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a(n) = (2/3)*n*(n^3 - 6*n^2 + 11*n - 3).
4

%I #24 Feb 20 2024 09:40:14

%S 0,2,4,6,24,90,252,574,1136,2034,3380,5302,7944,11466,16044,21870,

%T 29152,38114,48996,62054,77560,95802,117084,141726,170064,202450,

%U 239252,280854,327656,380074,438540,503502,575424,654786,742084,837830,942552,1056794,1181116

%N a(n) = (2/3)*n*(n^3 - 6*n^2 + 11*n - 3).

%H Seiichi Manyama, <a href="/A319575/b319575.txt">Table of n, a(n) for n = 0..10000</a>

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

%F a(n) = [x^4] JacobiTheta3(x)^n.

%F a(n) = A319574(n,4).

%F From _Colin Barker_, Oct 02 2018: (Start)

%F G.f.: 2*x*(1 + x)*(1 - 4*x + 7*x^2) / (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) for n>4. (End)

%p a := n -> (2/3)*n*(n^3 - 6*n^2 + 11*n - 3):

%p seq(a(n), n=0..38);

%t A319575[n_] := 2/3*n*(n^3-6*n^2+11*n-3); Array[A319575, 50, 0] (* or *)

%t LinearRecurrence[{5, -10, 10, -5, 1}, {0, 2, 4, 6, 24}, 50] (* _Paolo Xausa_, Feb 20 2024 *)

%o (PARI) concat(0, Vec(2*x*(1 + x)*(1 - 4*x + 7*x^2) / (1 - x)^5 + O(x^40))) \\ _Colin Barker_, Oct 02 2018

%Y Cf. A000012 (m=0), A005843 (m=1), A046092 (m=2), A130809 (m=3), this sequence (m=4), A319576 (m=5), A319577 (m=6).

%Y Column n=4 of A122141.

%Y Cf. A319574.

%K nonn,easy

%O 0,2

%A _Peter Luschny_, Oct 01 2018