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a(n) = T(n, 3), where T(n, k) = Sum_{i=0..n} i^k * binomial(n, i) * (1/2)^(n-k).
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%I #32 Nov 21 2025 06:53:19

%S 0,4,20,54,112,200,324,490,704,972,1300,1694,2160,2704,3332,4050,4864,

%T 5780,6804,7942,9200,10584,12100,13754,15552,17500,19604,21870,24304,

%U 26912,29700,32674,35840,39204,42772,46550,50544,54760,59204,63882,68800

%N a(n) = T(n, 3), where T(n, k) = Sum_{i=0..n} i^k * binomial(n, i) * (1/2)^(n-k).

%C A mean of binomials as might occur as the Expectation of random variables.

%H Andrew Howroyd, <a href="/A366151/b366151.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

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

%F a(n) = [x^n] (2*x*(2 + 2*x - x^2))/(x - 1)^4.

%F a(n) = n! * [x^n] exp(x)*(x^3 + 6*x^2 + 4*x).

%F From _Amiram Eldar_, Nov 21 2025: (Start)

%F Sum_{n>=1} 1/a(n) = Pi^2/18 - 11/54.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/36 - 2*log(2)/9 + 5/54. (End)

%p a := n -> n^2*(n + 3): seq(a(n), n = 0..35);

%t A366151[n_] := n^2*(n + 3); Array[A366151, 50, 0] (* _Paolo Xausa_, Nov 14 2025 *)

%o (PARI) a(n) = n^2*(n+3); \\ _Amiram Eldar_, Nov 21 2025

%Y T(n, 0) = A000012; T(n, 1) = A001477; T(n, 2) = A002378; T(n, 3) = this sequence.

%Y T(1, n) = A011782; T(2, n) = A063376(n) (with offset 0); T(n, n) = A072034(n).

%K nonn,easy

%O 0,2

%A _Peter Luschny_, Oct 27 2023