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%I #14 Oct 13 2015 15:10:01
%S 8,38,111,256,511,924,1554,2472,3762,5522,7865,10920,14833,19768,
%T 25908,33456,42636,53694,66899,82544,100947,122452,147430,176280,
%U 209430,247338,290493,339416,394661,456816,526504,604384,691152,787542
%N Coefficient of x^5 in (1-x-x^2)^(-n).
%C The coefficient of x^5 in (1-x-x^2)^(-n) is the coefficient of x^5 in (1 + x + 2x^2 + 3x^3 + 5x^4 + 8x^5)^n. Using the multinomial theorem one then finds that a(n) = n(n+1)(n+2)(n^2 + 27n + 132)/5!
%C The inverse binomial transform yields 8,30,43,29,9,1,0,0,... (0 continued) - _R. J. Mathar_, May 23 2008
%D Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%F a(n) = n(n+1)(n+2)(n^2 + 27n + 132)/5!
%F O.g.f.: x(3x-4)(x-2)/(1-x)^6. - _R. J. Mathar_, May 23 2008
%t a[n_] := n(n + 1)(n + 2)(n^2 + 27n + 132)/5! Do[Print[n, " ", a[n]], {n, 1, 25}]
%t LinearRecurrence[{6,-15,20,-15,6,-1},{8,38,111,256,511,924},40] (* _Harvey P. Dale_, Oct 13 2015 *)
%o (PARI) a(n)=binomial(n+2,3)*(n^2+27*n+132)/20 \\ _Charles R Greathouse IV_, Jul 29 2011
%Y Cf. A000096, A006503, A006504.
%K nonn,easy
%O 1,1
%A _Sergio Falcon_, May 22 2008
%E More terms from _R. J. Mathar_, May 23 2008
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