A286354 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of Product_{j>=1} (1 - x^j)^k.
1, 1, 0, 1, -1, 0, 1, -2, -1, 0, 1, -3, -1, 0, 0, 1, -4, 0, 2, 0, 0, 1, -5, 2, 5, 1, 1, 0, 1, -6, 5, 8, 0, 2, 0, 0, 1, -7, 9, 10, -5, 0, -2, 1, 0, 1, -8, 14, 10, -15, -4, -7, 0, 0, 0, 1, -9, 20, 7, -30, -6, -10, 0, -2, 0, 0, 1, -10, 27, 0, -49, 0, -5, 8, 0, -2, 0, 0, 1, -11, 35, -12, -70, 21, 11, 25, 9, 0, 1, 0, 0
Offset: 0
Examples
A(3,2) = 2 because we have [2, 1], [2', 1], [2, 1'], [2', 1'] (number of partitions of 3 into an even number of distinct parts with 2 types of each part), [3], [3'] (number of partitions of 3 into an odd number of distinct parts with 2 types of each part) and 4 - 2 = 2. Square array begins: 1, 1, 1, 1, 1, 1, ... 0, -1, -2, -3, -4, -5, ... 0, -1, -1, 0, 2, 5, ... 0, 0, 2, 5, 8, 10, ... 0, 0, 1, 0, -5, -15, ... 0, 1, 2, 0, -4, -6, ...
Links
Crossrefs
Programs
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Maple
A:= proc(n, k) option remember; `if`(n=0, 1, -k* add(numtheory[sigma](j)*A(n-j, k), j=1..n)/n) end: seq(seq(A(n, d-n), n=0..d), d=0..12); # Alois P. Heinz, Jun 21 2018 -
Mathematica
Table[Function[k, SeriesCoefficient[Product[(1 - x^i)^k , {i, Infinity}], {x, 0, n}]][j - n], {j, 0, 12}, {n, 0, j}] // Flatten Table[Function[k, SeriesCoefficient[QPochhammer[x, x, Infinity]^k, {x, 0, n}]][j - n], {j, 0, 12}, {n, 0, j}] // Flatten Table[Function[k, SeriesCoefficient[Sum[(-1)^i*x^(i*(3*i + 1)/2), {i, -Infinity, Infinity}]^k, {x, 0, n}]][j - n], {j, 0, 12}, {n, 0, j}] // Flatten
Formula
G.f. of column k: Product_{j>=1} (1 - x^j)^k.
G.f. of column k: (Sum_{j=-inf..inf} (-1)^j*x^(j*(3*j+1)/2))^k.
Column k is the Euler transform of period 1 sequence [-k, -k, -k, ...].
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