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a(n) = n^2*(n+1)*(n+2)^2/6.
1

%I #22 Oct 24 2023 23:15:23

%S 0,3,32,150,480,1225,2688,5292,9600,16335,26400,40898,61152,88725,

%T 125440,173400,235008,312987,410400,530670,677600,855393,1068672,

%U 1322500,1622400,1974375,2384928,2861082

%N a(n) = n^2*(n+1)*(n+2)^2/6.

%H Delbert L. Johnson, <a href="/A004256/b004256.txt">Table of n, a(n) for n = 0..19999</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

%F From _Harvey P. Dale_, May 26 2015: (Start)

%F G.f.: (3*x+14*x^2+3*x^3)/(x-1)^6.

%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); a(0)=0, a(1)=3, a(2)=32, a(3)=150, a(4)=480, a(5)=1225. (End)

%F From _Amiram Eldar_, Nov 02 2021: (Start)

%F Sum_{n>=1} 1/a(n) = 3/8.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 69/8 - 12*log(2). (End)

%t Table[(n^2 (n+1)(n+2)^2)/6,{n,0,40}] (* or *) LinearRecurrence[{6,-15,20,-15,6,-1},{0,3,32,150,480,1225},40] (* _Harvey P. Dale_, May 26 2015 *)

%o (Magma) [n^2*(n+1)*(n+2)^2/6 : n in [0..50]]; // _Wesley Ivan Hurt_, Jan 22 2022

%Y Equals (n+2) * A002417.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_