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a(n) = n^4 + 3*n^3 + 8*n^2 + 9*n + 2.
7

%I #20 Sep 08 2022 08:46:16

%S 2,23,92,263,614,1247,2288,3887,6218,9479,13892,19703,27182,36623,

%T 48344,62687,80018,100727,125228,153959,187382,225983,270272,320783,

%U 378074,442727,515348,596567,687038,787439,898472,1020863,1155362,1302743,1463804,1639367

%N a(n) = n^4 + 3*n^3 + 8*n^2 + 9*n + 2.

%H Vincenzo Librandi, <a href="/A270868/b270868.txt">Table of n, a(n) for n = 0..1000</a>

%H Andrew Misseldine, <a href="http://arxiv.org/abs/1508.03757">Counting Schur Rings over Cyclic Groups</a>, arXiv preprint arXiv:1508.03757 [math.RA], 2015 (page 19, 5th row; page 21, 4th row).

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F G.f.: (2+13*x-3*x^2+13*x^3-x^4)/(1-x)^5.

%F a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5).

%F From _G. C. Greubel_, Apr 01 2016: (Start)

%F a(2*m) == 0 (mod 2).

%F a(4*m + 2) == 0 (mod 4).

%F E.g.f.: (2 +21*x +24*x^2 +9*x^3 +x^4)*exp(x). (End)

%F a(n)+a(n+2)-2*a(n+1) = 6*A033816(n+1). - _Wesley Ivan Hurt_, Apr 02 2016

%p A270868:=n->n^4 + 3*n^3 + 8*n^2 + 9*n + 2: seq(A270868(n), n=0..50); # _Wesley Ivan Hurt_, Apr 01 2016

%t Table[n^4 + 3 n^3 + 8 n^2 + 9 n + 2, {n, 0, 40}]

%o (Magma) [n^4+3*n^3+8*n^2+9*n+2: n in [0..40]];

%Y Cf. A033816, A270867.

%K nonn,easy

%O 0,1

%A _Vincenzo Librandi_, Apr 01 2016