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%I #25 Aug 24 2024 15:30:38
%S -427,1,365,2969,10117,24113,47261,81865,130229,194657,277453,380921,
%T 507365,659089,838397,1047593,1288981,1564865,1877549,2229337,2622533,
%U 3059441,3542365,4073609,4655477,5290273,5980301,6727865,7535269,8404817,9338813,10339561
%N a(n) = 384*n^3 - 1184*n^2 + 1228*n - 427.
%H Vincenzo Librandi, <a href="/A272131/b272131.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_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F O.g.f.: (-427 + 1709*x - 2201*x^2 + 3223*x^3)/(1-x)^4.
%F E.g.f.: (-427 + 428*x - 32*x^2 + 384*x^3)*exp(x).
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3.
%F See page 7 in Brent's paper:
%F a(n) = (2*n-1)^2*A272129(n) - 4*(n-1)^2*A272129(n-1).
%F A272132(n) = (2*n-1)^2*a(n) - 4*(n-1)^2*a(n-1).
%F n*a(n) = 1 + 3^7*(n-1)/(n+1) + 5^7*((n-1)*(n-2))/((n+1)*(n+2)) + ... for n >= 1. See A245244. - _Peter Bala_, Jan 19 2019
%p [384*n^3-1184*n^2+1228*n-427$n=0..35]; # _Muniru A Asiru_, Jan 28 2019
%t Table[384 n^3 - 1184 n^2 + 1228 n - 427, {n, 0, 40}]
%t LinearRecurrence[{4,-6,4,-1},{-427,1,365,2969},40] (* _Harvey P. Dale_, Aug 24 2024 *)
%o (Magma) [384*n^3 - 1184*n^2 + 1228*n - 427: n in [0..50]];
%o (PARI) lista(nn) = for(n=0, nn, print1(384*n^3-1184*n^2+1228*n-427, ", ")); \\ _Altug Alkan_, Apr 26 2016
%Y Cf. A272129, A272132, A245244.
%K sign,easy
%O 0,1
%A _Vincenzo Librandi_, Apr 26 2016