

A208954


a(n) = n^4*(n1)*(n+1)/12.


1



0, 4, 54, 320, 1250, 3780, 9604, 21504, 43740, 82500, 146410, 247104, 399854, 624260, 945000, 1392640, 2004504, 2825604, 3909630, 5320000, 7130970, 9428804, 12313004, 15897600, 20312500, 25704900, 32240754, 40106304, 49509670, 60682500, 73881680, 89391104
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OFFSET

1,2


COMMENTS

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.
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+n1 and the second row k+n, k+n+1, k+n+2 ...k+2*n1;
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)*(n1)*(n+1)/12 (I thank Professor Daniel Cass for deducing this formula from the data presented.)


LINKS

Table of n, a(n) for n=1..32.
Index entries for linear recurrences with constant coefficients, signature (7,21,35,35,21,7,1).


FORMULA

G.f.: 2*x^2*(1+x)*(2*x^2+11*x+2) / (x1)^7 .  R. J. Mathar, Dec 17 2012
From Amiram Eldar, Jan 10 2022: (Start)
Sum_{n>=2} 1/a(n) = 33  2*Pi^2  2*Pi^4/15.
Sum_{n>=2} (1)^n/a(n) = 7*Pi^4/60 + Pi^2  21. (End)


EXAMPLE

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.


MATHEMATICA

Table[(n^4 (n1)(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 *)


CROSSREFS

Sequence in context: A095210 A156469 A001545 * A269507 A302942 A292305
Adjacent sequences: A208951 A208952 A208953 * A208955 A208956 A208957


KEYWORD

nonn,easy


AUTHOR

J. M. Bergot, May 31 2012


STATUS

approved



