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Search: id:A100583
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| A100583 |
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a(n) = 3*n^3+(9/2)*n^2+n+(1/4)*(-1)^n-1/4. |
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+0
2
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| 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
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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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
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LINKS
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Author?, WisFaq (Dutch)
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FORMULA
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a(n) = 4*Sum{i=1 to n}[i^2 + (n+1-i)*(n+1-round(i/2))].
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CROSSREFS
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Sequence in context: A046329 A046377 A075816 this_sequence A036464 A000938 A165618
Adjacent sequences: A100580 A100581 A100582 this_sequence A100584 A100585 A100586
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KEYWORD
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nonn,new
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AUTHOR
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Floor van Lamoen (fvlamoen(AT)hotmail.com), Nov 30 2004
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EXTENSIONS
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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. - njas, Nov 01 2009
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