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%I #26 Dec 05 2015 19:17:36
%S 1,1,1,2,0,1,2,2,0,1,4,1,1,0,1,4,3,2,1,0,1,7,3,2,1,1,0,1,8,6,2,3,1,1,
%T 0,1,12,5,6,2,2,1,1,0,1,14,11,5,4,3,2,1,1,0,1,21,11,8,5,4,2,2,1,1,0,1,
%U 24,17,11,9,4,5,2,2,1,1,0,1,34,20,15,9,8,4,4,2,2,1,1,0,1,41,30,18,14,9,7,5,4,2,2,1,1,0,1
%N Triangle T(n,k) = number of partitions of n in which the least part occurs exactly k times, k=1..n.
%C Reversed rows converge to A002865. - _Joerg Arndt_, Jul 07 2014
%C T(n,k) is the number of partitions of n for which the difference between the two largest distinct parts is k (in partitions having only 1 part, we assume that 0 is also a part). This follows easily from the definition by taking the conjugate partitions. Example: T(7,2) = 3 because we have [3,1,1,1,1], [3,3,1], and [4,2,1]. - _Emeric Deutsch_, Dec 05 2015.
%H Alois P. Heinz, <a href="/A096144/b096144.txt">Rows n = 1..141, flattened</a>
%F G.f. for k-th column: sum(m>=1, x^(k*m)/prod(i>=m+1, 1-x^i ) ).
%e Triangle starts:
%e 01: 1
%e 02: 1 1
%e 03: 2 0 1
%e 04: 2 2 0 1
%e 05: 4 1 1 0 1
%e 06: 4 3 2 1 0 1
%e 07: 7 3 2 1 1 0 1
%e 08: 8 6 2 3 1 1 0 1
%e 09: 12 5 6 2 2 1 1 0 1
%e 10: 14 11 5 4 3 2 1 1 0 1
%e 11: 21 11 8 5 4 2 2 1 1 0 1
%e 12: 24 17 11 9 4 5 2 2 1 1 0 1
%e 13: 34 20 15 9 8 4 4 2 2 1 1 0 1
%e 14: 41 30 18 14 9 7 5 ...
%e T(7,2)=3 because we have: 5+1+1, 3+2+2, 3+2+1+1. - _Geoffrey Critzer_, Jun 20 2014
%p b:= proc(n, i) option remember; `if`(i=1, x^n,
%p `if`(irem(n, i, 'k')=0, x^k, 0)+
%p add(b(n-i*j, i-1), j=0..(n-1)/i))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n$2)):
%p seq(T(n), n=1..20); # _Alois P. Heinz_, Jun 20 2014
%t nn=20;Table[Take[Map[Drop[#,1]&,Drop[CoefficientList[Series[Sum[y x^k/(1-y x^k) Product[1/(1- x^j),{j,k+1,nn}],{k,1,nn}],{x,0,nn}],{x,y}],1]][[i]],i],{i,1,nn}]//Grid (* _Geoffrey Critzer_, Jun 20 2014 *)
%Y Cf. A002865 (first column), A096373 (second column), A000041 (row sums).
%Y T(2n,n) gives A232697(n). - _Alois P. Heinz_, Jun 20 2014
%K easy,nonn,tabl
%O 1,4
%A _Vladeta Jovovic_, Jul 24 2004
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