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%I #20 Sep 03 2019 15:23:46
%S 1,1,0,1,1,0,1,2,1,0,1,3,3,2,0,1,4,6,6,2,0,1,5,10,13,9,3,0,1,6,15,24,
%T 24,14,4,0,1,7,21,40,51,42,22,5,0,1,8,28,62,95,100,73,32,6,0,1,9,36,
%U 91,162,206,190,120,46,8,0,1,10,45,128,259,384,425,344,192,66,10,0
%N 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.
%C A(n,k) is the number of partitions of n into distinct parts (or odd parts) with k types of each part.
%H Seiichi Manyama, <a href="/A286335/b286335.txt">Antidiagonals n = 0..139, flattened</a>
%H N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>
%F G.f. of column k: Product_{j>=1} (1 + x^j)^k.
%F A(n,k) = Sum_{i=0..k} binomial(k,i) * A308680(n,k-i). - _Alois P. Heinz_, Aug 29 2019
%e A(3,2) = 6 because we have [3], [3'], [2, 1], [2', 1], [2, 1'] and [2', 1'] (partitions of 3 into distinct parts with 2 types of each part).
%e Also A(3,2) = 6 because we have [3], [3'], [1, 1, 1], [1, 1, 1'], [1, 1', 1'] and [1', 1', 1'] (partitions of 3 into odd parts with 2 types of each part).
%e Square array begins:
%e 1, 1, 1, 1, 1, 1, ...
%e 0, 1, 2, 3, 4, 5, ...
%e 0, 1, 3, 6, 10, 15, ...
%e 0, 2, 6, 13, 24, 40, ...
%e 0, 2, 9, 24, 51, 95, ...
%e 0, 3, 14, 42, 100, 206, ...
%p b:= proc(n, i, k) option remember; `if`(n=0, 1, `if`(i<1, 0, add(
%p (t-> b(t, min(t, i-1), k)*binomial(k, j))(n-i*j), j=0..n/i)))
%p end:
%p A:= (n, k)-> b(n$2, k):
%p seq(seq(A(n, d-n), n=0..d), d=0..12); # _Alois P. Heinz_, Aug 29 2019
%t Table[Function[k, SeriesCoefficient[Product[(1 + x^i)^k , {i, Infinity}], {x, 0, n}]][j - n], {j, 0, 11}, {n, 0, j}] // Flatten
%Y Columns k=0-32 give: A000007, A000009, A022567-A022596.
%Y Rows n=0-2 give: A000012, A001477, A000217.
%Y Main diagonal gives A270913.
%Y Antidiagonal sums give A299106.
%Y Cf. A144064, A286352, A308680.
%K nonn,tabl
%O 0,8
%A _Ilya Gutkovskiy_, May 07 2017
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