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%I #43 May 09 2019 15:02:17
%S 1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,2,2,1,1,1,1,2,2,1,2,1,1,1,3,2,2,1,2,1,
%T 1,1,3,3,3,2,2,2,1,1,1,5,4,4,3,2,2,2,1,1,1,5,5,4,3,3,3,2,2,1,1,1,7,6,
%U 5,5,4,3,3,2,2,1,1,1,8,7,6,6,4,4,4,3,2,2,1,1
%N Triangle read by rows: T(n,k) is the number of partitions of n into distinct parts, one of which is k (1<=k<=n).
%C Row sums yield A015723. T(n,1)=A025147(n-1); T(n,2)=A015744(n-2); T(n,3)=A015745(n-3); T(n,4)=A015746(n-4); T(n,5)=A015750(n-5). - _Emeric Deutsch_, Mar 29 2006
%C Number of parts of size k in all partitions of n into distinct parts. Number of partitions of n-k into distinct parts not including a part of size k. - _Franklin T. Adams-Watters_, Jan 24 2012
%H Mircea Merca, <a href="/A015716/b015716.txt">Table of n, a(n) for n = 1..7260</a>
%F G.f.: G(t,x) = Product_{j>=1} (1+x^j) * Sum_{i>=1} t^i*x^i/(1+x^i). - _Emeric Deutsch_, Mar 29 2006
%F From _Mircea Merca_, Feb 28 2014: (Start)
%F a(n) = A238450(n) + A238451(n).
%F T(n,k) = Sum_{j=1..floor(n/k)} (-1)^(j-1)*A000009(n-j*k).
%F G.f.: for column k: q^k/(1+q^k)*(-q;q)_{inf}. (End)
%e T(8,3)=2 because we have [5,3] and [4,3,1].
%e Triangle begins:
%e n/k 1 2 3 4 5 6 7 8 9 10
%e 01: 1
%e 02: 0 1
%e 03: 1 1 1
%e 04: 1 0 1 1
%e 05: 1 1 1 1 1
%e 06: 2 2 1 1 1 1
%e 07: 2 2 1 2 1 1 1
%e 08: 3 2 2 1 2 1 1 1
%e 09: 3 3 3 2 2 2 1 1 1
%e 10: 5 4 4 3 2 2 2 1 1 1
%e ...
%e The strict integer partitions of 6 are {(6), (5,1), (4,2), (3,2,1)}, with multiset union {1,1,2,2,3,4,5,6}, with multiplicities (2,2,1,1,1,1), which is row n = 6. - _Gus Wiseman_, May 07 2019
%p g:=product(1+x^j,j=1..50)*sum(t^i*x^i/(1+x^i),i=1..50): gser:=simplify(series(g,x=0,18)): for n from 1 to 14 do P[n]:=sort(coeff(gser,x^n)) od: for n from 1 to 14 do seq(coeff(P[n],t,j),j=1..n) od; # yields sequence in triangular form - _Emeric Deutsch_, Mar 29 2006
%p seq(seq(coeff(x^k*(product(1+x^j, j=1..n))/(1+x^k), x, n), k=1..n), n=1..13); # _Mircea Merca_, Feb 28 2014
%t z = 15; d[n_] := d[n] = Select[IntegerPartitions[n], DeleteDuplicates[#] == # &]; p[n_, k_] := p[n, k] = d[n][[k]]; s[n_] := s[n] = Flatten[Table[p[n, k], {k, 1, PartitionsQ[n]}]]; t[n_, k_] := Count[s[n], k]; u = Table[t[n, k], {n, 1, z}, {k, 1, n}]; TableForm[u] (* A015716 as a triangle *)
%t v = Flatten[u] (* A015716 as a sequence *)
%t (* _Clark Kimberling_, Mar 14 2014 *)
%Y Cf. A015723, A015744, A015745, A015746, A015750, A025147, A027293, A066633.
%Y Cf. A006128, A022629, A066189, A246867, A325504, A325506, A325513.
%K nonn,tabl
%O 1,16
%A _Clark Kimberling_
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