cp's OEIS Frontend

A208954 a(n) = n^4*(n-1)*(n+1)/12.

%I A208954 #19 Jan 10 2022 03:17:57
%S A208954 0,4,54,320,1250,3780,9604,21504,43740,82500,146410,247104,399854,
%T A208954 624260,945000,1392640,2004504,2825604,3909630,5320000,7130970,
%U A208954 9428804,12313004,15897600,20312500,25704900,32240754,40106304,49509670,60682500,73881680,89391104
%N A208954 a(n) = n^4*(n-1)*(n+1)/12.
%C A208954 The product of a 2 X n matrix and a n X 2 matrix will give a constant result when the entries are the same consecutive numbers for each matrix. These constants for n are listed in the sequence.
%C A208954 Let k be the least consecutive number for the entries in a 2xn matrix. The first row will have entries k, k+1, k+2...k+n-1 and the second row k+n, k+n+1, k+n+2 ...k+2*n-1;
%C A208954 Its n X 2 matrix will have its first column the first row of the 2 X n and its second column the second row.  The product will yield a 2x2 determinant having a value of (n^4)*(n-1)*(n+1)/12 (I thank Professor Daniel Cass for deducing this formula from the data presented.)
%H A208954 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).
%F A208954 G.f.: -2*x^2*(1+x)*(2*x^2+11*x+2) / (x-1)^7 . - _R. J. Mathar_, Dec 17 2012
%F A208954 From _Amiram Eldar_, Jan 10 2022: (Start)
%F A208954 Sum_{n>=2} 1/a(n) = 33 - 2*Pi^2 - 2*Pi^4/15.
%F A208954 Sum_{n>=2} (-1)^n/a(n) = 7*Pi^4/60 + Pi^2 - 21. (End)
%e A208954 For n=4 and k=-3 you get the 2x4 matrix with first row -3,-2,-1,0 and the second row 1,2,3,4.  Multiplying it by its 4x2 matrix will give 320.  If n=4 and k=151, the same 320 results.
%t A208954 Table[(n^4 (n-1)(n+1))/12,{n,40}] (* or *) LinearRecurrence[ {7,-21,35,-35,21,-7,1},{0,4,54,320,1250,3780,9604},40] (* _Harvey P. Dale_, Oct 01 2013 *)
%K A208954 nonn,easy
%O A208954 1,2
%A A208954 _J. M. Bergot_, May 31 2012