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%I #15 Jun 22 2024 14:11:32
%S 1,1,1,1,1,2,8,29,85,211,464,938,1808,3459,6826,14198,30960,69143,
%T 154433,340006,734561,1561313,3286129,6900097,14542101,30855957,
%U 65908862,141395972,303745077,651763377,1395140215,2978858672
%N Binomial transform of the expansion of 1/(1-x^5-x^6).
%C In general, the binomial transform of 1/(1-x^r-x^(r+1)) is given by (1-x)^r/((1-x)^(r+1)-x^r), with a(n) = Sum_{k=0..floor((n+1)/2)} binomial(n+k,(r+1)k) = Sum_{k=0..floor((r+1)n/r)} binomial(k,(r+1)n-r*k).
%C Number of compositions of 6*n into parts 5 and 6. - _Seiichi Manyama_, Jun 22 2024
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,7,-1).
%F G.f.: (1-x)^5/((1-x)^6-x^5).
%F a(n) = 6a(n-1)-15a(n-2)+20a(n-3)-15a(n-4)+7a(n-5)-a(n-6).
%F a(n) = Sum_{k=0..floor((n+1)/2)} binomial(n+k, 6k).
%F a(n) = Sum_{k=0..floor(6n/5)} binomial(k, 6n-5k).
%F a(n) = A017837(6*n). - _Seiichi Manyama_, Jun 22 2024
%Y Cf. A017837, A099099, A099131.
%K easy,nonn
%O 0,6
%A _Paul Barry_, May 09 2005