%I #42 Jan 01 2021 03:29:20
%S 0,0,0,0,0,0,0,0,0,0,1,1,2,4,8,16,32,64,128,256,512,1024,2047,4093,
%T 8184,16364,32720,65424,130816,261568,523008,1045760,2091008,4180992,
%U 8359937,16715781,33423378,66830392,133628064,267190704,534250592,1068239616
%N Fibonacci 11-step numbers.
%H G. C. Greubel, <a href="/A168082/b168082.txt">Table of n, a(n) for n = 1..1000</a>
%H Martin Burtscher, Igor Szczyrba, and RafaĆ Szczyrba, <a href="http://www.emis.de/journals/JIS/VOL18/Szczyrba/sz3.pdf">Analytic Representations of the n-anacci Constants and Generalizations Thereof</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
%H Kai Wang, <a href="https://www.researchgate.net/publication/344295426_IDENTITIES_FOR_GENERALIZED_ENNEANACCI_NUMBERS">Identities for generalized enneanacci numbers</a>, Generalized Fibonacci Sequences (2020).
%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,1,1,1,1,1,1,1,1,1).
%F From _Joerg Arndt_, Sep 22 2020: (Start)
%F a(n) = Sum_{k=1..11} a(n-k).
%F G.f.: x^11/(1 - Sum_{k=1..11} x^k ).
%F a(n) = 2*a(n-1) - a(n-12). (End)
%F Another form of the g.f. f: f(z) = (z^(k-1)-z^(k))/(1-2*z+z^(k+1)) with k=11. a(n) = Sum_((-1)^i*binomial(n-10-11*i,i)*2^(n-10-12*i), i=0..floor((n-10)/12))-Sum_((-1)^i*binomial(n-11-11*i,i)*2^(n-11-12*i), i=0..floor((n-11)/12)) with Sum_(alpha(i),i=m..n) = 0 for m>n. - _Richard Choulet_, Feb 22 2010
%p a:= proc(n) option remember; `if`(n<11, 0,
%p `if`(n=11, 1, add(a(n-j), j=1..11)))
%p end:
%p seq(a(n), n=1..50); # _Alois P. Heinz_, Sep 23 2020
%t With[{nn=11},LinearRecurrence[Table[1,{nn}],Join[Table[0,{nn-1}],{1}],50]] (* _Harvey P. Dale_, Aug 17 2013 *)
%K nonn,easy
%O 1,13
%A _Vladimir Joseph Stephan Orlovsky_, Nov 18 2009