A058395 Square array read by antidiagonals. Based on triangular numbers (A000217) with each term being the sum of 2 consecutive terms in the previous row.
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
Offset: 0
Examples
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, ...
Crossrefs
Programs
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Maple
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
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]]; Table[T[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Apr 13 2023 *)
Formula
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.
Comments