 2003
G.f. for k-th column: x^(2*k-1)/(1-x)^(k+1).
Identities for the Fibonacci polynomials F(n, x):
F(m+n+1, x) = F(m+1, x)*F(n+1, x) + x*F(m, x)F(n, x).
F(n, x)^2-F(n-1, x)*F(n+1, x) = (-x)^(n-1).
The degree of F(n, x) is floor((n-1)/2) and F(2p, x) = F(p, x) times a polynomial of equal degree which is 1 mod p.
From Roger L. Bagula, Feb 20 2009: (Start)
p(x,n) = Sum_{m=0..floor((n+1)/2)} binomial(n-m+1, m)*x^m;
p(x,n) = p(x, n - 1) + x*p(x, n - 2). (End)
T(n, k) = A102541(2*n+2, 2*k+1) + A102541(2*n+1, 2*k) - A102541(2*n+3, 2*k+1), n >= 0 and 0 <= k <= floor(n/2). - Johannes W. Meijer, Aug 26 2013
G.f.: 1/(1-x-y*x^2) = R(0)/2, where R(k) = 1 + 1/(1 - (2*k+1+ x*y)*x/((2*k+2+ x*y)*x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
O.g.f. G(x,t) = x/(1-x-tx^2) = x + x^2 + (1+t) x^3 + (1+2t) x^4 + ... has the inverse Ginv(x,t) = -[1+x-sqrt[(1+x)^2 + 4tx^2]]/(2tx) = x - x^2 + (1-t) x^3 + (-1+3t) x^4 + ..., an o.g.f. for the signed Motzkin polynomials of A055151, consistent with A134264 with h_0 = 1, h_1 = -1, h_2 = -t, and h_n = 0 otherwise. - Tom Copeland, Jan 21 2016
O.g.f. H(x,t) = x (1+tx)/ [1-x(1+tx)] = x + (1+t) x^2 + (1+2t) x^3 + ... = -L[Cinv(-tx)/t], where L(x) = x/(1+x) with inverse Linv(x) = x/(1-x) and Cinv(x) = x (1-x) is the inverse of C(x) = (1-sqrt(1-4x))/2, the o.g.f. of the shifted Catalan numbers A000108. Then Hinv(x,t) = -C[t Linv(-x)]/t = [-1 + sqrt(1+4tx/(1+x))]/2t = x - (1+t) x^2 + (1+2t+2t^2) x^3 - (1+3t+6t^2+5t^3) x^4 + ..., which is signed A098474, reverse of A124644. - Tom Copeland, Jan 25 2016
T(n, k) = GegenbauerC(k, (n+1)/2-k, 1). - Peter Luschny, May 10 2016

A001608 Perrin sequence (or Ondrej Such sequence): a(n) = a(n-2) + a(n-3) with a(0) = 3, a(1) = 0, a(2) = 2.

Original entry on oeis.org

3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17, 22, 29, 39, 51, 68, 90, 119, 158, 209, 277, 367, 486, 644, 853, 1130, 1497, 1983, 2627, 3480, 4610, 6107, 8090, 10717, 14197, 18807, 24914, 33004, 43721, 57918, 76725, 101639, 134643, 178364, 236282, 313007, 414646, 549289
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Has been called the skiponacci sequence or skiponacci numbers. - N. J. A. Sloane, May 24 2013
For n >= 3, also the numbers of maximal independent vertex sets, maximal matchings, minimal edge covers, and minimal vertex covers in the n-cycle graph C_n. - Eric W. Weisstein, Mar 30 2017 and Aug 03 2017
With the terms indexed as shown, has property that p prime => p divides a(p). The smallest composite n such that n divides a(n) is 521^2. For quotients a(p)/p, where p is prime, see A014981.
Asymptotically, a(n) ~ r^n, with r=1.3247179572447... the inverse of the real root of 1-x^2-x^3=0 (see A060006). If n>9 then a(n)=round(r^n). - Ralf Stephan, Dec 13 2002
The recursion can be used to compute a(-n). The result is -A078712(n). - T. D. Noe, Oct 10 2006
For n>=3, a(n) is the number of maximal independent sets in a cycle of order n. - Vincent Vatter, Oct 24 2006
Pisano period lengths are given in A104217. - R. J. Mathar, Aug 10 2012
From Roman Witula, Feb 01 2013: (Start)
Let r1, r2 and r3 denote the roots of x^3 - x - 1. Then the following identity holds:
a(k*n) + (a(k))^n - (a(k) - r1^k)^n - (a(k) - r2^k)^n - (a(k) - r3^k)^n
= 0 for n = 0, 1, 2,
= 6 for n = 3,
= 12*a(k) for n = 4,
= 10*[2*(a(k))^2 - a(-k)] for n = 5,
= 30*a(k)*[(a(k))^2 - a(-k)] for n = 6,
= 7*[6*(a(k))^4 - 9*a(-k)*(a(k))^2 + 2*(a(-k))^2 - a(k)] for n = 7,
= 56*a(k)*[((a(k))^2 - a(-k))^2 - a(k)/2] for n = 8,
where a(-k) = -A078712(k) and the formula (5.40) from the paper of Witula and Slota is used. (End)
The parity sequence of a(n) is periodic with period 7 and has the form (1,0,0,1,0,1,1). Hence we get that a(n) and a(2*n) are congruent modulo 2. Similarly we deduce that a(n) and a(3*n) are congruent modulo 3. Is it true that a(n) and a(p*n) are congruent modulo p for every prime p? - Roman Witula, Feb 09 2013
The trinomial x^3 - x - 1 divides the polynomial x^(3*n) - a(n)*x^(2*n) + ((a(n)^2 - a(2*n))/2)*x^n - 1 for every n>=1. For example, for n=3 we obtain the factorization x^9 - 3*x^6 + 2*x^3 - 1 = (x^3 - x - 1)*(x^6 + x^4 - 2*x^3 + x^2 - x + 1). Sketch of the proof: Let p,s,t be roots of the Perrin polynomial x^3 - x - 1. Then we have (a(n))^2 = (p^n + s^n + t^n)^2 = a(2*n) + 2*a(n)*x^n -2*x^n + 2/x^n for every x = p,s,t, i.e., x^(3*n) - a(n)*x^(2*n) + ((a(n)^2 - a(2*n))/2)*x^n - 1 = 0 for every x = p,s,t, which finishes the proof. By discussion of the power(a(n))^3 = (p^n + s^n + t^n)^3 it can be deduced that the trinomial x^3 - x - 1 divides the polynomial 2*x^(4*n) - a(n)*x^(3*n) - a(2*n)*x^(2*n) + ((a(n)^3 - a(3*n) - 3)/3)*x^n - a(n) = 0. Co-author of these divisibility relations is also my young student Szymon Gorczyca (13 years old as of 2013). - Roman Witula, Feb 09 2013
The sum of powers of the real root and complex roots of x^3-x-1=0 as expressed as powers of the plastic number r, (see A060006). Let r0=1, r1=r, r2=1+r^(-1) and c0=2, c1=-r and c3 = r^(-5) then a(n) = r(n-2)+r(n-3) + c(n-2)+c(n-3). Example: a(5) = 1 + r^(-1) + 1 + r + 2 - r + r^(-5) = 4 + r^(-1) + r^(-5) = 5. - Richard Turk, Jul 14 2016
Also the number of minimal total dominating sets in the n-sun graph. - Eric W. Weisstein, Apr 27 2018
Named after the French engineer François Olivier Raoul Perrin (1841-1910). - Amiram Eldar, Jun 05 2021
a(p) = p*A127687(p) for p prime. - Robert FERREOL, Apr 09 2024

Examples

			From _Roman Witula_, Feb 01 2013: (Start)
We note that if a + b + c = 0 then:
1) a^3 + b^3 + c^3 = 3*a*b*c,
2) a^4 + b^4 + c^4 = 2*((a^2 + b^2 + c^2)/2)^2,
3) (a^5 + b^5 + c^5)/5 = (a^3 + b^3 + c^3)/3 * (a^2 +
    b^2  + c^2)/2,
4) (a^7 + b^7 + c^7)/7 = (a^5 + b^5 + c^5)/5 * (a^2 + b^2 + c^2)/2 = 2*(a^3 + b^3 + c^3)/3 * (a^4 + b^4 + c^4)/4,
5) (a^7 + b^7 + c^7)/7 * (a^3 + b^3 + c^3)/3 = ((a^5 + b^5 + c^5)/5)^2.
Hence, by the Binet formula for a(n) we obtain the relations: a(3) = 3, a(4) = 2*(a(2)/2)^2 = 2, a(5)/5 = a(3)/3 * a(2)/2, i.e., a(5) = 5, and similarly that a(7) = 7. (End)

References

  • Olivier Bordellès, Thèmes d'Arithmétique, Ellipses, 2006, Exercice 4.11, p. 127.0
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.2.2.
  • Dmitry Fomin, On the properties of a certain recursive sequence, Mathematics and Informatics Quarterly, Vol. 3 (1993), pp. 50-53.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 70.
  • Manfred Schroeder, Number Theory in Science and Communication, 3rd ed., Springer, 1997.
  • A. G. Shannon, P. G. Anderson and A. F. Horadam, Properties of Cordonnier, Perrin and Van der Laan numbers, International Journal of Mathematical Education in Science and Technology, Volume 37:7 (2006), 825-831. See Q_n.
  • 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

Closely related to A182097.
Cf. A000931, bisection A109377.
Cf. A013998 (Unrestricted Perrin pseudoprimes).
Cf. A018187 (Restricted Perrin pseudoprimes).

Programs

  • GAP
    a:=[3,0,2];; for n in [4..20] do a[n]:=a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Jul 12 2018
  • Haskell
    a001608 n = a000931_list !! n
a001608_list = 3 : 0 : 2 : zipWith (+) a001608_list (tail a001608_list)
-- Reinhard Zumkeller, Feb 10 2011
  • Magma
    I:=[3,0,2]; [n le 3 select I[n] else Self(n-2) +Self(n-3): n in [1..50]]; // G. C. Greubel, Mar 18 2019
  • Maple
    A001608 :=proc(n) option remember; if n=0 then 3 elif n=1 then 0 elif n=2 then 2 else procname(n-2)+procname(n-3); fi; end proc;
[seq(A001608(n),n=0..50)]; # N. J. A. Sloane, May 24 2013
  • Mathematica
    LinearRecurrence[{0, 1, 1}, {3, 0, 2}, 50] (* Harvey P. Dale, Jun 26 2011 *)
per = Solve[x^3 - x - 1 == 0, x]; f[n_] := Floor @ Re[N[ per[[1, -1, -1]]^n + per[[2, -1, -1]]^n + per[[3, -1, -1]]^n]]; Array[f, 46, 0] (* Robert G. Wilson v, Jun 29 2010 *)
a[n_] := n*Sum[Binomial[k, n-2*k]/k, {k, 1, n/2}]; a[0]=3; Table[a[n] , {n, 0, 45}] (* Jean-François Alcover, Oct 04 2012, after Vladimir Kruchinin *)
CoefficientList[Series[(3 - x^2)/(1 - x^2 - x^3), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 03 2015 *)
Table[RootSum[-1 - # + #^3 &, #^n &], {n, 0, 20}] (* Eric W. Weisstein, Mar 30 2017 *)
RootSum[-1 - # + #^3 &, #^Range[0, 20] &] (* Eric W. Weisstein, Dec 30 2017 *)
  • PARI
    a(n)=if(n<0,0,polsym(x^3-x-1,n)[n+1])
  • PARI
    A001608_list(n) = polsym(x^3-x-1,n) \\ Joerg Arndt, Mar 10 2019
  • Python
    A001608_list, a, b, c = [3, 0, 2], 3, 0, 2
for _ in range(100):
    a, b, c = b, c, a+b
    A001608_list.append(c) # Chai Wah Wu, Jan 27 2015
  • Sage
    ((3-x^2)/(1-x^2-x^3)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Mar 18 2019

Formula

G.f.: (3 - x^2)/(1 - x^2 - x^3). - Simon Plouffe in his 1992 dissertation
a(n) = r1^n + r2^n + r3^n where r1, r2, r3 are three roots of x^3-x-1=0.
a(n-1) + a(n) + a(n+1) = a(n+4), a(n) - a(n-1) = a(n-5). - Jon Perry, Jun 05 2003
From Gary W. Adamson, Feb 01 2004: (Start)
a(n) = trace(M^n) where M is the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 1 0], the companion matrix of the characteristic polynomial of this sequence, P = X^3 - X - 1.
M^n * [3, 0, 2] = [a(n), a(n+1), a(n+2)]; e.g., M^7 * [3, 0, 2] = [7, 10, 12].
a(n) = 2*A000931(n+3) + A000931(n). (End)
a(n) = 3*p(n) - p(n-2) = 2*p(n) + p(n-3), with p(n) := A000931(n+3), n >= 0. - Wolfdieter Lang, Jun 21 2010
From Francesco Daddi, Aug 03 2011: (Start)
a(0) + a(1) + a(2) + ... + a(n) = a(n+5) - 2.
a(0) + a(2) + a(4) + ... + a(2*n) = a(2*n+3).
a(1) + a(3) + a(5) + ... + a(2*n+1) = a(2*n+4) - 2. (End)
From Francesco Daddi, Aug 04 2011: (Start)
a(0) + a(3) + a(6) + a(9) + ... + a(3*n) = a(3*n+2) + 1.
a(0) + a(5) + a(10) + a(15) + ... + a(5*n) = a(5*n+1)+3.
a(0) + a(7) + a(14) + a(21) + ... + a(7*n) = (a(7*n) + a(7*n+1) + 3)/2. (End)
a(n) = n*Sum_{k=1..floor(n/2)} binomial(k,n-2*k)/k, n > 0, a(0)=3. - Vladimir Kruchinin, Oct 21 2011
(a(n)^3)/2 + a(3n) - 3*a(n)*a(2n)/2 - 3 = 0. - Richard Turk, Apr 26 2017
2*a(4n) - 2*a(n) - 2*a(n)*a(3n) - a(2n)^2 + a(n)^2*a(2n) = 0. - Richard Turk, May 02 2017
a(n)^4 + 6*a(4n) - 4*a(3n)*a(n) - 3*a(2n)^2 - 12a(n) = 0. - Richard Turk, May 02 2017
a(n+5)^2 + a(n+1)^2 - a(n)^2 = a(2*(n+5)) + a(2*(n+1)) - a(2*n). - Aleksander Bosek, Mar 04 2019
From Aleksander Bosek, Mar 18 2019: (Start)
a(n+12) = a(n) + 2*a(n+4) + a(n+11);
a(n+16) = a(n) + 4*a(n+9) + a(n+13);
a(n+18) = a(n) + 2*a(n+6) + 5*a(n+12);
a(n+21) = a(n) + 2*a(n+12) + 6*a(n+14);
a(n+27) = a(n) + 3*a(n+9) + 4*a(n+22). (End)
a(n) = Sum_{j=0..floor((n-g)/(2*g))} 2*n/(n-2*(g-2)*j-(g-2)) * Hypergeometric2F1([-(n-2g*j-g)/2, -(2j+1)], [1], 1), g = 3 and n an odd integer. - Richard Turk, Oct 14 2019
E.g.f.: exp(r1*x) + exp(r2*x) + exp(r3*x), where r1, r2, r3 are three roots of x^3 - x - 1 = 0. - Fabian Pereyra, Nov 02 2024

Extensions

Additional comments from Mike Baker, Oct 11 2005
Definition edited by Chai Wah Wu, Jan 27 2015
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A001687 a(n) = a(n-2) + a(n-5).

Original entry on oeis.org

0, 1, 0, 1, 0, 1, 1, 1, 2, 1, 3, 2, 4, 4, 5, 7, 7, 11, 11, 16, 18, 23, 29, 34, 45, 52, 68, 81, 102, 126, 154, 194, 235, 296, 361, 450, 555, 685, 851, 1046, 1301, 1601, 1986, 2452, 3032, 3753, 4633, 5739, 7085, 8771, 10838, 13404, 16577, 20489, 25348, 31327
Offset: 0

Views

Author

N. J. A. Sloane, following a suggestion from Robert G. Wilson v

Keywords

Comments

a(n+1) is the number of compositions of n into parts 2 and 5. [Joerg Arndt, Mar 15 2013]

References

  • 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. A005686.

Programs

  • Maple
    A001687:=-z/(-1+z**2+z**5); # Simon Plouffe in his 1992 dissertation
  • Mathematica
    CoefficientList[Series[x/(1-x^2-x^5),{x,0,60}],x] (* or *) Nest[ Append[#,#[[-5]]+#[[-2]]]&, {0,1,0,1,0}, 60]  (* Harvey P. Dale, Apr 06 2011 *)
LinearRecurrence[{0, 1, 0, 0, 1}, {0, 1, 0, 1, 0}, 100] (* T. D. Noe, Aug 09 2012 *)
  • Maxima
    a(n):=sum(if mod(n-5*k,3)=0 then binomial(k,(5*k-n)/3) else 0,k,1,n); /* Vladimir Kruchinin, May 24 2011 */
  • PARI
    a(n)=if(n<0,polcoeff(x^4/(1+x^3-x^5)+x^-n*O(x),-n),polcoeff(x/(1-x^2-x^5)+x^n*O(x),n)) /* Michael Somos, Jul 15 2004 */

Formula

G.f.: x/(1-x^2-x^5).
G.f. A(x) satisfies 1+x^4*A(x) = 1/(1-x^5-x^7-x^9-....). - Jon Perry, Jul 04 2004
G.f.: -x/( x^5 - 1 + 5*x^2/Q(0) ) where Q(k) = x + 5 + k*(x+1) - x*(k+1)*(k+6)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Mar 15 2013

A007387 Number of 3rd-order maximal independent sets in cycle graph.

Original entry on oeis.org

0, 2, 3, 2, 5, 2, 7, 2, 9, 7, 11, 14, 13, 23, 20, 34, 34, 47, 57, 67, 91, 101, 138, 158, 205, 249, 306, 387, 464, 592, 713, 898, 1100, 1362, 1692, 2075, 2590, 3175, 3952, 4867, 6027, 7457, 9202, 11409, 14069, 17436, 21526, 26638, 32935, 40707, 50371, 62233
Offset: 1

Views

Author

N. J. A. Sloane, Mira Bernstein

Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. Yanco and A. Bagchi, "K-th order maximal independent sets in path and cycle graphs," J. Graph Theory, submitted, 1994.

Links

Crossrefs

Cf. A001608, A007388, A007389.

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 50); [0] cat Coefficients(R!( x^2*(2+3*x+2*x^3-3*x^6)/(1-x^2-x^5) )); // G. C. Greubel, Oct 19 2019
  • Maple
    seq(coeff(series(x^2*(2+3*x+2*x^3-3*x^6)/(1-x^2-x^5), x, n+1), x, n), n = 1..50); # G. C. Greubel, Oct 19 2019
  • Mathematica
    Rest[CoefficientList[Series[x^2*(2+3*x+2*x^3-3*x^6)/(1-x^2-x^5), {x, 0, 50}], x]] (* Harvey P. Dale, Oct 23 2011 *)
  • PARI
    my(x='x+O('x^50)); concat([0], Vec(x^2*(2+3*x+2*x^3-3*x^6)/(1-x^2-x^5))) \\ G. C. Greubel, Oct 19 2019
  • Sage
    def A007387_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x^2*(2+3*x+2*x^3-3*x^6)/(1-x^2-x^5)).list()
a=A007387_list(50); a[1:] # G. C. Greubel, Oct 19 2019

Formula

For n >= 9: a(n) = a(n-2) + a(n-5) per A133394. - G. Reed Jameson (Reedjameson(AT)yahoo.com), Dec 13 2007, Dec 16 2007
G.f.: x^2*(2 + 3*x + 2*x^3 - 3*x^6)/(1 - x^2 - x^5). - R. J. Mathar, Oct 30 2009
a(n) = Sum_{j=0..floor((n-g)/(2*g))} (2*n/(n-2*(g-2)*j-(g-2))) * Hypergeometric2F1([-(n-2g*j-g)/2,-(2j+1)], [1], 1), with g = 5, n >= g, and n an odd integer. - Richard Turk, Oct 14 2019

Extensions

More terms from Harvey P. Dale, Oct 23 2011

A369813 Expansion of 1/(1 - x^2 - x^7).

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 1, 1, 1, 2, 1, 3, 1, 4, 2, 5, 4, 6, 7, 7, 11, 9, 16, 13, 22, 20, 29, 31, 38, 47, 51, 69, 71, 98, 102, 136, 149, 187, 218, 258, 316, 360, 452, 509, 639, 727, 897, 1043, 1257, 1495, 1766, 2134, 2493, 3031, 3536, 4288, 5031, 6054, 7165, 8547, 10196, 12083, 14484, 17114, 20538, 24279, 29085, 34475
Offset: 0

Views

Author

Seiichi Manyama, Feb 02 2024

Keywords

Comments

Number of compositions of n into parts 2 and 7.

Links

Crossrefs

Cf. A005709, A017847, A369814, A369815, A369816.
Cf. A007380.

Programs

  • Mathematica
    CoefficientList[Series[1/(1-x^2-x^7),{x,0,80}],x] (* or *) LinearRecurrence[{0,1,0,0,0,0,1},{1,0,1,0,1,0,1},80] (* Harvey P. Dale, Dec 04 2024 *)
  • PARI
    my(N=70, x='x+O('x^N)); Vec(1/(1-x^2-x^7))
  • PARI
    a(n) = sum(k=0, n\7, ((n-5*k)%2==0)*binomial((n-5*k)/2, k));

Formula

a(n) = a(n-2) + a(n-7).
a(n) = A007380(n-7) for n >= 8.

A007389 7th-order maximal independent sets in cycle graph.

Original entry on oeis.org

0, 2, 3, 2, 5, 2, 7, 2, 9, 2, 11, 2, 13, 2, 15, 2, 17, 11, 19, 22, 21, 35, 23, 50, 25, 67, 36, 86, 58, 107, 93, 130, 143, 155, 210, 191, 296, 249, 403, 342, 533, 485, 688, 695, 879, 991, 1128, 1394, 1470, 1927, 1955, 2615, 2650, 3494, 3641, 4622, 5035, 6092, 6962, 8047
Offset: 1

Views

Author

N. J. A. Sloane, Mira Bernstein

Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. Yanco and A. Bagchi, K-th order maximal independent sets in path and cycle graphs, J. Graph Theory, submitted, 1994, apparently unpublished.

Links

Crossrefs

Cf. A001608, A007387, A007388.

Formula

Empirical g.f.: x^2*(7*x^14 + 5*x^12 + 3*x^10 - 2*x^7 - 2*x^5 - 2*x^3 - 3*x - 2) / (x^9 + x^2 - 1). - Colin Barker, Mar 29 2014
Theorem: a(n) = Sum_{j=0..floor((n-g)/(2*g))} (2*n/(n-2*(g-2)*j-(g-2))) * Hypergeometric2F1([-(n-2g*j-g)/2,-(2j+1)], [1], 1), g = 9, n >= g and n an odd integer. - Richard Turk, Oct 14 2019 For proof see attached text file.

A007388 5th-order maximal independent sets in cycle graph.

Original entry on oeis.org

0, 2, 3, 2, 5, 2, 7, 2, 9, 2, 11, 2, 13, 9, 15, 18, 17, 29, 19, 42, 28, 57, 46, 74, 75, 93, 117, 121, 174, 167, 248, 242, 341, 359, 462, 533, 629, 781, 871, 1122, 1230, 1584, 1763, 2213, 2544, 3084, 3666, 4314, 5250, 6077, 7463, 8621, 10547
Offset: 1

Views

Author

N. J. A. Sloane, Mira Bernstein

Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. Yanco and A. Bagchi, K-th order maximal independent sets in path and cycle graphs, J. Graph Theory, submitted, 1994, apparently unpublished.

Links

Crossrefs

Cf. A001608, A007387, A007389.

Formula

Empirical g.f.: x^2*(5*x^10+3*x^8-2*x^5-2*x^3-3*x-2) / (x^7+x^2-1). - Colin Barker, Mar 29 2014
For n >= 13: a(n) = a(n-2) + a(n-7). - Sean A. Irvine, Jan 02 2018
a(n) = Sum_{j=0..floor((n-g)/(2*g))} (2*n/(n-2*(g-2)*j-(g-2))) * Hypergeometric2F1([-(n-2g*j-g)/2,-(2j+1)], [1], 1), g = 7, n >= g and n an odd integer. - Richard Turk, Oct 14 2019

Extensions

Typo in data (242 was inadvertently repeated) fixed by Colin Barker, Mar 29 2014
More terms from Sean A. Irvine, Jan 02 2018
Showing 1-10 of 21 results. Next

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