A015716 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).
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, 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, 5, 5, 4, 3, 3, 2, 2, 1, 1, 1, 8, 7, 6, 6, 4, 4, 4, 3, 2, 2, 1, 1
Offset: 1
Examples
T(8,3)=2 because we have [5,3] and [4,3,1]. Triangle begins: n/k 1 2 3 4 5 6 7 8 9 10 01: 1 02: 0 1 03: 1 1 1 04: 1 0 1 1 05: 1 1 1 1 1 06: 2 2 1 1 1 1 07: 2 2 1 2 1 1 1 08: 3 2 2 1 2 1 1 1 09: 3 3 3 2 2 2 1 1 1 10: 5 4 4 3 2 2 2 1 1 1 ... 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
Links
- Mircea Merca, Table of n, a(n) for n = 1..7260
Crossrefs
Programs
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Maple
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 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
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Mathematica
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 *) v = Flatten[u] (* A015716 as a sequence *) (* Clark Kimberling, Mar 14 2014 *)
Formula
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
From Mircea Merca, Feb 28 2014: (Start)
T(n,k) = Sum_{j=1..floor(n/k)} (-1)^(j-1)*A000009(n-j*k).
G.f.: for column k: q^k/(1+q^k)*(-q;q)_{inf}. (End)
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