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A058395
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Square array read by antidiagonals. Based on triangular numbers (A000217) with each term being the sum of 2 consecutive terms in the previous row.
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4
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1, 0, 1, 3, 1, 1, 0, 3, 2, 1, 6, 3, 4, 3, 1, 0, 6, 6, 6, 4, 1, 10, 6, 9, 10, 9, 5, 1, 0, 10, 12, 15, 16, 13, 6, 1, 15, 10, 16, 21, 25, 25, 18, 7, 1, 0, 15, 20, 28, 36, 41, 38, 24, 8, 1, 21, 15, 25, 36, 49, 61, 66, 56, 31, 9, 1, 0, 21, 30, 45, 64, 85, 102, 104, 80, 39, 10, 1, 28, 21, 36, 55, 81, 113, 146, 168, 160, 111, 48, 11, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,4
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COMMENTS
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Changing the formula by replacing T(2n, 0) = T(n, 3) with T(2n, 0) = T(n, m) for some other value of m would change the generating function to the coefficient of x^n in expansion of (1 + x)^k / (1 - x^2)^m. This would produce A058393, A058394, A057884 (and effectively A007318).
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LINKS
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FORMULA
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T(n, k) = T(n-1, k-1) + T(n, k-1) with T(0, k) = 1, T(2*n, 0) = T(n, 3) and T(2*n + 1, 0) = 0. Coefficient of x^n in expansion of (1 + x)^k / (1 - x^2)^3.
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EXAMPLE
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The array T(n, k) starts:
[0] 1, 0, 3, 0, 6, 0, 10, 0, 15, 0, ...
[1] 1, 1, 3, 3, 6, 6, 10, 10, 15, 15, ...
[2] 1, 2, 4, 6, 9, 12, 16, 20, 25, 30, ...
[3] 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
[4] 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, ...
[5] 1, 5, 13, 25, 41, 61, 85, 113, 145, 181, ...
[6] 1, 6, 18, 38, 66, 102, 146, 198, 258, 326, ...
[7] 1, 7, 24, 56, 104, 168, 248, 344, 456, 584, ...
[8] 1, 8, 31, 80, 160, 272, 416, 592, 800, 1040, ...
[9] 1, 9, 39, 111, 240, 432, 688, 1008, 1392, 1840, ...
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MAPLE
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gf := n -> (1 + x)^n / (1 - x^2)^3: ser := n -> series(gf(n), x, 20):
seq(lprint([n], seq(coeff(ser(n), x, k), k = 0..9)), n = 0..9); # Peter Luschny, Apr 12 2023
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MATHEMATICA
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T[0, k_] := If[OddQ[k], 0, (k+2)(k+4)/8];
T[n_, k_] := T[n, k] = If[k == 0, 1, T[n-1, k-1] + T[n-1, k]];
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CROSSREFS
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The triangle A055252 also appears in half of the array.
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KEYWORD
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AUTHOR
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STATUS
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approved
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