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 A000045 Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1. (Formerly M0692 N0256) 4654

%I M0692 N0256

%S 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,

%T 10946,17711,28657,46368,75025,121393,196418,317811,514229,832040,

%U 1346269,2178309,3524578,5702887,9227465,14930352,24157817,39088169,63245986,102334155

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

%C Also sometimes called Lamé's sequence.

%C F(n+2) = number of binary sequences of length n that have no consecutive 0's.

%C F(n+2) = number of subsets of {1,2,...,n} that contain no consecutive integers.

%C F(n+1) = number of tilings of a 2 X n rectangle by 2 X 1 dominoes.

%C 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

%C F(n) = number of compositions of n+1 with no part equal to 1. [Cayley, Grimaldi]

%C 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).

%C For Fibonacci search see Knuth, Vol. 3; Horowitz and Sahni; etc.

%C F(n) is the diagonal sum of the entries in Pascal's triangle at 45 degrees slope. - _Amarnath Murthy_, Dec 29 2001

%C 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

%C F(n+1) = number of (3412,132)-, (3412,213)- and (3412,321)-avoiding involutions in S_n.

%C This is also the Horadam sequence (0,1,1,1). - _Ross La Haye_, Aug 18 2003

%C 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

%C Number of meaningful differential operations of the k-th order on the space R^3. - _Branko Malesevic_, Mar 02 2004

%C 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.

%C 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

%C 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

%C 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]

%C 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

%C 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

%C 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]

%C 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

%C 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

%C Lengths of successive words (starting with a) under the substitution: {a -> ab, b -> a}. - _Jeroen F.J. Laros_, Jan 22 2005

%C 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)

%C a(n) = Sum_{k=0..n} abs(A108299(n, k)). - _Reinhard Zumkeller_, Jun 01 2005

%C a(n) = A001222(A000304(n)).

%C F(n+2) = Sum_{k=0..n} binomial(floor((n+k)/2),k), row sums of A046854. - _Paul Barry_, Mar 11 2003

%C Number of order ideals of the "zig-zag" poset. See vol. 1, ch. 3, prob. 23 of Stanley. - _Mitch Harris_, Dec 27 2005

%C 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

%C 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

%C 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

%C Number of Kekulé structures in polyphenanthrenes. See the paper by Lukovits and Janezic for details. - _Parthasarathy Nambi_, Aug 22 2006

%C 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

%C 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 super-diagonal 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

%C 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

%C 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

%C F(n+2) = The number of Khalimsky-continuous functions with a two-point codomain. - Shiva Samieinia (shiva(AT)math.su.se), Oct 04 2007

%C From Kauffman and Lopes, Proposition 8.2, p. 21: "The sequence of the determinants of the Fibonacci sequence of rational knots is the Fibonacci sequence (of numbers)." - _Jonathan Vos Post_, Oct 26 2007

%C This is a_1(n) in the Doroslovacki reference.

%C Let phi = (sqrt(5)+1)/2 = 1.6180339...; then phi^n = (1/phi)*a(n) + a(n+1). Example: phi^4 = 6.8541019... = (0.6180339...)*3 + 5. Also phi = 1/1 + 1/2 + 1/(2*5) + 1/(5*13) + 1/(13*34) + 1/(34*89) + ... - _Gary W. Adamson_, Dec 15 2007

%C 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

%C a(n) = the number of different ways to run up a staircase with n steps, taking steps of odd sizes where the order is relevant and there is no other restriction on the number or the size of each step taken. - _Mohammad K. Azarian_, May 21 2008

%C Equals row sums of triangle A144152. - _Gary W. Adamson_, Sep 12 2008

%C Except for the initial term, the numerator of the convergents to the recursion x = 1/(x+1). - _Cino Hilliard_, Sep 15 2008

%C 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

%C 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

%C Fp == 5^((p-1)/2) mod p, p = prime [Schroeder, p. 90]. - _Gary W. Adamson_ & _Alexander R. Povolotsky_, Feb 21 2009

%C (Ln)^2 - 5*(Fn)^2 = 4*(-1)^n. Example: 11^2 - 5*5 = -4. - _Gary W. Adamson_, Mar 11 2009

%C 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 (smbelcas(AT)toroidalsnark.net), Jul 04 2009

%C (F(n),F(n+4)) satisfies the Diophantine equation: X^2 + Y^2 - 7XY = 9*(-1)^n. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 06 2009

%C (F(n),F(n+2)) satisfies the Diophantine equation: X^2 + Y^2 - 3XY = (-1)^n. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 08 2009

%C a(n+2) = A083662(A131577(n)). - _Reinhard Zumkeller_, Sep 26 2009

%C 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

%C 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

%C (F(n-1) + F(n+1))^2 - 5F(n-2)*F(n+2) = 9*(-1)^n. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Mar 31 2010

%C 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

%C 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

%C 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

%C As n-> inf., (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

%C From _Hieronymus Fischer_, Oct 20 2010: (Start)

%C 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.

%C For all numbers 1 <= k < F(n), the inequality |k*phi-round(k*phi)| > |F(n)*phi-round(F(n)*phi)| holds.

%C F(n)*phi - round(F(n)*phi) = -((-phi)^(-n)), for n > 1.

%C Fract(1/2 + F(n)*phi) = 1/2 -(-phi)^(-n), for n > 1.

%C Fract(F(n)*phi) = (1/2)*(1 + (-1)^n) - (-phi)^(-n), n > 1.

%C Inverse: n = -log_phi |1/2 - fract(1/2 + F(n)*phi)|.

%C (End)

%C F(A001177(n)*k) mod n = 0, for any integer k. - _Gary Detlefs_, Nov 27 2010

%C F(n+k)^2 - F(n)^2 = F(k)*F(2n+k), for even k. - _Gary Detlefs_, Dec 04 2010

%C F(n+k)^2 + F(n)^2 = F(k)*F(2n+k), for odd k. - _Gary Detlefs_, Dec 04 2010

%C F(n) = round(phi* F(n-1)) for n > 1. - _Joseph P. Shoulak_, Jan 13 2012

%C For n > 0: a(n) = length of n-th row in Wythoff array A003603. - _Reinhard Zumkeller_, Jan 26 2012

%C From _Bridget Tenner_, Feb 22 2012: (Start)

%C The number of free permutations of [n].

%C The number of permutations of [n] for which s_k in supp(w) implies s_{k+-1} not in supp(w).

%C The number of permutations of [n] in which every decomposition into length(w) reflections is actually composed of simple reflections. (End)

%C The sequence F(n+1)^(1/n) is increasing. The sequence F(n+2)^(1/n) is decreasing. - _Thomas Ordowski_, Apr 19 2012

%C 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

%C From _Ravi Kumar Davala_, Jan 30 2014: (Start)

%C Proof of Ratushnyak's first conjecture: For n > 1, F(n+2)^2 - F(n)*F(n+1) + (-1)^n = 2F(n+1)^2.

%C Consider: F(n+2)^2 - F(n)*F(n+1) - 2F(n+1)^2

%C = F(n+2)^2 - F(n+1)^2 - F(n+1)^2 - F(n)*F(n+1)

%C = (F(n+2) + F(n+1))*(F(n+2) - F(n+1)) - F(n+1)*(F(n+1) + F(n))

%C = F(n+3)*F(n) - F(n+1)*F(n+2) = -(-1)^n.

%C Proof of second conjecture: L(n) stands for Lucas number sequence from A000032.

%C Consider the fact that

%C L(2n+1)^2 = L(4n+2) - 2

%C (F(2n) + F(2n+2))^2 = F(4n+1) + F(4n+3) - 2

%C (F(2n) + F(2n+2))^2 = (Sum_{k = 2..2n} F(2k)) + F(4n+3).

%C (End)

%C 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

%C 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

%C 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

%C 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

%C 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

%C Central terms of triangles in A162741 and A208245, n > 0. - _Reinhard Zumkeller_, Jul 28 2013

%C 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

%C 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

%C A010056(a(n)) = 1. - _Reinhard Zumkeller_, Oct 10 2013

%C 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

%C ( 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

%C 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

%C 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]

%C 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

%C 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

%C 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

%C a(4)=3 and a(6)=8 are the only Fibonacci numbers that are of the form prime+1. - _Emmanuel Vantieghem_, Oct 02 2014

%C 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

%C 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

%C Lim_{n -> oo} (log F(n+1)/log F(n))^n = e. - _Thomas Ordowski_, Oct 06 2014

%C 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

%C a(n-1) counts closed walks on the graph G(1-vertex;l-loop,2-loop). - _David Neil McGrath_, Nov 26 2014

%C From _Tom Copeland_, Nov 02 2014: (Start)

%C 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).

%C 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)].

%C 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).

%C BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)].

%C 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).

%C 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].

%C (End)

%C 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

%C 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

%C F(n+1) equals the number of binary words of length n avoiding runs of zeros of odd lengths. - _Milan Janjic_, Jan 28 2015

%C From _Russell Jay Hendel_, Apr 12 2015: (Start)

%C We prove Conjecture 1 of Rashid listed in the Formula section.

%C 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".

%C 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).

%C 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).

%C 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.

%C (End)

%C 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

%C 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

%C F(n) divides F(nk). Proved by Marjorie Bicknell and Verner E Hoggatt Jr. - _Juhani Heino_, Aug 24 2015

%C 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

%C 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

%C 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

%C 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

%C 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

%C 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

%C (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

%C From _Clark Kimberling_, Jun 13 2016: (Start)

%C 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*.

%C 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.

%C Let T(r) be the tree obtained by substituting r for x.

%C 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)

%C 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

%C Limit of the matrix power M^k shown in A163733, Sep 14 2016; as k-->inf. results in a single column vector equal to the Fibonacci sequence. - _Gary W. Adamson_, Sep 19 2016

%C 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

%C Also the number of independent vertex sets and vertex covers in the (n-2)-path graph. - _Eric W. Weisstein_, Sep 22 2017

%C 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

%C 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

%C 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

%C F(n+1) is the number of fixed points of the Foata transformation on S_n. - _Kevin Long_, Oct 17 2018

%C 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

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%H Rigoberto Flórez, Robinson A. Higuita, Antara Mukherjee, <a href="https://arxiv.org/abs/1804.02481">The Geometry of some Fibonacci Identities in the Hosoya Triangle</a>, arXiv:1804.02481 [math.NT], 2018.

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%H J. Huang, <a href="https://arxiv.org/abs/1405.1636">Hecke algebras with independent parameters</a>, arXiv:1405.1636 [math.RT], 2014; Journal of Algebraic Combinatorics 43 (2016) 521-551.

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%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibtable.html">Fibonacci numbers with tables of F(0)-F(500).</a>

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%H Leonardo of Pisa [Leonardo Pisano], <a href="/A000045/a000045.jpg">Illustration of initial terms</a>, from Liber Abaci [The Book of Calculation], 1202 (photo by David Singmaster).

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%H H. Mishima, Factorizations of Fibonacci numbers <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha108.htm">n=1..100</a>, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha109.htm">n=101..200</a>, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha110.htm">n=201..300</a>, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha111.htm">n=301..400</a>, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha112.htm">n=401..480</a>

%H P. Moree, <a href="http://arXiv.org/abs/math.CO/0311205">Convoluted convolved Fibonacci numbers</a>, arXiv:math/0311205 [math.CO], 2003.

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%H Mariana Nagy, Simon R. Cowell, Valeriu Beiu, <a href="https://arxiv.org/abs/1902.05944">Survey of Cubic Fibonacci Identities - When Cuboids Carry Weight</a>, arXiv:1902.05944 [math.HO], 2019.

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%H Arulalan Rajan, R. Vittal Rao, Ashok Rao and H. S. Jamadagni, <a href="http://arxiv.org/abs/1205.5398">Fibonacci Sequence, Recurrence Relations, Discrete Probability Distributions and Linear Convolution</a>, arXiv preprint arXiv:1205.5398 [math.PR], 2012. - From _N. J. A. Sloane_, Oct 23 2012

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%H Marc Renault, <a href="http://www.math.temple.edu/~renault/fibonacci/thesis.html">Properties of the Fibonacci sequence under various moduli</a>, MSc Thesis, Wake Forest U, 1996.

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%H E. S. Rowland, <a href="http://people.hofstra.edu/Eric_Rowland/investigations/fibonacci.html">Fibonacci Sequence Calculator up to n=1474</a>

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%H Robert Walker, <a href="http://www.youtube.com/watch?v=Wx4ZfuMl-FI&amp;NR=1">Inharmonic "Golden Rhythmicon" - Fibonacci Sequence in Pairs Approaching Golden Ratio - With Bounce</a>

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%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Matching.html">Matching</a>

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%H Wikipedia, <a href="http://www.wikipedia.org/wiki/Fibonacci_number">Fibonacci number</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Cassini_and_Catalan_identities">Cassini and Catalan identities</a>

%H Willem's Fibonacci site, <a href="http://home.zonnet.nl/LeonardEuler/fiboe.htm">Fibonacci</a>

%H Mike Winkler, <a href="http://arxiv.org/abs/1412.0519">On the structure and the behaviour of Collatz 3n + 1 sequences - Finite subsequences and the role of the Fibonacci sequence</a>, arXiv:1412.0519 [math.GM], 2014.

%H Roman Witula, Damian Slota and Edyta Hetmaniok, <a href="http://ami.ektf.hu/uploads/papers/finalpdf/AMI_41_from255to263.pdf">Bridges between different known integer sequences</a>, Annales Mathematicae et Informaticae, 41 (2013) pp. 255-263.

%H R. Yanco, <a href="/A007380/a007380.pdf">Letter and Email to N. J. A. Sloane, 1994</a>

%H R. Yanco and A. Bagchi, <a href="/A007380/a007380_1.pdf">K-th order maximal independent sets in path and cycle graphs</a>, Unpublished manuscript, 1994. (Annotated scanned copy)

%H Donovan Young, <a href="https://www.emis.de/journals/JIS/VOL21/Young/young2.html">The Number of Domino Matchings in the Game of Memory</a>, J. Int. Seq., Vol. 21 (2018), Article 18.8.1.

%H Aimei Yu and Xuezheng Lv, <a href="http://dx.doi.org/10.1007/s10910-006-9088-7">The Merrifield-Simmons indices and Hosoya indices of trees with k pendant vertices</a>, J. Math. Chem., Vol. 41 (2007), pp. 33-43. See page 35.

%H Tianping Zhang and Yuankui Ma, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Zhang/zhang56.html">On Generalized Fibonacci Polynomials and Bernoulli Numbers</a>, Journal of Integer Sequences, Vol. 8 (2005), Article 05.5.3.

%H <a href="/index/Cor#core">Index entries for "core" sequences</a>

%H <a href="/index/Di#divseq">Index to divisibility sequences</a>

%H <a href="/index/Par#partN">Index entries for related partition-counting sequences</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,1).

%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>

%H <a href="/index/Be#Benford">Index entries for sequences related to Benford's law</a>

%F G.f.: x / (1 - x - x^2).

%F G.f.: Sum_{n>=0} x^n * Product_{k=1..n} (k + x)/(1 + k*x). - _Paul D. Hanna_, Oct 26 2013

%F F(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)).

%F Alternatively, F(n) = ((1/2+sqrt(5)/2)^n - (1/2-sqrt(5)/2)^n)/sqrt(5).

%F F(n) = F(n-1) + F(n-2) = -(-1)^n F(-n).

%F F(n) = round(phi^n/sqrt(5)).

%F F(n+1) = Sum_{j=0..floor(n/2)} binomial(n-j, j).

%F 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

%F E.g.f.: (2/sqrt(5))*exp(x/2)*sinh(sqrt(5)*x/2). - _Len Smiley_, Nov 30 2001

%F [0 1; 1 1]^n [0 1] = [F(n); F(n+1)]

%F x | F(n) ==> x | F(kn).

%F 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

%F 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

%F From _Kurmang. Aziz. Rashid_, Feb 21 2004: (Start)

%F 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.]

%F Conjecture 2: for n >= 0, (F(n+2)*F(n+3)) - (F(n+1)*F(n+4)) + (-1)^n = 0.

%F [Two more conjectures removed by _Peter Luschny_, Nov 17 2017 ]

%F Theorem 1: for n >= 0, (F(n+3)^ 2 - F(n+1)^ 2)/F(n+2) = (F(n+3)+ F(n+1)).

%F Theorem 2: for n >= 0, F(n+10) = 11*F(n+5) + F(n).

%F Theorem 3: for n >= 6, F(n) = 4*F(n-3) + F(n-6). (End)

%F 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

%F 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

%F 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

%F |2*F(n) - 9*F(n+1)| = 4*A000032(n) + A000032(n+1). - _Creighton Dement_, Aug 13 2004

%F For x > Phi, Sum_{n>=0} F(n)/x^n = x/(x^2 - x - 1) - _Gerald McGarvey_, Oct 27 2004

%F 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

%F a(n-1) = Sum_{k=0..n} (-1)^k*binomial(n-ceiling(k/2), floor(k/2)). - _Benoit Cloitre_, May 05 2005

%F F(n+1) = Sum_{k=0..n} binomial((n+k)/2, (n-k)/2)(1+(-1)^(n-k))/2. - _Paul Barry_, Aug 28 2005

%F 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 F(n) = 2^-(n-1)*Sum_{k=0..floor((n-1)/2)} binomial(n,2*k+1)*5^k. - _Hieronymus Fischer_, Feb 07 2006

%F a(n) = (b(n+1) + b(n-1))/n where {b(n)} is the sequence A001629. - _Sergio Falcon_, Nov 22 2006

%F 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

%F From _Miklos Kristof_, Mar 19 2007: (Start)

%F Let L(n) = A000032(n) = Lucas numbers. Then:

%F For a >= b and odd b, F(a+b) + F(a-b) = L(a)*F(b).

%F For a >= b and even b, F(a+b) + F(a-b) = F(a)*L(b).

%F For a >= b and odd b, F(a+b) - F(a-b) = F(a)*L(b).

%F For a >= b and even b, F(a+b) - F(a-b) = L(a)*F(b).

%F F(n+m) + (-1)^m*F(n-m) = F(n)*L(m);

%F F(n+m) - (-1)^m*F(n-m) = L(n)*F(m);

%F 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 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 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 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)

%F A corollary to Kristof 2007 is 2*F(a+b) = F(a)*L(b) + L(a)*F(b). - _Graeme McRae_, Apr 24 2014

%F 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 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 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

%F 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 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

%F 1 = 1/(1*2) + 1/(1*3) + 1/(2*5) + 1/(3*8) + 1/(5*13) + ... = 1/2 + 1/3 + 1/10 + 1/24 + 1/65 + 1/168 + ...; where A059929 = (0, 2, 3, 10, 24, 65, 168, ...). - _Gary W. Adamson_, Mar 16 2008

%F 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

%F F(n) = (1/(n-1)!) * (n^(n-1) - (C(n-2,0) + 4*C(n-2,1) + 3*C(n-2,2))*n^(n-2) + (10*C(n-3,0) + 49*C(n-3,1) + 95*C(n-3,2) + 83*C(n-3,3) + 27*C(n-3,4))*n^(n-3) - (90*C(n-4,0) + 740*C(n-4,1) + 2415*C(n-4,2) + 4110*C(n-4,3) + 3890*C(n-4,4) + 1950*C(n-4,5) + 405*C(n-4,6))*n^(n-4) + ... ). - _André F. Labossière_, Nov 24 2004

%F a(n+1) = Sum_{k=0..n} A109466(n,k)*(-1)^(n-k). -_Philippe Deléham_, Oct 26 2008

%F 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

%F 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 (smbelcas(AT)toroidalsnark.net), Jul 04 2009

%F 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

%F 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

%F Lim_{k->infinity} 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

%F 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

%F 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

%F a(n) = 2*a(n-2) + a(n-3), n > 2. - _Gary Detlefs_, Sep 08 2010

%F a(2^n) = Product_{i=0..n-1} A000032(2^i). - _Vladimir Shevelev_, Nov 28 2010

%F a(n)^2 - a(n-1)^2 = a(n+1)*a(n-2), see A121646.

%F a(n) = sqrt((-1)^k*(a(n+k)^2 - a(k)*a(2n+k))), for any k. - _Gary Detlefs_, Dec 03 2010

%F F(2*n) = F(n+2)^2 - F(n+1)^2 - 2*F(n)^2. - _Richard R. Forberg_, Jun 04 2011

%F (-1)^(n+1) = F(n)^2 + F(n)*F(1+n) - F(1+n)^2.

%F F(n) = -F(n+2)(-2 + (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). - _Artur Jasinski_, Nov 17 2011

%F F(n) = 1 + Sum_{x=1..n-2} F(x). - _Joseph P. Shoulak_, Feb 05 2012

%F F(n) = 4*F(n-2) - 2*F(n-3) - F(n-6). - _Gary Detlefs_, Apr 01 2012

%F F(n) = round(phi^(n+1)/(phi+2)). - _Thomas Ordowski_, Apr 20 2012

%F From _Sergei N. Gladkovskii_, Jun 03 2012: (Start)

%F 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).

%F Let E(x) be the e.g.f., i.e.,

%F 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

%F 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).

%F (End)

%F From _Hieronymus Fischer_, Nov 30 2012: (Start)

%F 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).

%F 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 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.

%F Since S(j,n) = binomial(n-2+j,j-1), the formula above equals the well-known binomial formula, essentially. (End)

%F G.f. A(x) = x / (1 - x / (1 - x / (1 + x))). - _Michael Somos_, Jan 04 2013

%F Sum_{n >= 1} (-1)^(n-1)/(a(n)*a(n+1)) = 1/phi (phi=golden ratio). - _Vladimir Shevelev_, Feb 22 2013

%F From _Vladimir Shevelev_, Feb 24 2013: (Start)

%F (1) Expression a(n+1) via a(n): a(n+1) = (a(n) + sqrt(5*(a(n))^2 + 4*(-1)^n))/2;

%F (2) Sum_{k=1...n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);

%F (3) a(n)/a(n+1) = 1/phi + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)

%F 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:= Chebyshef 2nd kind). - _Bill Gosper_, Mar 04 2013

%F 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

%F 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

%F 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

%F 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

%F 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

%F 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) = ceil(b(n-1)), if n is odd, with convergence. - _Richard R. Forberg_, Jan 19 2014

%F 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 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 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 F(n) = round(1/(Sum_{j>=n+2} 1/F(j))). - _Richard R. Forberg_, Aug 14 2014

%F a(n) = hypergeometric([-n/2+1/2, -n/2+1], [-n+1], -4) for n >= 2. - _Peter Luschny_, Sep 19 2014

%F 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

%F 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

%F 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 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 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 F(n+1) = ceiling( (1/phi)*Sum_{k=0..n} F(k) ). - _Tom Edgar_, Sep 10 2015

%F a(n) = (L(n-3) + L(n+3))/10 where L(n)=A000032(n). - _J. M. Bergot_, Nov 25 2015

%F From _Bob Selcoe_, Mar 27 2016 (Start):

%F F(n) = (F(2n+k+1) - F(n+1)*F(n+k+1))/F(n+k), k >= 0.

%F Thus when k=0: F(n) = sqrt(F(2n+1) - F(n+1)^2).

%F F(n) = cbrt(F(3n) - F(n+1)^3 + F(n-1)^3).

%F 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).

%F (End)

%F 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

%F From _Ilya Gutkovskiy_, Aug 03 2016: (Start)

%F a(n) = F_n(1), where F_n(x) are the Fibonacci polynomials.

%F Inverse binomial transform of A001906.

%F 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)

%F 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 For n > 0, a(n) = b(n+1) where b(n) = Sum_{k=1..n} b(n-k)*A000931(k-1), b(0) = 1. - _J. Conrad_, Apr 19 2017

%F 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 F(n) = Sum_{k=0..floor((n-1)/2)} ( (n-k-1)! / ((n-2k-1)! * k!) ). - _Zhandos Mambetaliyev_, Nov 08 2017

%F 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

%F From _Isaac Saffold_, Jul 19 2018: (Start)

%F Let [a/p] denote the Legendre symbol. Then, for an odd prime p:

%F F(p+n) == [5/p]*F([5/p]+n) (mod p), if [5/p] = 1 or -1.

%F F(p+n) == 3*F(n) (mod p), if [5/p] = 0 (i.e. p = 5).

%F 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)

%F a(n) = A094718(4, n). a(n) = A101220(0, j, n).

%F a(n) = A090888(0, n+1) = A118654(0, n+1) = A118654(1, n-1) = A109754(0, n) = A109754(1, n-1), for n > 0.

%F a(n) = (L(n-3) + L(n-2) + L(n-1) + L(n))/5 with L(n)=A000032(n). - _Art Baker_, Jan 04 2019

%e 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 entries in the sequence. - _Cino Hilliard_, Sep 15 2008

%e From _Joerg Arndt_, May 21 2013: (Start)

%e 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:

%e 01: [ 2 1 2 1 1 ]

%e 02: [ 2 1 3 1 ]

%e 03: [ 2 1 4 ]

%e 04: [ 3 1 2 1 ]

%e 05: [ 3 1 3 ]

%e 06: [ 3 2 2 ]

%e 07: [ 4 1 2 ]

%e 08: [ 4 2 1 ]

%e 09: [ 4 3 ]

%e 10: [ 5 1 1 ]

%e 11: [ 5 2 ]

%e 12: [ 6 1 ]

%e 13: [ 7 ]

%e 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:

%e 01: [ 1 2 1 2 ]

%e 02: [ 1 3 1 1 ]

%e 03: [ 1 3 2 ]

%e 04: [ 1 4 1 ]

%e 05: [ 1 5 ]

%e 06: [ 2 2 1 1 ]

%e 07: [ 2 3 1 ]

%e 08: [ 2 4 ]

%e 09: [ 3 2 1 ]

%e 10: [ 3 3 ]

%e 11: [ 4 2 ]

%e 12: [ 5 1 ]

%e 13: [ 6 ]

%e (End)

%e 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

%e 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

%p A000045 := proc(n) combinat[fibonacci](n); end;

%p 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

%p spec := [B, {B=Sequence(Set(Z, card>1))}, unlabeled ]: seq(combstruct[count](spec, size=n), n=1..39); # _Zerinvary Lajos_, Apr 04 2008

%p # The following Maple command isFib(n) yields true or false depending on whether n is a Fibonacci number or not.

%p 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

%t Table[Fibonacci[k], {k, 0, 50}] (* _Mohammad K. Azarian_, Jul 11 2015 *)

%t 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 (smbelcas(AT)toroidalsnark.net), Jul 04 2009 *)

%t LinearRecurrence[{1, 1}, {0, 1}, 40] (* _Harvey P. Dale_, Aug 03 2014 *)

%t Fibonacci[Range[0, 20]] (* _Eric W. Weisstein_, Sep 22 2017 *)

%t CoefficientList[Series[-(x/(-1 + x + x^2)), {x, 0, 20}], x] (* _Eric W. Weisstein_, Sep 22 2017 *)

%o (Axiom) [fibonacci(n) for n in 0..50]

%o (MAGMA) [Fibonacci(n): n in [0..38]];

%o (MAGMA) [0,1] cat [n: n in [1..50000000] | IsSquare(5*n^2-4) or IsSquare(5*n^2+4)]; // _Vincenzo Librandi_, Nov 19 2014

%o (Maxima) makelist(fib(n),n,0,100); /* _Martin Ettl_, Oct 21 2012 */

%o (PARI) a(n) = fibonacci(n)

%o (PARI) a(n)=my(phi=quadgen(5));(phi^n-(-1/phi)^n)/(2*phi-1) \\ _Charles R Greathouse IV_, Jun 17 2012

%o (PARI) a(n)=polcoeff(sum(m=0, n, x^m*prod(k=1, m, k+x +x*O(x^n))/prod(k=1, m, 1+k*x +x*O(x^n))), n) \\ _Paul D. Hanna_, Oct 26 2013

%o (Python) # _Jaap Spies_, Jan 05 2007 (Change leading dots to blanks.)

%o def fib():

%o ... """ Generates the Fibonacci numbers, starting with 0 """

%o ... x, y = 0, 1

%o ... while 1:

%o ....... yield x

%o ....... x, y = y, x+y

%o .

%o f = fib()

%o a = [f.next() for i in range(100)]

%o .

%o def A000045(n):

%o ... """ Returns Fibonacci number with index n, offset 0,4 """

%o ... return a[n]

%o ................

%o def A000045_list(N):

%o ... """ Returns a list of the first n Fibonacci numbers """

%o ... return a[:N]

%o .

%o (Python) # As b-file:

%o from gmpy2 import fib

%o for n in xrange(100): print str(n)+" "+str(fib(n)) # _Bruno Berselli_, Dec 06 2016

%o (Sage) ## Demonstration program from Jaap Spies:

%o a = sloane.A000045; ## choose sequence

%o print a ## This returns the name of the sequence.

%o print a(38) ## This returns the 38th number of the sequence.

%o print a.list(39) ## This returns a list of the first 39 numbers.

%o (Sage) # Alternatively:

%o a = BinaryRecurrenceSequence(1,1); print [a(n) for n in (0..19)]

%o # Closed form integer formula with F(1) = 0 from Paul Hankin (use only for fun).

%o F = lambda n: (4<<(n-1)*(n+2))//((4<<2*(n-1))-(2<<(n-1))-1)&((2<<(n-1))-1)

%o print [F(n) for n in (0..19)] # _Peter Luschny_, Aug 28 2016

%o (Sage) [i for i in fibonacci_sequence(0, 40)] # _Bruno Berselli_, Jun 26 2014

%o -- which also has other versions.

%o fib :: Int -> Integer

%o fib n = fibs !! n

%o .. where

%o .... fibs = 0 : 1 : zipWith (+) fibs (tail fibs)

%o {- Example of use: map fib [0..38] _Gerald McGarvey_, Sep 29 2009 -}

%o (Julia)

%o function fib(n)

%o F = BigInt[1 1; 1 0]

%o Fn = F^n

%o Fn[2, 1]

%o end

%o println([fib(n) for n in 0:38]) # _Peter Luschny_, Feb 23 2017

%o (GAP)

%o 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

%o (Scheme)

%o ;; The following definition uses macro definec for the memoization (caching) of the results. See http://oeis.org/wiki/Memoization#Scheme

%o (definec (A000045 n) (if (< n 2) n (+ (A000045 (- n 1)) (A000045 (- n 2))))) ;; _Antti Karttunen_, Oct 06 2017

%Y Cf. A039834 (signed Fibonacci numbers), A001690 (complement), A000213, A000288, A000322, A000383, A060455, A030186, A020695, A020701, A071679, A099731, A100492, A094216, A094638, A000108, A101399, A101400, A001611, A000071, A157725, A001911, A157726, A006327, A157727, A157728, A157729, A167616, A059929, A144152, A152063, A114690, A003893, A000032, A060441, A000930, A003269, A000957, A057078, A007317, A091867, A104597, A249548, A262342, A001060, A022095, A072649.

%Y First row of arrays A103323, A172236, A234357. Second row of arrays A099390, A048887, and A092921 (k-generalized Fibonacci numbers).

%Y Cf. A001175 (Pisano periods), A001177 (Entry points), A001176 (number of zeros in a fundamental period).

%Y Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074.

%Y Boustrophedon transforms: A000738, A000744.

%Y Powers: A103323, A105317, A254719.

%Y Numbers of prime factors: A022307 and A038575.

%Y Cf. A163733.

%K nonn,core,nice,easy,hear,changed

%O 0,4

%A _N. J. A. Sloane_, 1964

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Last modified April 19 10:39 EDT 2019. Contains 322255 sequences. (Running on oeis4.)