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%I #31 Jan 17 2020 14:57:13
%S 1,1,1,1,1,2,1,1,2,3,1,1,2,4,5,1,1,2,5,8,7,1,1,2,7,17,19,11,1,1,2,9,
%T 36,62,43,15,1,1,2,13,88,253,258,98,22,1,1,2,19,218,1104,1886,1050,
%U 220,30,1,1,2,27,550,5082,15772,14800,4365,504,42,1,1,2,40,1413,24119,140549,241582,118238,18012,1116,56
%N Array read by antidiagonals upwards: T(n,k) = number of partitions of k^n into n-th powers (n>=1, k>=0).
%H Alois P. Heinz, <a href="/A259799/b259799.txt">Antidiagonals n = 1..16, flattened</a>
%H H. L. Fisher, <a href="/A027601/a027601.pdf">Letter to N. J. A. Sloane, Mar 16 1989</a>
%e The array begins:
%e 1, 1, 2, 3, 5, 7, 11, 15, 22, 30, ...
%e 1, 1, 2, 4, 8, 19, 43, 98, 220, 504, ...
%e 1, 1, 2, 5, 17, 62, 258, 1050, 4365, 18012, ...
%e 1, 1, 2, 7, 36, 253, 1886, 14800, 118238, ...
%e 1, 1, 2, 9, 88, 1104, 15772, 241582, ...
%e ...
%p b:= proc(n, i, k) option remember; `if`(n=0 or i=1, 1,
%p `if`(i=2, 1+iquo(n, i^k), b(n, i-1, k)+
%p `if`(i^k>n, 0, b(n-i^k, i, k))))
%p end:
%p T:= (n, k)-> b(k^n, k, n):
%p seq(seq(T(d-k, k), k=0..d-1), d=1..12); # _Alois P. Heinz_, Jul 10 2015
%t b[n_, i_, k_] := b[n, i, k] = If[n==0 || i==1, 1, If[i==2, 1+Quotient[n, i^k], b[n, i-1, k] + If[i^k>n, 0, b[n-i^k, i, k]]]]; T[n_, k_] := b[k^n, k, n]; Table[ Table[ T[d-k, k], {k, 0, d-1}], {d, 1, 12}] // Flatten (* _Jean-François Alcover_, Jul 15 2015, after _Alois P. Heinz_ *)
%Y Rows: A000041, A037444, A259792-A259795.
%Y Columns: A259796, A027601, A259797, A259798.
%Y T(n,n) gives A331402.
%K nonn,tabl
%O 1,6
%A _N. J. A. Sloane_, Jul 06 2015
%E More terms from _Alois P. Heinz_, Jul 10 2015
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