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a(n) = 32*n^2 - 56*n + 25.
4

%I #24 Sep 08 2022 08:46:16

%S 25,1,41,145,313,545,841,1201,1625,2113,2665,3281,3961,4705,5513,6385,

%T 7321,8321,9385,10513,11705,12961,14281,15665,17113,18625,20201,21841,

%U 23545,25313,27145,29041,31001,33025,35113,37265,39481,41761,44105,46513,48985

%N a(n) = 32*n^2 - 56*n + 25.

%C Subsequence of A001844.

%H Vincenzo Librandi, <a href="/A272129/b272129.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_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F O.g.f.: (25 - 74*x + 113*x^2)/(1-x)^3.

%F E.g.f.: (25 - 24*x + 32*x^2)*exp(x).

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

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

%p [32*n^2-56*n+25$n=0..40]; # _Muniru A Asiru_, Jan 28 2019

%t Table[32 n^2 - 56 n + 25, {n, 0, 40}]

%t LinearRecurrence[{3,-3,1},{25,1,41},50] (* _Harvey P. Dale_, Jul 03 2018 *)

%o (Magma) [32*n^2 - 56*n + 25: n in [0..50]];

%o (PARI) lista(nn) = for(n=0, nn, print1(32*n^2-56*n+25, ", ")); \\ _Altug Alkan_, Apr 26 2016

%Y Cf. A001844, A272126, A272127, A272128, A272131, A272132, A272133, A245244.

%K nonn,easy

%O 0,1

%A _Vincenzo Librandi_, Apr 26 2016