OFFSET
0,8
LINKS
Alois P. Heinz, Antidiagonals n = 0..40, flattened
FORMULA
See program.
For k>1: A(n,k) = [x^(k^n)] 1/Product_{j>=0} (1-x^(k^j)).
EXAMPLE
A(2,3) = 5, because there are 5 partitions of 3^2=9 into powers of 3: [1,1,1,1,1,1,1,1,1], [1,1,1,1,1,1,3], [1,1,1,3,3], [3,3,3], [9].
Square array A(n,k) begins:
1, 1, 1, 1, 1, 1, ...
1, 1, 2, 2, 2, 2, ...
1, 1, 4, 5, 6, 7, ...
1, 1, 10, 23, 46, 82, ...
1, 1, 36, 239, 1086, 3707, ...
1, 1, 202, 5828, 79326, 642457, ...
MAPLE
b:= proc(n, j, k) local nn;
nn:= n+1;
if n<0 then 0
elif j=0 or n=0 or k<=1 then 1
elif j=1 then nn
elif n>=j then (nn-j) *binomial(nn, j) *add(binomial(j, h)
/(nn-j+h) *b(j-h-1, j, k) *(-1)^h, h=0..j-1)
else b(n, j, k):= b(n-1, j, k) +b(k*n, j-1, k)
fi
end:
A:= (n, k)-> b(1, n, k):
seq(seq(A(n, d-n), n=0..d), d=0..13);
MATHEMATICA
b[n_, j_, k_] := Module[{nn = n+1}, Which[n < 0, 0, j == 0 || n == 0 || k <= 1, 1, j == 1, nn, n >= j, (nn-j)*Binomial[nn, j]*Sum[Binomial[j, h]/(nn-j+h)* b[j-h-1, j, k]*(-1)^h, {h, 0, j-1}], True, b[n, j, k] = b[n-1, j, k] + b[k*n, j-1, k] ] ]; a[n_, k_] := b[1, n, k]; Table[Table[a[n, d-n], {n, 0, d}], {d, 0, 13}] // Flatten (* Jean-François Alcover, Dec 12 2013, translated from Maple *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Alois P. Heinz, Oct 11 2008
EXTENSIONS
Edited by Alois P. Heinz, Jan 12 2011
STATUS
approved