%I #22 Nov 10 2020 12:19:03
%S 1,0,2,0,1,2,0,1,1,3,0,0,3,2,2,0,0,2,4,1,4,0,0,1,6,3,3,2,0,0,1,6,4,6,
%T 1,4,0,0,0,6,7,8,3,3,3,0,0,0,5,7,14,4,6,2,4,0,0,0,5,7,18,7,9,5,3,2,0,
%U 0,0,3,10,22,9,14,6,6,1,6,0,0,0,2,9,26,15,19,11,9,3,5,2
%N Triangle read by rows: T(n,k) = number of partitions of n where the maximum number of objects in partitions of any given size is k.
%C Every column is eventually 0; the last row with a nonzero value in column k is A024916(k). T(A024916(k)-i, k) <= P(i), where P is the partition function (A000041); equality holds for 0 <= i <= k. The partition represented by the last number in column k is row k of A010766.
%H Alois P. Heinz, <a href="/A134979/b134979.txt">Rows n = 1..200, flattened</a>
%e For the partition [3,2^2], there are 3 objects in the part of size 3 and 4 objects in the parts of size 2, so this partition is counted towards T(7,4).
%e Triangle T(n,k) begins:
%e 1;
%e 0, 2;
%e 0, 1, 2;
%e 0, 1, 1, 3;
%e 0, 0, 3, 2, 2;
%e 0, 0, 2, 4, 1, 4;
%e 0, 0, 1, 6, 3, 3, 2;
%e 0, 0, 1, 6, 4, 6, 1, 4;
%e 0, 0, 0, 6, 7, 8, 3, 3, 3;
%e 0, 0, 0, 5, 7, 14, 4, 6, 2, 4;
%e 0, 0, 0, 5, 7, 18, 7, 9, 5, 3, 2;
%e 0, 0, 0, 3, 10, 22, 9, 14, 6, 6, 1, 6;
%e ...
%p b:= proc(n, i, m) option remember; `if`(n=0 or i=1, x^
%p max(m, n), add(b(n-i*j, i-1, max(m, i*j)), j=0..n/i))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n$2, 0)):
%p seq(T(n), n=1..20); # _Alois P. Heinz_, Feb 07 2020
%t b[n_, i_, m_] := b[n, i, m] = If[n == 0 || i == 1, x^Max[m, n], Sum[b[n - i j, i - 1, Max[m, i j]], {j, 0, n/i}]];
%t T[n_] := Table[Coefficient[b[n, n, 0], x, i], {i, 1, n}];
%t Array[T, 20] // Flatten (* _Jean-François Alcover_, Nov 10 2020, after _Alois P. Heinz_ *)
%Y Cf. A008284, A091602, A000041 (row sums), A000005 (main diagonal), A032741 (2nd diagonal), A010766.
%Y Column sums give A332233.
%K nonn,look,tabl
%O 1,3
%A _Franklin T. Adams-Watters_, Feb 04 2008