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A107025 Binomial transform of the expansion of 1/(1-x^5-x^6). 1

%I #6 Jun 13 2015 00:51:49

%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)}.

%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); a(n)=6a(n-1)-15a(n-2)+20a(n-3)-15a(n-4)+7a(n-5)-a(n-6); a(n)=sum{k=0..floor((n+1)/2), binomial(n+k, 6k)}; a(n)=sum{k0..floor(6n/5), binomial(k, 6n-5k)}.

%Y Cf. A017837

%K easy,nonn

%O 0,6

%A _Paul Barry_, May 09 2005

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