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%I #40 Sep 04 2020 13:53:47
%S 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,
%T 6,7,8,9,10,11,12,13,14,15,16,18,21,25,30,36,43,51,60,70,81,93,106,
%U 120,135,151,169,190,215,245,281,324,375,435,505,586,679
%N Expansion of 1/(1 - x^15 - x^16 - ...).
%C 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
%C Number of compositions of n into parts >= 15. - _Ilya Gutkovskiy_, May 23 2017
%H Alois P. Heinz, <a href="/A017909/b017909.txt">Table of n, a(n) for n = 0..1000</a>
%H I. M. Gessel, Ji Li, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Gessel/gessel6.html">Compositions and Fibonacci identities</a>, J. Int. Seq. 16 (2013) 13.4.5
%H <a href="/index/Rec#order_15">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).
%F G.f.: (x-1)/(x-1+x^15). - _Alois P. Heinz_, Aug 04 2008
%F 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
%p 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
%t 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 *)
%t CoefficientList[Series[(x-1)/(x-1+x^15),{x,0,100}],x] (* _Harvey P. Dale_, Sep 04 2020 *)
%o (PARI) Vec((x-1)/(x-1+x^15)+O(x^99)) \\ _Charles R Greathouse IV_, Sep 26 2012
%K nonn,easy
%O 0,31
%A _N. J. A. Sloane_.
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