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%I #25 Sep 08 2022 08:46:16
%S 12465,1,3281,68385,388849,1305665,3307281,7029601,13255985,22917249,
%T 37091665,57004961,84030321,119688385,165647249,223722465,295877041,
%U 384221441,491013585,618658849,769710065,946867521,1152978961,1391039585,1664192049,1975726465
%N a(n) = 6144*n^4 - 29184*n^3 + 52416*n^2 - 41840*n + 12465.
%H Vincenzo Librandi, <a href="/A272132/b272132.txt">Table of n, a(n) for n = 0..1000</a>
%H Richard P. Brent, <a href="http://arxiv.org/abs/1407.3533">Generalising Tuenter's binomial sums</a>, arXiv:1407.3533 [math.CO], 2014 (page 16).
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F O.g.f.: (12465 - 62324*x + 127926*x^2 - 72660*x^3 + 142049*x^4)/(1-x)^5.
%F E.g.f.: (12465 - 12464*x + 7872*x^2 + 7680*x^3 + 6144*x^4)*exp(x).
%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
%F See page 7 in Brent's paper:
%F a(n) = (2*n-1)^2*A272131(n) - 4*(n-1)^2*A272131(n-1).
%F A272133(n) = (2*n-1)^2*a(n) - 4*(n-1)^2*a(n-1).
%F n*a(n) = 1 + 3^9*(n-1)/(n+1) + 5^9*((n-1)*(n-2))/((n+1)*(n+2)) + ... for n >= 1. See A245244. - _Peter Bala_, Jan 19 2019
%p [6144*n^4-29184*n^3+52416*n^2-41840*n+12465$n=0..30]; # _Muniru A Asiru_, Jan 28 2019
%t Table[6144 n^4 - 29184 n^3 + 52416 n^2 - 41840 n + 12465, {n, 0, 40}]
%t LinearRecurrence[{5,-10,10,-5,1},{12465,1,3281,68385,388849},30] (* _Harvey P. Dale_, Aug 06 2022 *)
%o (Magma) [6144*n^4 - 29184*n^3 + 52416*n^2 - 41840*n + 12465: n in [0..40]];
%o (PARI) lista(nn) = for(n=0, nn, print1(6144*n^4-29184*n^3+52416*n^2-41840*n+12465, ", ")); \\ _Altug Alkan_, Apr 26 2016
%Y Cf. A272131, A272133, A245244.
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, Apr 26 2016