A286653 Square array A(n,k), n>=0, k>=1, read by antidiagonals, where column k is the expansion of Product_{j>=1} (1 - x^(k*j))/(1 - x^j).
1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 2, 0, 1, 1, 2, 2, 2, 0, 1, 1, 2, 3, 4, 3, 0, 1, 1, 2, 3, 4, 5, 4, 0, 1, 1, 2, 3, 5, 6, 7, 5, 0, 1, 1, 2, 3, 5, 6, 9, 9, 6, 0, 1, 1, 2, 3, 5, 7, 10, 12, 13, 8, 0, 1, 1, 2, 3, 5, 7, 10, 13, 16, 16, 10, 0, 1, 1, 2, 3, 5, 7, 11, 14, 19, 22, 22, 12, 0
Offset: 0
Examples
Square array begins: 1, 1, 1, 1, 1, 1, ... 0, 1, 1, 1, 1, 1, ... 0, 1, 2, 2, 2, 2, ... 0, 2, 2, 3, 3, 3, ... 0, 2, 4, 4, 5, 5, ... 0, 3, 5, 6, 6, 7, ...
Links
- Seiichi Manyama, Antidiagonals n = 0..139, flattened
- Eric Weisstein's World of Mathematics, Partition Function b_k
- Index entries for sequences related to partitions
Crossrefs
Programs
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Maple
b:= proc(n, i, k) option remember; `if`(n=0, [1, 0], `if`(k*i*(i+1)/2
[0, l[1]*j]+l)(b(n-i*j, i-1, k)), j=0..min(n/i, k)))) end: A:= (n, k)-> b(n$2, k-1)[1]: seq(seq(A(n, 1+d-n), n=0..d), d=0..16); # Alois P. Heinz, Oct 17 2018 -
Mathematica
Table[Function[k, SeriesCoefficient[Product[(1 - x^(i k))/(1 - x^i), {i, Infinity}], {x, 0, n}]][j - n + 1], {j, 0, 12}, {n, 0, j}] // Flatten Table[Function[k, SeriesCoefficient[QPochhammer[x^k, x^k]/QPochhammer[x, x], {x, 0, n}]][j - n + 1], {j, 0, 12}, {n, 0, j}] // Flatten
Formula
G.f. of column k: Product_{j>=1} (1 - x^(k*j))/(1 - x^j).
Comments