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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A122932 a(n) = A000085(n) - A000079(n-1).

Original entry on oeis.org

0, 0, 0, 2, 10, 44, 168, 636, 2364, 8984, 34672, 138104, 564408, 2382288, 10333152, 46173968, 211733776, 997182752, 4809439296, 23758139808, 119951644320, 618882541760, 3257839688320, 17492182188992, 95680426983360
Offset: 1

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Author

Alford Arnold, Sep 22 2006

Keywords

Comments

Also sums values in A117506 but not in A007318.

Examples

			Row five of A117506 is 1 5 9 5 10 16 5 10 9 5 1.
Row five of A007318 is 1 5 10 10 5 1.
So included values are 9 5 16 5 9;
therefore a(5) = 44 = 76 - 32.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[(2k)!/k!/2^k Binomial[n, 2k], {k, 0, n/2}] - 2^(n - 1) // FullSimplify, {n, 26}] (* Robert G. Wilson v, Sep 27 2006 *)  (* or *)
    Table[HypergeometricU[ -(n/2), 1/2, -(1/2)]/(-(1/2))^(-(-n/2)) - 2^(n - 1), {n, 26}] (* Robert G. Wilson v, Sep 27 2006 *)  (* or *)
    (* first do *) Needs["DiscreteMath`Combinatorica`"] (* then *) Table[NumberOfTableaux[M[Star[n+1]]] - 2^(n - 1), {n, 26}] (* Robert G. Wilson v, Sep 27 2006 *)

Formula

a(n) = A000085(n) - A000079(n) for n>0.

Extensions

More terms from Robert G. Wilson v, Sep 27 2006

A132807 A000108(n) + A000079(n) - 1.

Original entry on oeis.org

1, 2, 5, 12, 29, 73, 195, 556, 1685, 5373
Offset: 0

Views

Author

Gary W. Adamson, Aug 31 2007

Keywords

Examples

			a(3) = 12 = C(3) + 2^3 - 1 = 5 + 8 = 1.
		

Crossrefs

Formula

a(n) = C(n) + 2^n - 1.

A221180 Erroneous version of A000079.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8092, 16184, 32368, 64736, 129472, 258944, 517888, 1035776, 2071552, 4143104, 8286208, 16572416, 33144832, 66289664, 132579328, 265158656, 530317312, 1060634624, 2121269248, 4242538496, 8485076992, 16970153984, 33940307968
Offset: 0

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Author

N. J. A. Sloane, Jan 13 2013

Keywords

Comments

Included in accordance with the OEIS policy of listing published but incorrect sequences. The entries for 2^13 on are incorrect.

Crossrefs

Cf. A000079.

A226643 Numbers n such that the binary XOR of the divisors of n (A178910) is a binary palindrome (A006995) and not a power of 2 (A000079).

Original entry on oeis.org

81, 162, 169, 324, 338, 648, 676, 1296, 1352, 2401, 2592, 2704, 3249, 4802, 5184, 5408, 6498, 9604, 10368, 10816, 12996, 19208, 20736, 21632, 25992, 38416, 41472, 43264, 51984, 76832, 82944, 86528, 103968, 112225, 153664, 165888, 173056, 194481
Offset: 1

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Author

Robert G. Wilson v, Aug 18 2013

Keywords

Comments

A takeoff of A227843.

Crossrefs

Programs

  • Mathematica
    f[n_] := Fold[ BitXor[#1, #2] &, 0, Divisors@ n]; palQ[n_Integer, base_Integer] := Module[{idn = IntegerDigits[n, base]}, idn == Reverse@ idn]; fQ[n_] := palQ[ f@ n, 2] && ! IntegerQ@ Log2@ n; Select[ Range@ 200000, fQ]

A230435 Triangle by rows, A001047 convolved with A000079.

Original entry on oeis.org

1, 2, 5, 4, 10, 19, 8, 20, 38, 65, 16, 40, 76, 130, 211, 32, 80, 152, 260, 422, 665, 64, 160, 304, 520, 844, 1330, 2059, 128, 320, 608, 1040, 1688, 2660, 4118, 6305, 256, 640, 1216, 2080, 3376, 5320, 8236, 12610, 19171
Offset: 0

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Author

Christopher Tompkins, Nov 18 2013

Keywords

Comments

Generated from Running Total of each row of A036561.
Left edge is A000079 (offset 0): (1, 2, 4, 8, 16, 32, 64, ...)
Right edge is A001047 (offset 1): (1, 5, 19, 65, 211, 665, ...)
Row sums are A066810 (offset 2): (1, 7, 33, 131, 473, 1611, ...)

Examples

			The start of the sequence as a triangle read by rows:
  1;
  2,  5;
  4,  10, 19;
  8,  20, 38,  65;
  16, 40, 76,  130, 211;
  32, 80, 152, 260, 422, 665;
  ...
		

Crossrefs

Programs

  • Mathematica
    T[n_,k_]:=Sum[3^j*2^(n-j),{j,0,k}];Flatten[Table[T[n,k],{n,0,8},{k,0,n}]] (* Detlef Meya, Dec 20 2023 *)

Formula

T(n,k) = Sum_{j=0..k} 3^j*2^(n-j). - Detlef Meya, Dec 20 2023
T(n,k) = 2^n*(3*(3/2)^k-2). - Alois P. Heinz, Dec 20 2023

A375476 a(3*n)=A001045(n+1), a(3*n+1)=A084214(n), a(3*n+2)=A000079(n) for n >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 3, 4, 4, 5, 6, 8, 11, 14, 16, 21, 26, 32, 43, 54, 64, 85, 106, 128, 171, 214, 256, 341, 426, 512, 683, 854, 1024, 1365, 1706, 2048, 2731, 3414, 4096, 5461, 6826, 8192, 10923, 13654, 16384, 21845, 27306, 32768, 43691, 54614, 65536
Offset: 0

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Author

Paul Curtz, Aug 18 2024

Keywords

Crossrefs

Formula

G.f. 1 + x - x^2*(1+x+x^2+x^3+2*x^4+3*x^5) / ( (1+x)*(2*x^3-1)*(x^2-x+1) ).

A375723 Square array read by antidiagonals, where the top row is the powers of 2 (A000079) and the other numbers are the sum of the neighbors in the preceding row.

Original entry on oeis.org

1, 2, 3, 4, 7, 10, 8, 14, 24, 34, 16, 28, 49, 83, 117, 32, 56, 98, 171, 288, 405, 64, 112, 196, 343, 597, 1002, 1407, 128, 224, 392, 686, 1200, 2085, 3492, 4899, 256, 448, 784, 1372, 2401, 4198, 7285, 12184, 17083, 512, 896, 1568, 2744, 4802, 8403, 14686, 25463, 42546, 59629
Offset: 0

Views

Author

WG Zeist, Aug 25 2024

Keywords

Comments

Each number in the top row of the array is determined by the pre-defined sequence (in this case, the powers of 2, A000079). Each number in lower rows is the sum of the numbers vertically or diagonally above it (so, the number at the left end of each row is the sum of two numbers, and all other numbers the sum of three). This is the same method as for constructing A217536, which has the top row be the nonnegative integers instead; other similar arrays are described in the comments of that sequence.
The main diagonal is the powers of 7 (A000420), and all numbers above or to the right of the main diagonal are multiples of powers of 2 and powers of 7. Specifically, the number in row m and column n, for n >= m, is 2^(n-m) * 7^m. Above the main diagonal, all numbers in the same column have the same final digit in base 10, and all numbers are 7/2 times the number immediately above.
More broadly, for any similarly constructed array with the powers of x as the top row, then the main diagonal will be the powers of (x^2 + x + 1) and the numbers above the main diagonal will be x^(n-m) * (x^2 + x + 1)^m (see also A062105, which can be interpreted as a similar array with the powers of 1 in the top row, and A020474 with the powers of 0).

Examples

			The array starts:
  1  2  4   8
  3  7  14  28
  10 24 49  98
  34 83 171 343
		

Crossrefs

The main diagonal gives A000420 (powers of 7). The first column gives A059738.

Formula

T(m+1,n) = sum(T(m,k), |k-n| <= 1) (and T(0,n)=2^n), m, n >= 0.

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

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
		

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  • 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.
  • Clifford A. Pickover, The Math Book, From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publ., NY, 2009, page 274.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 55-58, 255-260.
  • Alfred S. Posamentier and I. Lehmann, The Fabulous Fibonacci Numbers, Prometheus Books, Amherst, NY 2007.
  • Paulo Ribenboim, The New Book of Prime Number Records, Springer, 1996.
  • Paulo Ribenboim, My Numbers, My Friends: Popular Lectures on Number Theory, Springer-Verlag, NY, 2000, p. 3.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 45, 59.
  • J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, Princeton, NJ, 1978.
  • A. M. Robert, A Course in p-adic Analysis, Springer-Verlag, 2000; p. 213.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 288.
  • Manfred R. Schroeder, "Number Theory in Science and Communication", 5th ed., Springer-Verlag, 2009
  • L. E. Sigler, Fibonacci's Liber Abaci, Springer, 2003, pp. 404-405 and [26] on p. 627.
  • Simson, [No first name given], An explanation of an obscure passage in Albert Girard's Commentary ..., Phil. Trans. Royal Soc., 10 (1753), 430-433.
  • Parmanand Singh, The so-called fibonacci numbers in ancient and medieval India, Historia Mathematica, vol. 12 (1985), 229-244.
  • 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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  • D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, pp. 61-67, Penguin Books 1987.
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 53.
  • R. Witula, D. Slota, delta-Fibonacci Numbers, Appl. Anal. Discrete Math., 3 (2009), 310-329.

Crossrefs

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.
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

A000041 a(n) is the number of partitions of n (the partition numbers).

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 31185, 37338, 44583, 53174, 63261, 75175, 89134, 105558, 124754, 147273, 173525
Offset: 0

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Comments

Also number of nonnegative solutions to b + 2c + 3d + 4e + ... = n and the number of nonnegative solutions to 2c + 3d + 4e + ... <= n. - Henry Bottomley, Apr 17 2001
a(n) is also the number of conjugacy classes in the symmetric group S_n (and the number of irreducible representations of S_n).
Also the number of rooted trees with n+1 nodes and height at most 2.
Coincides with the sequence of numbers of nilpotent conjugacy classes in the Lie algebras gl(n). A006950, A015128 and this sequence together cover the nilpotent conjugacy classes in the classical A,B,C,D series of Lie algebras. - Alexander Elashvili, Sep 08 2003
Number of distinct Abelian groups of order p^n, where p is prime (the number is independent of p). - Lekraj Beedassy, Oct 16 2004
Number of graphs on n vertices that do not contain P3 as an induced subgraph. - Washington Bomfim, May 10 2005
Numbers of terms to be added when expanding the n-th derivative of 1/f(x). - Thomas Baruchel, Nov 07 2005
Sequence agrees with expansion of Molien series for symmetric group S_n up to the term in x^n. - Maurice D. Craig (towenaar(AT)optusnet.com.au), Oct 30 2006
Also the number of nonnegative integer solutions to x_1 + x_2 + x_3 + ... + x_n = n such that n >= x_1 >= x_2 >= x_3 >= ... >= x_n >= 0, because by letting y_k = x_k - x_(k+1) >= 0 (where 0 < k < n) we get y_1 + 2y_2 + 3y_3 + ... + (n-1)y_(n-1) + nx_n = n. - Werner Grundlingh (wgrundlingh(AT)gmail.com), Mar 14 2007
Let P(z) := Sum_{j>=0} b_j z^j, b_0 != 0. Then 1/P(z) = Sum_{j>=0} c_j z^j, where the c_j must be computed from the infinite triangular system b_0 c_0 = 1, b_0 c_1 + b_1 c_0 = 0 and so on (Cauchy products of the coefficients set to zero). The n-th partition number arises as the number of terms in the numerator of the expression for c_n: The coefficient c_n of the inverted power series is a fraction with b_0^(n+1) in the denominator and in its numerator having a(n) products of n coefficients b_i each. The partitions may be read off from the indices of the b_i. - Peter C. Heinig (algorithms(AT)gmx.de), Apr 09 2007
A sequence of positive integers p = p_1 ... p_k is a descending partition of the positive integer n if p_1 + ... + p_k = n and p_1 >= ... >= p_k. If formally needed p_j = 0 is appended to p for j > k. Let P_n denote the set of these partition for some n >= 1. Then a(n) = 1 + Sum_{p in P_n} floor((p_1-1)/(p_2+1)). (Cf. A000065, where the formula reduces to the sum.) Proof in Kelleher and O'Sullivan (2009). For example a(6) = 1 + 0 + 0 + 0 + 0 + 1 + 0 + 0 + 1 + 1 + 2 + 5 = 11. - Peter Luschny, Oct 24 2010
Let n = Sum( k_(p_m) p_m ) = k_1 + 2k_2 + 5k_5 + 7k_7 + ..., where p_m is the m-th generalized pentagonal number (A001318). Then a(n) is the sum over all such pentagonal partitions of n of (-1)^(k_5+k_7 + k_22 + ...) ( k_1 + k_2 + k_5 + ...)! /( k_1! k_2! k_5! ...), where the exponent of (-1) is the sum of all the k's corresponding to even-indexed GPN's. - Jerome Malenfant, Feb 14 2011
From Jerome Malenfant, Feb 14 2011: (Start)
The matrix of a(n) values
a(0)
a(1) a(0)
a(2) a(1) a(0)
a(3) a(2) a(1) a(0)
....
a(n) a(n-1) a(n-2) ... a(0)
is the inverse of the matrix
1
-1 1
-1 -1 1
0 -1 -1 1
....
-d_n -d_(n-1) -d_(n-2) ... -d_1 1
where d_q = (-1)^(m+1) if q = m(3m-1)/2 = the m-th generalized pentagonal number (A001318), = 0 otherwise. (End)
Let k > 0 be an integer, and let i_1, i_2, ..., i_k be distinct integers such that 1 <= i_1 < i_2 < ... < i_k. Then, equivalently, a(n) equals the number of partitions of N = n + i_1 + i_2 + ... + i_k in which each i_j (1 <= j <= k) appears as a part at least once. To see this, note that the partitions of N of this class must be in 1-to-1 correspondence with the partitions of n, since N - i_1 - i_2 - ... - i_k = n. - L. Edson Jeffery, Apr 16 2011
a(n) is the number of distinct degree sequences over all free trees having n + 2 nodes. Take a partition of the integer n, add 1 to each part and append as many 1's as needed so that the total is 2n + 2. Now we have a degree sequence of a tree with n + 2 nodes. Example: The partition 3 + 2 + 1 = 6 corresponds to the degree sequence {4, 3, 2, 1, 1, 1, 1, 1} of a tree with 8 vertices. - Geoffrey Critzer, Apr 16 2011
a(n) is number of distinct characteristic polynomials among n! of permutations matrices size n X n. - Artur Jasinski, Oct 24 2011
Conjecture: starting with offset 1 represents the numbers of ordered compositions of n using the signed (++--++...) terms of A001318 starting (1, 2, -5, -7, 12, 15, ...). - Gary W. Adamson, Apr 04 2013 (this is true by the pentagonal number theorem, Joerg Arndt, Apr 08 2013)
a(n) is also number of terms in expansion of the n-th derivative of log(f(x)). In Mathematica notation: Table[Length[Together[f[x]^n * D[Log[f[x]], {x, n}]]], {n, 1, 20}]. - Vaclav Kotesovec, Jun 21 2013
Conjecture: No a(n) has the form x^m with m > 1 and x > 1. - Zhi-Wei Sun, Dec 02 2013
Partitions of n that contain a part p are the partitions of n - p. Thus, number of partitions of m*n - r that include k*n as a part is A000041(h*n-r), where h = m - k >= 0, n >= 2, 0 <= r < n; see A111295 as an example. - Clark Kimberling, Mar 03 2014
a(n) is the number of compositions of n into positive parts avoiding the pattern [1, 2]. - Bob Selcoe, Jul 08 2014
Conjecture: For any j there exists k such that all primes p <= A000040(j) are factors of one or more a(n) <= a(k). Growth of this coverage is slow and irregular. k = 1067 covers the first 102 primes, thus slower than A000027. - Richard R. Forberg, Dec 08 2014
a(n) is the number of nilpotent conjugacy classes in the order-preserving, order-decreasing and (order-preserving and order-decreasing) injective transformation semigroups. - Ugbene Ifeanyichukwu, Jun 03 2015
Define a segmented partition a(n,k, ) to be a partition of n with exactly k parts, with s(j) parts t(j) identical to each other and distinct from all the other parts. Note that n >= k, j <= k, 0 <= s(j) <= k, s(1)t(1) + ... + s(j)t(j) = n and s(1) + ... + s(j) = k. Then there are up to a(k) segmented partitions of n with exactly k parts. - Gregory L. Simay, Nov 08 2015
(End)
From Gregory L. Simay, Nov 09 2015: (Start)
The polynomials for a(n, k, ) have degree j-1.
a(n, k, ) = 1 if n = 0 mod k, = 0 otherwise
a(rn, rk, ) = a(n, k, )
a(n odd, k, ) = 0
Established results can be recast in terms of segmented partitions:
For j(j+1)/2 <= n < (j+1)(j+2)/2, A000009(n) = a(n, 1, <1>) + ... + a(n, j, ), j < n
a(n, k, ) = a(n - j(j-1)/2, k)
(End)
a(10^20) was computed using the NIST Arb package. It has 11140086260 digits and its head and tail sections are 18381765...88091448. See the Johansson 2015 link. - Stanislav Sykora, Feb 01 2016
Satisfies Benford's law [Anderson-Rolen-Stoehr, 2011]. - N. J. A. Sloane, Feb 08 2017
The partition function p(n) is log-concave for all n>25 [DeSalvo-Pak, 2014]. - Michel Marcus, Apr 30 2019
a(n) is also the dimension of the n-th cohomology of the infinite real Grassmannian with coefficients in Z/2. - Luuk Stehouwer, Jun 06 2021
Number of equivalence relations on n unlabeled nodes. - Lorenzo Sauras Altuzarra, Jun 13 2022
Equivalently, number of idempotent mappings f from a set X of n elements into itself (i.e., satisfying f o f = f) up to permutation (i.e., f~f' :<=> There is a permutation sigma in Sym(X) such that f' o sigma = sigma o f). - Philip Turecek, Apr 17 2023
Conjecture: Each integer n > 2 different from 6 can be written as a sum of finitely many numbers of the form a(k) + 2 (k > 0) with no summand dividing another. This has been verified for n <= 7140. - Zhi-Wei Sun, May 16 2023
a(n) is also the number of partitions of n*(n+3)/2 into n distinct parts. - David García Herrero, Aug 20 2024
a(n) is also the number of non-isomorphic sigma algebras on {1,...,n}. A000110(n) counts all sigma algebras on {1,...,n}. Every sigma algebra on a finite set X is exactly the collection of all unions of its atoms (its minimal nonempty members), and those atoms partition X. An isomorphism of sigma algebras must map atoms to atoms, so the isomorphism class of a sigma algebra is determined by the multiset of its atom-sizes, which is an integer partition of n. - Matthew Azar, Jul 18 2025

Examples

			a(5) = 7 because there are seven partitions of 5, namely: {1, 1, 1, 1, 1}, {2, 1, 1, 1}, {2, 2, 1}, {3, 1, 1}, {3, 2}, {4, 1}, {5}. - _Bob Selcoe_, Jul 08 2014
G.f. = 1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 7*x^5 + 11*x^6 + 15*x^7 + 22*x^8 + ...
G.f. = 1/q + q^23 + 2*q^47 + 3*q^71 + 5*q^95 + 7*q^119 + 11*q^143 + 15*q^167 + ...
From _Gregory L. Simay_, Nov 08 2015: (Start)
There are up to a(4)=5 segmented partitions of the partitions of n with exactly 4 parts. They are a(n,4, <4>), a(n,4,<3,1>), a(n,4,<2,2>), a(n,4,<2,1,1>), a(n,4,<1,1,1,1>).
The partition 8,8,8,8 is counted in a(32,4,<4>).
The partition 9,9,9,5 is counted in a(32,4,<3,1>).
The partition 11,11,5,5 is counted in a(32,4,<2,2>).
The partition 13,13,5,1 is counted in a(32,4,<2,1,1>).
The partition 14,9,6,3 is counted in a(32,4,<1,1,1,1>).
a(n odd,4,<2,2>) = 0.
a(12, 6, <2,2,2>) = a(6,3,<1,1,1>) = a(6-3,3) = a(3,3) = 1. The lone partition is 3,3,2,2,1,1.
(End)
		

References

  • George E. Andrews, The Theory of Partitions, Addison-Wesley, Reading, Mass., 1976.
  • George E. Andrews and K. Ericksson, Integer Partitions, Cambridge University Press 2004.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 307.
  • R. Ayoub, An Introduction to the Analytic Theory of Numbers, Amer. Math. Soc., 1963; Chapter III.
  • Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem, Mathematics and Computer Education Journal, Vol. 31, No. 1, pp. 24-28, Winter 1997.
  • 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. Zentralblatt MATH, Zbl 1071.05501.
  • Bruce C. Berndt, Ramanujan's Notebooks Part V, Springer-Verlag.
  • B. C. Berndt, Number Theory in the Spirit of Ramanujan, Chap. I Amer. Math. Soc. Providence RI 2006.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 999.
  • J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 183.
  • Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 411.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 94-96.
  • L. E. Dickson, History of the Theory of Numbers, Vol.II Chapter III pp. 101-164, Chelsea NY 1992.
  • N. J. Fine, Basic Hypergeometric Series and Applications, Amer. Math. Soc., 1988; p. 37, Eq. (22.13).
  • H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958, p. 90.
  • G. H. Hardy and S. Ramanujan, Asymptotic formulas in combinatorial analysis, Proc. London Math. Soc., 17 (1918), 75-.
  • G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, Cambridge, University Press, 1940, pp. 83-100, 113-131.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers (Fifth edition), Oxford Univ. Press (Clarendon), 1979, 273-296.
  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.2.1.4, p. 396.
  • D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section XIV.1, p. 491.
  • S. Ramanujan, Collected Papers, Chap. 25, Cambridge Univ. Press 1927 (Proceedings of the Camb. Phil. Soc., 19 (1919), pp. 207-213).
  • S. Ramanujan, Collected Papers, Chap. 28, Cambridge Univ. Press 1927 (Proceedings of the London Math. Soc., 2, 18(1920)).
  • S. Ramanujan, Collected Papers, Chap. 30, Cambridge Univ. Press 1927 (Mathematische Zeitschrift, 9 (1921), pp. 147-163).
  • S. Ramanujan, Collected Papers, Ed. G. H. Hardy et al., Cambridge 1927; Chelsea, NY, 1962. See Table IV on page 308.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 122.
  • J. E. Roberts, Lure of the Integers, pp. 168-9 MAA 1992.
  • 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).
  • R. E. Tapscott and D. Marcovich, "Enumeration of Permutational Isomers: The Porphyrins", Journal of Chemical Education, 55 (1978), 446-447.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 286-289, 297-298, 303.
  • Robert M. Young, "Excursions in Calculus", Mathematical Association of America, p. 367.

Crossrefs

Partial sums give A000070.
For successive differences see A002865, A053445, A072380, A081094, A081095.
Antidiagonal sums of triangle A092905. a(n) = A054225(n,0).
Boustrophedon transforms: A000733, A000751.
Cf. A167376 (complement), A061260 (multisets), A000700 (self-conjug), A330644 (not self-conj).

Programs

  • GAP
    List([1..10],n->Size(OrbitsDomain(SymmetricGroup(IsPermGroup,n),SymmetricGroup(IsPermGroup,n),\^))); # Attila Egri-Nagy, Aug 15 2014
    
  • Haskell
    import Data.MemoCombinators (memo2, integral)
    a000041 n = a000041_list !! n
    a000041_list = map (p' 1) [0..] where
       p' = memo2 integral integral p
       p _ 0 = 1
       p k m = if m < k then 0 else p' k (m - k) + p' (k + 1) m
    -- Reinhard Zumkeller, Nov 03 2015, Nov 04 2013
    
  • Julia
    # DedekindEta is defined in A000594
    A000041List(len) = DedekindEta(len, -1)
    A000041List(50) |> println # Peter Luschny, Mar 09 2018
  • Magma
    a:= func< n | NumberOfPartitions(n) >; [ a(n) : n in [0..10]];
    
  • Maple
    A000041 := n -> combinat:-numbpart(n): [seq(A000041(n), n=0..50)]; # Warning: Maple 10 and 11 give incorrect answers in some cases: A110375.
    spec := [B, {B=Set(Set(Z,card>=1))}, unlabeled ];
    [seq(combstruct[count](spec, size=n), n=0..50)];
    with(combstruct):ZL0:=[S,{S=Set(Cycle(Z,card>0))}, unlabeled]: seq(count(ZL0,size=n),n=0..45); # Zerinvary Lajos, Sep 24 2007
    G:={P=Set(Set(Atom,card>0))}: combstruct[gfsolve](G,labeled,x); seq(combstruct[count]([P,G,unlabeled],size=i),i=0..45); # Zerinvary Lajos, Dec 16 2007
    # Using the function EULER from Transforms (see link at the bottom of the page).
    1,op(EULER([seq(1,n=1..49)])); # Peter Luschny, Aug 19 2020
  • Mathematica
    Table[ PartitionsP[n], {n, 0, 45}]
    a[ n_] := SeriesCoefficient[ q^(1/24) / DedekindEta[ Log[q] / (2 Pi I)], {q, 0, n}]; (* Michael Somos, Jul 11 2011 *)
    a[ n_] := SeriesCoefficient[ 1 / Product[ 1 - x^k, {k, n}], {x, 0, n}]; (* Michael Somos, Jul 11 2011 *)
    CoefficientList[1/QPochhammer[q] + O[q]^100, q] (* Jean-François Alcover, Nov 25 2015 *)
    a[0] := 1; a[n_] := a[n] = Block[{k=1, s=0, i=n-1}, While[i >= 0, s=s-(-1)^k (a[i]+a[i-k]); k=k+1; i=i-(3 k-2)]; s]; Map[a, Range[0, 49]] (* Oliver Seipel, Jun 01 2024 after Euler *)
  • Maxima
    num_partitions(60,list); /* Emanuele Munarini, Feb 24 2014 */
    
  • MuPAD
    combinat::partitions::count(i) $i=0..54 // Zerinvary Lajos, Apr 16 2007
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( 1 / eta(x + x * O(x^n)), n))};
    
  • PARI
    /* The Hardy-Ramanujan-Rademacher exact formula in PARI is as follows (this is no longer necessary since it is now built in to the numbpart command): */
    Psi(n, q) = local(a, b, c); a=sqrt(2/3)*Pi/q; b=n-1/24; c=sqrt(b); (sqrt(q)/(2*sqrt(2)*b*Pi))*(a*cosh(a*c)-(sinh(a*c)/c))
    L(n, q) = if(q==1,1,sum(h=1,q-1,if(gcd(h,q)>1,0,cos((g(h,q)-2*h*n)*Pi/q))))
    g(h, q) = if(q<3,0,sum(k=1,q-1,k*(frac(h*k/q)-1/2)))
    part(n) = round(sum(q=1,max(5,0.5*sqrt(n)),L(n,q)*Psi(n,q)))
    /* Ralf Stephan, Nov 30 2002, fixed by Vaclav Kotesovec, Apr 09 2018 */
    
  • PARI
    {a(n) = numbpart(n)};
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( sum( k=1, sqrtint(n), x^k^2 / prod( i=1, k, 1 - x^i, 1 + x * O(x^n))^2, 1), n))};
    
  • PARI
    f(n)= my(v,i,k,s,t);v=vector(n,k,0);v[n]=2;t=0;while(v[1]1,i--;s+=i*(v[i]=(n-s)\i));t++);t \\ Thomas Baruchel, Nov 07 2005
    
  • PARI
    a(n)=if(n<0, 0, polcoeff(exp(sum(k=1, n, x^k/(1-x^k)/k, x*O(x^n))), n)) \\ Joerg Arndt, Apr 16 2010
    
  • Perl
    use ntheory ":all"; my @p = map { partitions($) } 0..100; say "[@p]"; # _Dana Jacobsen, Sep 06 2015
    
  • Python
    from sympy.functions.combinatorial.numbers import partition
    print([partition(i) for i in range(101)]) # Joan Ludevid, May 25 2025
    
  • Racket
    #lang racket
    ; SUM(k,-inf,+inf) (-1)^k p(n-k(3k-1)/2)
    ; For k outside the range (1-(sqrt(1-24n))/6 to (1+sqrt(1-24n))/6) argument n-k(3k-1)/2 < 0.
    ; Therefore the loops below are finite. The hash avoids repeated identical computations.
    (define (p n) ; Nr of partitions of n.
    (hash-ref h n
      (λ ()
       (define r
        (+
         (let loop ((k 1) (n (sub1 n)) (s 0))
          (if (< n 0) s
           (loop (add1 k) (- n (* 3 k) 1) (if (odd? k) (+ s (p n)) (- s (p n))))))
         (let loop ((k -1) (n (- n 2)) (s 0))
          (if (< n 0) s
           (loop (sub1 k) (+ n (* 3 k) -2) (if (odd? k) (+ s (p n)) (- s (p n))))))))
       (hash-set! h n r)
       r)))
    (define h (make-hash '((0 . 1))))
    ; (for ((k (in-range 0 50))) (printf "~s, " (p k))) runs in a moment.
    ; Jos Koot, Jun 01 2016
    
  • Sage
    [number_of_partitions(n) for n in range(46)]  # Zerinvary Lajos, May 24 2009
    
  • Sage
    @CachedFunction
    def A000041(n):
        if n == 0: return 1
        S = 0; J = n-1; k = 2
        while 0 <= J:
            T = A000041(J)
            S = S+T if is_odd(k//2) else S-T
            J -= k if is_odd(k) else k//2
            k += 1
        return S
    [A000041(n) for n in range(50)]  # Peter Luschny, Oct 13 2012
    
  • Sage
    # uses[EulerTransform from A166861]
    a = BinaryRecurrenceSequence(1, 0)
    b = EulerTransform(a)
    print([b(n) for n in range(50)]) # Peter Luschny, Nov 11 2020
    

Formula

G.f.: Product_{k>0} 1/(1-x^k) = Sum_{k>= 0} x^k Product_{i = 1..k} 1/(1-x^i) = 1 + Sum_{k>0} x^(k^2)/(Product_{i = 1..k} (1-x^i))^2.
G.f.: 1 + Sum_{n>=1} x^n/(Product_{k>=n} 1-x^k). - Joerg Arndt, Jan 29 2011
a(n) - a(n-1) - a(n-2) + a(n-5) + a(n-7) - a(n-12) - a(n-15) + ... = 0, where the sum is over n-k and k is a generalized pentagonal number (A001318) <= n and the sign of the k-th term is (-1)^([(k+1)/2]). See A001318 for a good way to remember this!
a(n) = (1/n) * Sum_{k=0..n-1} sigma(n-k)*a(k), where sigma(k) is the sum of divisors of k (A000203).
a(n) ~ 1/(4*n*sqrt(3)) * e^(Pi * sqrt(2n/3)) as n -> infinity (Hardy and Ramanujan). See A050811.
a(n) = a(0)*b(n) + a(1)*b(n-2) + a(2)*b(n-4) + ... where b = A000009.
From Jon E. Schoenfield, Aug 17 2014: (Start)
It appears that the above approximation from Hardy and Ramanujan can be refined as
a(n) ~ 1/(4*n*sqrt(3)) * e^(Pi * sqrt(2n/3 + c0 + c1/n^(1/2) + c2/n + c3/n^(3/2) + c4/n^2 + ...)), where the coefficients c0 through c4 are approximately
c0 = -0.230420145062453320665537
c1 = -0.0178416569128570889793
c2 = 0.0051329911273
c3 = -0.0011129404
c4 = 0.0009573,
as n -> infinity. (End)
From Vaclav Kotesovec, May 29 2016 (c4 added Nov 07 2016): (Start)
c0 = -0.230420145062453320665536704197233... = -1/36 - 2/Pi^2
c1 = -0.017841656912857088979502135349949... = 1/(6*sqrt(6)*Pi) - sqrt(3/2)/Pi^3
c2 = 0.005132991127342167594576391633559... = 1/(2*Pi^4)
c3 = -0.001112940489559760908236602843497... = 3*sqrt(3/2)/(4*Pi^5) - 5/(16*sqrt(6)*Pi^3)
c4 = 0.000957343284806972958968694349196... = 1/(576*Pi^2) - 1/(24*Pi^4) + 93/(80*Pi^6)
a(n) ~ exp(Pi*sqrt(2*n/3))/(4*sqrt(3)*n) * (1 - (sqrt(3/2)/Pi + Pi/(24*sqrt(6)))/sqrt(n) + (1/16 + Pi^2/6912)/n).
a(n) ~ exp(Pi*sqrt(2*n/3) - (sqrt(3/2)/Pi + Pi/(24*sqrt(6)))/sqrt(n) + (1/24 - 3/(4*Pi^2))/n) / (4*sqrt(3)*n).
(End)
a(n) < exp( (2/3)^(1/2) Pi sqrt(n) ) (Ayoub, p. 197).
G.f.: Product_{m>=1} (1+x^m)^A001511(m). - Vladeta Jovovic, Mar 26 2004
a(n) = Sum_{i=0..n-1} P(i, n-i), where P(x, y) is the number of partitions of x into at most y parts and P(0, y)=1. - Jon Perry, Jun 16 2003
G.f.: Product_{i>=1} Product_{j>=0} (1+x^((2i-1)*2^j))^(j+1). - Jon Perry, Jun 06 2004
G.f. e^(Sum_{k>0} (x^k/(1-x^k)/k)). - Franklin T. Adams-Watters, Feb 08 2006
a(n) = A114099(9*n). - Reinhard Zumkeller, Feb 15 2006
Euler transform of all 1's sequence (A000012). Weighout transform of A001511. - Franklin T. Adams-Watters, Mar 15 2006
a(n) = A027187(n) + A027193(n) = A000701(n) + A046682(n). - Reinhard Zumkeller, Apr 22 2006
A026820(a(n),n) = A134737(n) for n > 0. - Reinhard Zumkeller, Nov 07 2007
Convolved with A152537 gives A000079, powers of 2. - Gary W. Adamson, Dec 06 2008
a(n) = A026820(n, n); a(n) = A108949(n) + A045931(n) + A108950(n) = A130780(n) + A171966(n) - A045931(n) = A045931(n) + A171967(n). - Reinhard Zumkeller, Jan 21 2010
a(n) = Tr(n)/(24*n-1) = A183011(n)/A183010(n), n>=1. See the Bruinier-Ono paper in the Links. - Omar E. Pol, Jan 23 2011
From Jerome Malenfant, Feb 14 2011: (Start)
a(n) = determinant of the n X n Toeplitz matrix:
1 -1
1 1 -1
0 1 1 -1
0 0 1 1 -1
-1 0 0 1 1 -1
. . .
d_n d_(n-1) d_(n-2)...1
where d_q = (-1)^(m+1) if q = m(3m-1)/2 = p_m, the m-th generalized pentagonal number (A001318), otherwise d_q = 0. Note that the 1's run along the diagonal and the -1's are on the superdiagonal. The (n-1) row (not written) would end with ... 1 -1. (End)
Empirical: let F*(x) = Sum_{n=0..infinity} p(n)*exp(-Pi*x*(n+1)), then F*(2/5) = 1/sqrt(5) to a precision of 13 digits.
F*(4/5) = 1/2+3/2/sqrt(5)-sqrt(1/2*(1+3/sqrt(5))) to a precision of 28 digits. These are the only values found for a/b when a/b is from F60, Farey fractions up to 60. The number for F*(4/5) is one of the real roots of 25*x^4 - 50*x^3 - 10*x^2 - 10*x + 1. Note here the exponent (n+1) compared to the standard notation with n starting at 0. - Simon Plouffe, Feb 23 2011
The constant (2^(7/8)*GAMMA(3/4))/(exp(Pi/6)*Pi^(1/4)) = 1.0000034873... when expanded in base exp(4*Pi) will give the first 52 terms of a(n), n>0, the precision needed is 300 decimal digits. - Simon Plouffe, Mar 02 2011
a(n) = A035363(2n). - Omar E. Pol, Nov 20 2009
G.f.: A(x)=1+x/(G(0)-x); G(k) = 1 + x - x^(k+1) - x*(1-x^(k+1))/G(k+1); (continued fraction Euler's kind, 1-step ). - Sergei N. Gladkovskii, Jan 25 2012
Convolution of A010815 with A000712. - Gary W. Adamson, Jul 20 2012
G.f.: 1 + x*(1 - G(0))/(1-x) where G(k) = 1 - 1/(1-x^(k+1))/(1-x/(x-1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 22 2013
G.f.: Q(0) where Q(k) = 1 + x^(4*k+1)/( (x^(2*k+1)-1)^2 - x^(4*k+3)*(x^(2*k+1)-1)^2/( x^(4*k+3) + (x^(2*k+2)-1)^2/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 16 2013
a(n) = 24*spt(n) + 12*N_2(n) - Tr(n) = 24*A092269(n) + 12*A220908(n) - A183011(n), n >= 1. - Omar E. Pol, Feb 17 2013
a(n) = A066186(n)/n, n >= 1. - Omar E. Pol, Aug 16 2013
From Peter Bala, Dec 23 2013: (Start)
a(n-1) = Sum_{parts k in all partitions of n} mu(k), where mu(k) is the arithmetical Möbius function (see A008683).
Let P(2,n) denote the set of partitions of n into parts k >= 2. Then a(n-2) = -Sum_{parts k in all partitions in P(2,n)} mu(k).
n*( a(n) - a(n-1) ) = Sum_{parts k in all partitions in P(2,n)} k (see A138880).
Let P(3,n) denote the set of partitions of n into parts k >= 3. Then
a(n-3) = (1/2)*Sum_{parts k in all partitions in P(3,n)} phi(k), where phi(k) is the Euler totient function (see A000010). Using this result and Mertens's theorem on the average order of the phi function, we can find an approximate 3-term recurrence for the partition function: a(n) ~ a(n-1) + a(n-2) + (Pi^2/(3*n) - 1)*a(n-3). For example, substituting the values a(47) = 124754, a(48) = 147273 and a(49) = 173525 into the recurrence gives the approximation a(50) ~ 204252.48... compared with the true value a(50) = 204226. (End)
a(n) = Sum_{k=1..n+1} (-1)^(n+1-k)*A000203(k)*A002040(n+1-k). - Mircea Merca, Feb 27 2014
a(n) = A240690(n) + A240690(n+1), n >= 1. - Omar E. Pol, Mar 16 2015
From Gary W. Adamson, Jun 22 2015: (Start)
A production matrix for the sequence with offset 1 is M, an infinite n x n matrix of the following form:
a, 1, 0, 0, 0, 0, ...
b, 0, 1, 0, 0, 0, ...
c, 0, 0, 1, 0, 0, ...
d, 0, 0, 0, 1, 0, ...
.
.
... such that (a, b, c, d, ...) is the signed version of A080995 with offset 1: (1,1,0,0,-1,0,-1,...)
and a(n) is the upper left term of M^n.
This operation is equivalent to the g.f. (1 + x + 2x^2 + 3x^3 + 5x^4 + ...) = 1/(1 - x - x^2 + x^5 + x^7 - x^12 - x^15 + x^22 + ...). (End)
G.f.: x^(1/24)/eta(log(x)/(2 Pi i)). - Thomas Baruchel, Jan 09 2016, after Michael Somos (after Richard Dedekind).
a(n) = Sum_{k=-inf..+inf} (-1)^k a(n-k(3k-1)/2) with a(0)=1 and a(negative)=0. The sum can be restricted to the (finite) range from k = (1-sqrt(1-24n))/6 to (1+sqrt(1-24n))/6, since all terms outside this range are zero. - Jos Koot, Jun 01 2016
G.f.: (conjecture) (r(x) * r(x^2) * r(x^4) * r(x^8) * ...) where r(x) is A000009: (1, 1, 1, 2, 2, 3, 4, ...). - Gary W. Adamson, Sep 18 2016; Doron Zeilberger observed today that "This follows immediately from Euler's formula 1/(1-z) = (1+z)*(1+z^2)*(1+z^4)*(1+z^8)*..." Gary W. Adamson, Sep 20 2016
a(n) ~ 2*Pi * BesselI(3/2, sqrt(24*n-1)*Pi/6) / (24*n-1)^(3/4). - Vaclav Kotesovec, Jan 11 2017
G.f.: Product_{k>=1} (1 + x^k)/(1 - x^(2*k)). - Ilya Gutkovskiy, Jan 23 2018
a(n) = p(1, n) where p(k, n) = p(k+1, n) + p(k, n-k) if k < n, 1 if k = n, and 0 if k > n. p(k, n) is the number of partitions of n into parts >= k. - Lorraine Lee, Jan 28 2020
Sum_{n>=1} 1/a(n) = A078506. - Amiram Eldar, Nov 01 2020
Sum_{n>=0} a(n)/2^n = A065446. - Amiram Eldar, Jan 19 2021
From Simon Plouffe, Mar 12 2021: (Start)
Sum_{n>=0} a(n)/exp(Pi*n) = 2^(3/8)*Gamma(3/4)/(Pi^(1/4)*exp(Pi/24)).
Sum_{n>=0} a(n)/exp(2*Pi*n) = 2^(1/2)*Gamma(3/4)/(Pi^(1/4)*exp(Pi/12)).
[corrected by Vaclav Kotesovec, May 12 2023] (End)
[These are the reciprocals of phi(exp(-Pi)) (A259148) and phi(exp(-2*Pi)) (A259149), where phi(q) is the Euler modular function. See B. C. Berndt (RLN, Vol. V, p. 326), and formulas (13) and (14) in I. Mező, 2013. - Peter Luschny, Mar 13 2021]
a(n) = A000009(n) + A035363(n) + A006477(n). - R. J. Mathar, Feb 01 2022
a(n) = A008284(2*n,n) is also the number of partitions of 2n into n parts. - Ryan Brooks, Jun 11 2022
a(n) = A000700(n) + A330644(n). - R. J. Mathar, Jun 15 2022
a(n) ~ exp(Pi*sqrt(2*n/3)) / (4*n*sqrt(3)) * (1 + Sum_{r>=1} w(r)/n^(r/2)), where w(r) = 1/(-4*sqrt(6))^r * Sum_{k=0..(r+1)/2} binomial(r+1,k) * (r+1-k) / (r+1-2*k)! * (Pi/6)^(r-2*k) [Cormac O'Sullivan, 2023, pp. 2-3]. - Vaclav Kotesovec, Mar 15 2023

Extensions

Additional comments from Ola Veshta (olaveshta(AT)my-deja.com), Feb 28 2001
Additional comments from Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001

A000012 The simplest sequence of positive numbers: the all 1's sequence.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

N. J. A. Sloane, May 16 1994

Keywords

Comments

Number of ways of writing n as a product of primes.
Number of ways of writing n as a sum of distinct powers of 2.
Continued fraction for golden ratio A001622.
Partial sums of A000007 (characteristic function of 0). - Jeremy Gardiner, Sep 08 2002
An example of an infinite sequence of positive integers whose distinct pairwise concatenations are all primes! - Don Reble, Apr 17 2005
Binomial transform of A000007; inverse binomial transform of A000079. - Philippe Deléham, Jul 07 2005
A063524(a(n)) = 1. - Reinhard Zumkeller, Oct 11 2008
For n >= 0, let M(n) be the matrix with first row = (n n+1) and 2nd row = (n+1 n+2). Then a(n) = absolute value of det(M(n)). - K.V.Iyer, Apr 11 2009
The partial sums give the natural numbers (A000027). - Daniel Forgues, May 08 2009
From Enrique Pérez Herrero, Sep 04 2009: (Start)
a(n) is also tau_1(n) where tau_2(n) is A000005.
a(n) is a completely multiplicative arithmetical function.
a(n) is both squarefree and a perfect square. See A005117 and A000290. (End)
Also smallest divisor of n. - Juri-Stepan Gerasimov, Sep 07 2009
Also decimal expansion of 1/9. - Enrique Pérez Herrero, Sep 18 2009; corrected by Klaus Brockhaus, Apr 02 2010
a(n) is also the number of complete graphs on n nodes. - Pablo Chavez (pchavez(AT)cmu.edu), Sep 15 2009
Totally multiplicative sequence with a(p) = 1 for prime p. Totally multiplicative sequence with a(p) = a(p-1) for prime p. - Jaroslav Krizek, Oct 18 2009
n-th prime minus phi(prime(n)); number of divisors of n-th prime minus number of perfect partitions of n-th prime; the number of perfect partitions of n-th prime number; the number of perfect partitions of n-th noncomposite number. - Juri-Stepan Gerasimov, Oct 26 2009
For all n>0, the sequence of limit values for a(n) = n!*Sum_{k>=n} k/(k+1)!. Also, a(n) = n^0. - Harlan J. Brothers, Nov 01 2009
a(n) is also the number of 0-regular graphs on n vertices. - Jason Kimberley, Nov 07 2009
Differences between consecutive n. - Juri-Stepan Gerasimov, Dec 05 2009
From Matthew Vandermast, Oct 31 2010: (Start)
1) When sequence is read as a regular triangular array, T(n,k) is the coefficient of the k-th power in the expansion of (x^(n+1)-1)/(x-1).
2) Sequence can also be read as a uninomial array with rows of length 1, analogous to arrays of binomial, trinomial, etc., coefficients. In a q-nomial array, T(n,k) is the coefficient of the k-th power in the expansion of ((x^q -1)/(x-1))^n, and row n has a sum of q^n and a length of (q-1)*n + 1. (End)
The number of maximal self-avoiding walks from the NW to SW corners of a 2 X n grid.
When considered as a rectangular array, A000012 is a member of the chain of accumulation arrays that includes the multiplication table A003991 of the positive integers. The chain is ... < A185906 < A000007 < A000012 < A003991 < A098358 < A185904 < A185905 < ... (See A144112 for the definition of accumulation array.) - Clark Kimberling, Feb 06 2011
a(n) = A007310(n+1) (Modd 3) := A193680(A007310(n+1)), n>=0. For general Modd n (not to be confused with mod n) see a comment on A203571. The nonnegative members of the three residue classes Modd 3, called [0], [1], and [2], are shown in the array A088520, if there the third row is taken as class [0] after inclusion of 0. - Wolfdieter Lang, Feb 09 2012
Let M = Pascal's triangle without 1's (A014410) and V = a variant of the Bernoulli numbers A027641 but starting [1/2, 1/6, 0, -1/30, ...]. Then M*V = [1, 1, 1, 1, ...]. - Gary W. Adamson, Mar 05 2012
As a lower triangular array, T is an example of the fundamental generalized factorial matrices of A133314. Multiplying each n-th diagonal by t^n gives M(t) = I/(I-t*S) = I + t*S + (t*S)^2 + ... where S is the shift operator A129184, and T = M(1). The inverse of M(t) is obtained by multiplying the first subdiagonal of T by -t and the other subdiagonals by zero, so A167374 is the inverse of T. Multiplying by t^n/n! gives exp(t*S) with inverse exp(-t*S). - Tom Copeland, Nov 10 2012
The original definition of the meter was one ten-millionth of the distance from the Earth's equator to the North Pole. According to that historical definition, the length of one degree of latitude, that is, 60 nautical miles, would be exactly 111111.111... meters. - Jean-François Alcover, Jun 02 2013
Deficiency of 2^n. - Omar E. Pol, Jan 30 2014
Consider n >= 1 nonintersecting spheres each with surface area S. Define point p on sphere S_i to be a "public point" if and only if there exists a point q on sphere S_j, j != i, such that line segment pq INTERSECT S_i = {p} and pq INTERSECT S_j = {q}; otherwise, p is a "private point". The total surface area composed of exactly all private points on all n spheres is a(n)*S = S. ("The Private Planets Problem" in Zeitz.) - Rick L. Shepherd, May 29 2014
For n>0, digital roots of centered 9-gonal numbers (A060544). - Colin Barker, Jan 30 2015
Product of nonzero digits in base-2 representation of n. - Franklin T. Adams-Watters, May 16 2016
Alternating row sums of triangle A104684. - Wolfdieter Lang, Sep 11 2016
A fixed point of the run length transform. - Chai Wah Wu, Oct 21 2016
Length of period of continued fraction for sqrt(A002522) or sqrt(A002496). - A.H.M. Smeets, Oct 10 2017
a(n) is also the determinant of the (n+1) X (n+1) matrix M defined by M(i,j) = binomial(i,j) for 0 <= i,j <= n, since M is a lower triangular matrix with main diagonal all 1's. - Jianing Song, Jul 17 2018
a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = min(i,j) for 1 <= i,j <= n (see Xavier Merlin reference). - Bernard Schott, Dec 05 2018
a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = tau(gcd(i,j)) for 1 <= i,j <= n (see De Koninck & Mercier reference). - Bernard Schott, Dec 08 2020

Examples

			1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + ...)))) = A001622.
1/9 = 0.11111111111111...
From _Wolfdieter Lang_, Feb 09 2012: (Start)
Modd 7 for nonnegative odd numbers not divisible by 3:
A007310: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, ...
Modd 3:  1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, ...
(End)
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 186.
  • J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 692 pp. 90 and 297, Ellipses, Paris, 2004.
  • Xavier Merlin, Méthodix Algèbre, Exercice 1-a), page 153, Ellipses, Paris, 1995.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 277, 284.
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.
  • Paul Zeitz, The Art and Craft of Mathematical Problem Solving, The Great Courses, The Teaching Company, 2010 (DVDs and Course Guidebook, Lecture 6: "Pictures, Recasting, and Points of View", pp. 32-34).

Crossrefs

Programs

  • Haskell
    a000012 = const 1
    a000012_list = repeat 1 -- Reinhard Zumkeller, May 07 2012
    
  • Magma
    [1 : n in [0..100]];
    
  • Maple
    seq(1, i=0..150);
  • Mathematica
    Array[1 &, 50] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
  • Maxima
    makelist(1, n, 1, 30); /* Martin Ettl, Nov 07 2012 */
    
  • PARI
    {a(n) = 1};
    
  • Python
    print([1 for n in range(90)]) # Michael S. Branicky, Apr 04 2022

Formula

a(n) = 1.
G.f.: 1/(1-x).
E.g.f.: exp(x).
G.f.: Product_{k>=0} (1 + x^(2^k)). - Zak Seidov, Apr 06 2007
Completely multiplicative with a(p^e) = 1.
Regarded as a square array by antidiagonals, g.f. 1/((1-x)(1-y)), e.g.f. Sum T(n,m) x^n/n! y^m/m! = e^{x+y}, e.g.f. Sum T(n,m) x^n y^m/m! = e^y/(1-x). Regarded as a triangular array, g.f. 1/((1-x)(1-xy)), e.g.f. Sum T(n,m) x^n y^m/m! = e^{xy}/(1-x). - Franklin T. Adams-Watters, Feb 06 2006
Dirichlet g.f.: zeta(s). - Ilya Gutkovskiy, Aug 31 2016
a(n) = Sum_{l=1..n} (-1)^(l+1)*2*cos(Pi*l/(2*n+1)) = 1 identically in n >= 1 (for n=0 one has 0 from the undefined sum). From the Jolley reference, (429) p. 80. Interpretation: consider the n segments between x=0 and the n positive zeros of the Chebyshev polynomials S(2*n, x) (see A049310). Then the sum of the lengths of every other segment starting with the one ending in the largest zero (going from the right to the left) is 1. - Wolfdieter Lang, Sep 01 2016
As a lower triangular matrix, T = M*T^(-1)*M = M*A167374*M, where M(n,k) = (-1)^n A130595(n,k). Note that M = M^(-1). Cf. A118800 and A097805. - Tom Copeland, Nov 15 2016
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