%I #39 Jul 30 2015 16:50:31
%S 1,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,2,3,4,5,6,7,8,9,10,11,12,
%T 14,17,21,26,32,39,47,56,66,77,89,103,120,141,167,199,238,285,341,407,
%U 484,573,676,796,937,1104,1303,1541,1826,2167,2574,3058
%N Expansion of 1/(1 - x^11 - x^12 - ...).
%C a(n) = number of compositions of n in which each part is >= 11. - _Milan Janjic_, Jun 28 2010
%C a(n+21) equals the number of binary words of length n having at least 10 zeros between every two successive ones. - _Milan Janjic_, Feb 09 2015
%H Alois P. Heinz, <a href="/A017905/b017905.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).
%F G.f.: (x-1)/(x-1+x^11). - _Alois P. Heinz_, Aug 04 2008
%F For positive integers n and k such that k <= n <= 11*k, and 10 divides n-k, define c(n,k) = binomial(k,(n-k)/10), and c(n,k) = 0, otherwise. Then, for n>=1, a(n+11) = sum(c(n,k), k=1..n). - _Milan Janjic_, Dec 09 2011
%p a:= n-> (Matrix(11, (i,j)-> if (i=j-1) then 1 elif j=1 then [1, 0$9, 1][i] else 0 fi)^n)[11,11]: seq(a(n), n=0..70); # _Alois P. Heinz_, Aug 04 2008
%t LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, 80] (* _Vladimir Joseph Stephan Orlovsky_, Feb 17 2012 *)
%o (PARI) Vec((x-1)/(x-1+x^11)+O(x^99)) \\ _Charles R Greathouse IV_, Sep 26 2012
%K nonn,easy
%O 0,23
%A _N. J. A. Sloane_.