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a(n) = 122880*n^5 - 829440*n^4 + 2258688*n^3 - 3076288*n^2 + 2079892*n - 555731.
3

%I #25 Sep 08 2022 08:46:16

%S -555731,1,29525,1657129,16591741,80872529,269614501,711754105,

%T 1604794829,3229552801,5964902389,10302521801,16861638685,26403775729,

%U 39847496261,58283149849,82987617901,115439059265,157331655829,210590358121,277385630909,360148198801

%N a(n) = 122880*n^5 - 829440*n^4 + 2258688*n^3 - 3076288*n^2 + 2079892*n - 555731.

%H Vincenzo Librandi, <a href="/A272133/b272133.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_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

%F O.g.f.: (-555731 + 3334387*x - 8306446*x^2 + 12594614*x^3 - 1244143*x^4 + 8922919*x^5)/(1-x)^6.

%F E.g.f.: (-555731 + 555732*x - 263104*x^2 + 354048*x^3 + 399360*x^4 + 122880*x^5)*exp(x).

%F a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).

%F a(n) = (2*n-1)^2*A272132(n) - 4*(n-1)^2*A272132(n-1), see page 7 in Brent's paper.

%F n*a(n) = 1 + 3^11*(n-1)/(n+1) + 5^11*((n-1)*(n-2))/((n+1)*(n+2)) + ... for n >= 1. See A245244. - _Peter Bala_, Jan 19 2019

%p [122880*n^5-829440*n^4+2258688*n^3-3076288*n^2+2079892*n-555731$n=0..30]; # _Muniru A Asiru_, Jan 28 2019

%t Table[122880 n^5 - 829440 n^4 + 2258688 n^3 - 3076288 n^2 + 2079892 n - 555731, {n, 0, 40}]

%t LinearRecurrence[{6,-15,20,-15,6,-1},{-555731,1,29525,1657129,16591741,80872529},30] (* _Harvey P. Dale_, Feb 10 2021 *)

%o (Magma) [122880*n^5 - 829440*n^4 + 2258688*n^3 -3076288*n^2 + 2079892*n - 555731: n in [0..30]];

%o (PARI) lista(nn) = for(n=0, nn, print1(122880*n^5 - 829440*n^4 + 2258688*n^3 - 3076288*n^2 + 2079892*n - 555731, ", ")); \\ _Altug Alkan_, Apr 26 2016

%Y Cf. A272132, A245244.

%K sign,easy

%O 0,1

%A _Vincenzo Librandi_, Apr 26 2016