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A111490
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Sum of the element of the antidiagonals of the numerical array M(m,n) defined as follows. First row (M11, M12, ..., M1n): 1, 1, 1, 1, 1, 1, ... (all 1's). Second row (M21, M22, ..., M2n): 1, 2, 1, 2, 1, 2, ... (sequence 1, 2 repeated). Third row (M31, M32, ..., M3n): 1, 2, 3, 1, 2, 3, 1, 2, 3, ... (sequence 1, 2, 3 repeated). Fourth row (M41, M42, ..., M4n): 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, ... (sequence 1, 2, 3, 4 repeated). And so on.
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20
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1, 2, 4, 5, 9, 9, 15, 16, 21, 23, 33, 29, 41, 45, 51, 52, 68, 65, 83, 81, 91, 99, 121, 109, 128, 138, 152, 152, 180, 168, 198, 199, 217, 231, 253, 234, 270, 286, 308, 298, 338, 326, 368, 372, 384, 404, 450, 422, 463, 470, 500, 506, 558, 546, 584, 576, 610, 636
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Then the sequence is M(1, 2), M(1, 3)+M(2, 3), M(1, 4)+M(2, 4)+M(3, 4), etc. a(n) = Sum_{i=1..n} M(i, n+1).
The successive determinants of the arrays are the factorial numbers (A000142).- Robert G. Wilson v.
Also, a(n)=n+sum of remainders of n mod k, for k=1, 2, 3,..,n. - Juri-Stepan Gerasimov (2stepan(AT)rambler.ru), Aug 30 2009.
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FORMULA
| a(n)=A000027(n)+A004125(n). - Juri-Stepan Gerasimov (2stepan(AT)rambler.ru), Aug 30 2009.
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EXAMPLE
| Considering the 6X6 array:
1, 1, 1, 1, 1, 1
1, 2, 1, 2, 1, 2
1, 2, 3, 1, 2, 3
1, 2, 3, 4, 1, 2
1, 2, 3, 4, 5, 1
1, 2, 3, 4, 5, 6
The third element of the sequence is 1+2+1=4
The fifth element of the sequence is 1+2+3+2+1=9
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MATHEMATICA
| t = Table[Flatten@Table[Range@n, {m, Ceiling[99/n]}], {n, 99}]; f[n_] := Sum[ t[[i, n - i + 1]], {i, n}]; Array[f, 58] (* Robert G. Wilson v *)
(* to view table *) Table[Flatten@Table[Range@n, {m, Ceiling[40/n]}], {n, 10}] // TableForm
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CROSSREFS
| Sequence in context: A116919 A167511 A174112 * A079784 A189210 A188969
Adjacent sequences: A111487 A111488 A111489 * A111491 A111492 A111493
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KEYWORD
| easy,nonn
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AUTHOR
| Paolo P. Lava (paoloplava(AT)gmail.com) & Giorgio Balzarotti (paoloplava(AT)gmail.com), Nov 21 2005
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EXTENSIONS
| Edited and extended by Robert G. Wilson v (rgwv(at)rgwv.com), Nov 22 2005
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