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%I
%S 1,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,2,3,4,5,6,7,8,9,10,
%T 11,12,13,15,18,22,27,33,40,48,57,67,78,90,103,118,136,158,185,218,
%U 258,306,363,430,508,598,701,819,955,1113,1298,1516,1774
%N Expansion of 1/(1 - x^12 - x^13 - ...).
%C a(n) = number of compositions of n in which each part is >=12. - _Milan Janjic_, Jun 28 2010
%C a(n+23) equals the number of binary words of length n having at least 11 zeros between every two successive ones. - _Milan Janjic_, Feb 09 2015
%H Vincenzo Librandi, <a href="/A017906/b017906.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_12">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).
%F G.f.: (x-1)/(x-1+x^12). - _Alois P. Heinz_, Aug 04 2008
%F For positive integers n and k such that k <= n <= 12*k, and 11 divides n-k, define c(n,k) = binomial(k,(n-k)/11), and c(n,k) = 0, otherwise. Then, for n>=1, a(n+12) = sum(c(n,k), k=1..n). - _Milan Janjic_, Dec 09 2011
%p a:= n-> (Matrix(12, (i, j)-> `if`(i=j-1, 1, `if`(j=1, [1, 0$10, 1][i], 0)))^n)[12, 12]: seq(a(n), n=0..60); # _Alois P. Heinz_, Aug 04 2008
%t LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 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^12), {x, 0, 70}], x] (* _Vincenzo Librandi_, Jul 01 2013 *)
%o (PARI) Vec((x-1)/(x-1+x^12)+O(x^99)) \\ _Charles R Greathouse IV_, Sep 26 2012
%K nonn,easy
%O 0,25
%A _N. J. A. Sloane_.
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