

A100583


a(n) = 3*n^3+(9/2)*n^2+n+(1/4)*(1)^n1/4.


1



0, 8, 44, 124, 268, 492, 816, 1256, 1832, 2560, 3460, 4548, 5844, 7364, 9128, 11152, 13456, 16056, 18972, 22220, 25820, 29788, 34144, 38904, 44088, 49712, 55796, 62356, 69412, 76980, 85080, 93728, 102944, 112744, 123148, 134172, 145836
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OFFSET

0,2


COMMENTS

Conjectured to be the number of triangles in an n X n grid of squares with all diagonals drawn.
I contacted the author, Floor van Lamoen, but he is not aware of a proof that this sequence actually gives the number of triangles in an n X n grid of squares with diagonals. Robbert van der Kruk (robbertvdkruk(AT)live.nl), Oct 28 2009


LINKS

Table of n, a(n) for n=0..36.
Author?, WisFaq (Dutch)


FORMULA

a(n) = 4*Sum{i=1 to n}[i^2 + (n+1i)*(n+1round(i/2))].


CROSSREFS

Sequence in context: A046377 A075816 A188148 * A036464 A000938 A165618
Adjacent sequences: A100580 A100581 A100582 * A100584 A100585 A100586


KEYWORD

nonn


AUTHOR

Floor van Lamoen (fvlamoen(AT)hotmail.com), Nov 30 2004


EXTENSIONS

In view of Robbert van der Kruk's comment, I have used the first formula as the definition, and stated the number of triangles connection as a conjecture.  N. J. A. Sloane, Nov 01 2009


STATUS

approved



