A015441 Generalized Fibonacci numbers.

0, 1, 1, 7, 13, 55, 133, 463, 1261, 4039, 11605, 35839, 105469, 320503, 953317, 2876335, 8596237, 25854247, 77431669, 232557151, 697147165, 2092490071, 6275373061, 18830313487, 56482551853, 169464432775, 508359743893, 1525146340543, 4575304803901, 13726182847159

Offset: 0

Author

Olivier Gérard

Keywords

• nonn
• easy
• nice

Comments

a(n) is the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x,y) = xF(n-1)(x,y) + yF(n-2)(x,y), F(0)(x,y)=0, F(1)(x,y)=1, when y=6x^2. - Mario Catalani (mario.catalani(AT)unito.it), Dec 06 2002

For n>=1: number of length-(n-1) words with letters {0,1,2,3,4,5,6,7} where no two consecutive letters are nonzero, see fxtbook link below. - Joerg Arndt, Apr 08 2011

Starting with offset 1 and convolved with (1, 3, 3, 3, ...) = A003462: (1, 4, 13, 40, ...). - Gary W. Adamson, May 28 2009

a(n) is identical to its inverse binomial transform signed. Differences: A102901. - Paul Curtz, Feb 23 2010

The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=2, 7*a(n-2) equals the number of 7-colored compositions of n with all parts >=2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011

Pisano period lengths: 1, 1, 1, 2, 20, 1, 6, 2, 3, 20, 5, 2, 12, 6, 20, 4, 16, 3, 18, 20, ... - R. J. Mathar, Aug 10 2012

A015441 and A015518 are the only integer sequences (from the family of homogeneous linear recurrence relation of order 2 with positive integer coefficients with initial values a(0)=0 and a(1)=1) whose ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 14 2014

This is an autosequence of the first kind: the array of successive differences shows a main diagonal of zeros and the inverse binomial transform is identical to the sequence (with alternating signs). - Pointed out by Paul Curtz, Dec 05 2016

First two upper diagonals: A000400(n).

This is a variation on the Starhex honeycomb configuration A332243, see illustration in links. It is an alternating pattern of the 2nd iteration of the centered hexagonal numbers A003215 and centered 12-gonal 'Star' numbers A003154. - John Elias, Oct 06 2021

Examples

			G.f. = x + x^2 + 7*x^3 + 13*x^4 + 55*x^5 + 133*x^6 + 463*x^7 + 1261*x^8 + ...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
• Joerg Arndt, Matters Computational (The Fxtbook), p. 317-318
• A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
• John Elias, Illustration: A Starhex honeycomb variation
• Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
• H. Li and T. MacHenry, Permanents and Determinants, Weighted Isobaric Polynomials, and Integer Sequences, J. Int. Seq. 16 (2013) #13.3.5, example 48.
• F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.
• A. G. Shannon and J. V. Leyendekkers, The Golden Ratio family and the Binet equation, Notes on Number Theory and Discrete Mathematics, Vol. 21, No. 2, (2015), 35-42.
• Index entries for linear recurrences with constant coefficients, signature (1,6).

Crossrefs

Cf. A000079, A003462, A016153, A109466, A122117. Cf. A001656, A087451. Cf. A000400.

Programs

Magma

I:=[0,1]; [n le 2 select I[n] else Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

Maple

A015441:=n->(1/5)*((3^n)-((-2)^n)); seq(A015441(n), n=0..30); # Wesley Ivan Hurt, Mar 14 2014

Mathematica

a[n_]:=(MatrixPower[{{1,4},{1,-2}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]], {n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
LinearRecurrence[{1,6},{0,1},30] (* Harvey P. Dale, Apr 26 2011 *)
CoefficientList[Series[x/((1 + 2 x) (1 - 3 x)), {x, 0, 29}], x] (* Michael De Vlieger, Dec 05 2016 *)

PARI

{a(n) = (3^n - (-2)^n) / 5};

Sage

[lucas_number1(n,1,-6) for n in range(0, 27)] # Zerinvary Lajos, Apr 22 2009

Formula

G.f.: x/((1+2*x)*(1-3*x)).

a(n) = a(n-1) + 6*a(n-2).

a(n) = (1/5)*((3^n)-((-2)^n)). - henryk.wicke(AT)stud.uni-hannover.de

E.g.f.: (exp(3*x) - exp(-2*x))/5. - Paul Barry, Apr 20 2003

a(n+1) = Sum_{k=0..ceiling(n/2)} 6^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004

a(n) = (A000244(n) - A001045(n+1)(-1)^n - A001045(n)(-1)^n)/5. - Paul Barry, Apr 27 2004

The binomial transform of [1,1,7,13,55,133,463,...] is A122117. - Philippe Deléham, Oct 19 2006

a(n+1) = Sum_{k=0..n} A109466(n,k)*(-6)^(n-k). - Philippe Deléham, Oct 26 2008

a(n) = 3a(n-1) + (-1)^(n+1)*A000079(n-1). - Paul Curtz, Feb 23 2010

G.f.: Q(0) -1, where Q(k) = 1 + 6*x^2 + (k+2)*x - x*(k+1 + 6*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013

a(n) = (Sum_{1<=k<=n, k odd} binomial(n,k)*5^(k-1))/2^(n-1). - Vladimir Shevelev, Feb 05 2014

a(-n) = -(-1)^n * a(n) / 6^n for all n in Z. - Michael Somos, Mar 18 2014

From Peter Bala, Apr 01 2015: (Start)

Sum_{n >= 0} a(n+1)*x^n = exp( Sum_{n >= 1} A087451(n)*x^n/n ).

For k = 0, 1, 2, ... and for n >= 1, (5^k)*a(n) | a((5^k)*n).

The expansion of exp( Sum_{n >= 1} a(5*n)/(5*a(n))*x^n/n ) has integral coefficients. Cf. A001656. (End)

From Peter Bala, Jun 27 2025: (Start)

Sum_{n >= 1} (-6)^n/(a(n)*a(n+1)) = -2, since (-6)^n/(a(n)*a(n+1)) = (-2)^n/a(n) - (-2)^(n+1)/a(n+1) for n >= 1.

The following are examples of telescoping infinite products:

Product_{n >= 0} (1 + 6^n/a(2*n+2)) = 6, since (1 + 6^(2*n-1)/a(4*n))*(1 + 6^(2*n)/a(4*n+2)) = (6 - 4^(n+1)/b(n)) / (6 - 4^n/b(n-1)), where b(n) = (2*4^n + 3*9^n)/5 = A096951(n). Similarly,

Product_{n >= 1} (1 - 6^n/a(2*n+2)) = 3/13.

Product_{n >= 0} (1 + (-6)^n/a(2*n+2)) = 6/5.

Product_{n >= 1} (1 - (-6)^n/a(2*n+2)) = 15/13.

exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n. (End)