A017909 Expansion of 1/(1 - x^15 - x^16 - ...).

1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 21, 25, 30, 36, 43, 51, 60, 70, 81, 93, 106, 120, 135, 151, 169, 190, 215, 245, 281, 324, 375, 435, 505, 586, 679

Offset

0,31

Comments

a(n+29) equals the number of binary words of length n having at least 14 zeros between every two successive ones. - Milan Janjic, Feb 09 2015

Number of compositions of n into parts >= 15. - Ilya Gutkovskiy, May 23 2017

Links

Alois P. Heinz, Table of n, a(n) for n = 0..1000

I. M. Gessel, Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013) 13.4.5

Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

Formula

G.f.: (x-1)/(x-1+x^15). - Alois P. Heinz, Aug 04 2008

For positive integers n and k such that k <= n <= 15*k, and 14 divides n-k, define c(n,k) = binomial(k,(n-k)/14), and c(n,k) = 0, otherwise. Then, for n>=1, a(n+15) = sum(c(n,k), k=1..n). - Milan Janjic, Dec 09 2011

Maple

a:= n -> (Matrix(15, (i, j)-> if (i=j-1) then 1 elif j=1 then [1, 0$13, 1][i] else 0 fi)^n)[15, 15]: seq(a(n), n=0..80); # Alois P. Heinz, Aug 04 2008

Mathematica

LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2012 *)
CoefficientList[Series[(x-1)/(x-1+x^15), {x, 0, 100}], x] (* Harvey P. Dale, Sep 04 2020 *)

Prog

(PARI) Vec((x-1)/(x-1+x^15)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

Crossrefs

Sequence in context: A078510 A246100 A247250 * A345201 A316530 A296864

Adjacent sequences: A017906 A017907 A017908 * A017910 A017911 A017912

Keyword

nonn,easy

Author

N. J. A. Sloane.

Status

approved