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%I #43 Jul 13 2015 21:07:28
%S 1,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,2,3,4,5,6,7,8,9,
%T 10,11,12,13,14,16,19,23,28,34,41,49,58,68,79,91,104,118,134,153,176,
%U 204,238,279,328,386,454,533,624,728,846,980,1133,1309
%N Expansion of 1/(1 - x^13 - x^14 - ...).
%C a(n) = number of compositions of n in which each part is >= 13. - _Milan Janjic_, Jun 28 2010
%C a(n+25) equals the number of binary words of length n having at least 12 zeros between every two successive ones. - _Milan Janjic_, Feb 09 2015
%H Alois P. Heinz, <a href="/A017907/b017907.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_13">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).
%F G.f.: (x-1)/(x-1+x^13). - _Alois P. Heinz_, Aug 04 2008
%F For positive integers n and k such that k <= n <= 13*k, and 12 divides n-k, define c(n,k) = binomial(k,(n-k)/12), and c(n,k) = 0, otherwise. Then, for n>=1, a(n+13) = sum(c(n,k), k=1..n). - _Milan Janjic_, Dec 09 2011
%F a(0)=1, a(1)=0, a(2)=0, a(3)=0, a(4)=0, a(5)=0, a(6)=0, a(7)=0, a(8)=0, a(9)=0, a(10)=0, a(11)=0, a(12)=0, a(n)=a(n-1)+a(n-13). - _Harvey P. Dale_, Feb 07 2015
%p a:= n-> (Matrix(13, (i,j)-> if (i=j-1) then 1 elif j=1 then [1, 0$11, 1][i] else 0 fi)^n)[13,13]: 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, 1}, {1, 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^13),{x,0,70}],x] (* _Harvey P. Dale_, Feb 07 2015 *)
%o (PARI) Vec((x-1)/(x-1+x^13)+O(x^99)) \\ _Charles R Greathouse IV_, Sep 26 2012
%Y Column k=12 of A141539, k=13 of A220122. - _Alois P. Heinz_, Dec 09 2012
%K nonn,easy
%O 0,27
%A _N. J. A. Sloane_.
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