%I #25 Feb 24 2019 17:51:29
%S 1,0,1,1,0,1,1,1,0,1,2,1,1,0,1,2,2,1,1,0,1,4,2,2,1,1,0,1,4,4,2,2,1,1,
%T 0,1,7,4,4,2,2,1,1,0,1,8,7,4,4,2,2,1,1,0,1,12,8,7,4,4,2,2,1,1,0,1,14,
%U 12,8,7,4,4,2,2,1,1,0,1,21,14,12,8,7,4,4,2,2,1,1,0,1,24,21,14,12,8,7,4,4,2,2,1,1,0,1
%N Triangle read by rows: T(n,k) is the number of partitions of n having exactly k parts equal to 1 (n>=0, 0<=k<=n).
%C Row sums yield the partition numbers (A000041).
%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, not necessarily 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(6,2) = 2 because we have [3,1,1,1] and [4,2]. - _Emeric Deutsch_, Dec 05 2015
%H Alois P. Heinz, <a href="/A116598/b116598.txt">Rows n = 0..140, flattened</a>
%F G.f.: G(t,x) = 1/( (1-t*x)*prod(j>=2, 1-x^j ) ).
%F T(n,k) = p(n-k)-p(n-k-1) for k<n, where p(n) are the partition numbers (A000041).
%F T(n,0) = A002865(n).
%F Sum(k*T(n,k),k=0..n) = A000070(n-1) for n>=1.
%F Column k has g.f. x^k/prod(j>=2, 1-x^j ) (k>=0).
%e T(6,2) = 2 because we have [4,1,1] and [2,2,1,1].
%e Triangle starts:
%e 00: 1,
%e 01: 0, 1,
%e 02: 1, 0, 1,
%e 03: 1, 1, 0, 1,
%e 04: 2, 1, 1, 0, 1,
%e 05: 2, 2, 1, 1, 0, 1,
%e 06: 4, 2, 2, 1, 1, 0, 1,
%e 07: 4, 4, 2, 2, 1, 1, 0, 1,
%e 08: 7, 4, 4, 2, 2, 1, 1, 0, 1,
%e 09: 8, 7, 4, 4, 2, 2, 1, 1, 0, 1,
%e 10: 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1,
%e 11: 14, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1,
%e 12: 21, 14, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1,
%e 13: 24, 21, 14, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1,
%e 14: 34, 24, 21, 14, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1,
%e 15: 41, 34, 24, 21, 14, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1,
%e ...
%p with(combinat): T:=proc(n,k) if k<n then numbpart(n-k)-numbpart(n-k-1) elif k=n then 1 else 0 fi end: for n from 0 to 14 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
%t nn = 20; p = Product[1/(1 - x^i), {i, 2, nn}]; Prepend[CoefficientList[Table[Coefficient[Series[p /(1 - x y), {x, 0, nn}], x^n], {n, 1, nn}], y], 1] // Flatten (* _Geoffrey Critzer_, Jan 22 2012 *)
%Y Cf. A000041, A002865, A000070.
%K nonn,tabl
%O 0,11
%A _Emeric Deutsch_, Feb 18 2006