A117454 Triangle read by rows: T(n,k) is the number of partitions of n into distinct parts such that the difference between the largest and smallest parts is k (n>=1; 0<=k<=n-2 for n>=2).

1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 2, 0, 1, 1, 1, 0, 2, 0, 1, 1, 0, 1, 1, 2, 0, 1, 1, 1, 1, 1, 1, 2, 0, 1, 1, 0, 1, 2, 2, 1, 2, 0, 1, 1, 1, 0, 2, 2, 2, 1, 2, 0, 1, 1, 0, 2, 0, 3, 3, 2, 1, 2, 0, 1, 1, 1, 0, 2, 2, 3, 3, 2, 1, 2, 0, 1, 1, 0, 1, 2, 2, 3, 4, 3, 2, 1, 2, 0, 1, 1, 1, 1, 1, 3, 4, 3, 4, 3, 2, 1, 2, 0

Offset: 1

Author

Emeric Deutsch, Mar 18 2006

Keywords

• nonn
• tabf

Comments

Also number of partitions of n in which all integers smaller than the largest part occur and have k parts smaller than the largest part (n>=1, k>=0). Row 1 has one term; rows j (j>=2) have j-1 terms. Row sums yield A000009. sum(k*T(n,k),k=0..n-2)=A117455(n).

Examples

			T(12,5)=3 because we have [7,3,2],[6,5,1] and [6,3,2,1].
Triangle starts:
1;
1;
1,1;
1,0,1;
1,1,0,1;
1,0,2,0,1;
		

Crossrefs

Cf. A000009, A117455.

Programs

Maple

g:=sum(t^(i-1)*x^(i*(i+1)/2)/(1-x^i)/product(1-t*x^j,j=1..i-1),i=1..20): gser:=simplify(series(g,x=0,20)): for n from 1 to 16 do P[n]:=coeff(gser,x^n) od: 1; for n from 2 to 16 do seq(coeff(P[n],t,j),j=0..n-2) od; # yields sequence in triangular form

Mathematica

z = 20; d[n_] := d[n] = Select[IntegerPartitions[n], DeleteDuplicates[#] == # &]; p[n_, k_] := p[n, k] = d[n][[k]]; t = Table[Max[p[n, k]] - Min[p[n, k]], {n, 1, z}, {k, 1, PartitionsQ[n]}]; u = Table[Count[t[[n]], k], {n, 1, z}, {k, 0, n - 2}];
TableForm[u] (* A117454 as an array *)
Flatten[u]   (* A117454 as a sequence *)
(* Clark Kimberling, Mar 14 2014 *)

Formula

G.f.=G(t,x)=sum(t^(i-1)*x^(i(i+1)/2)/[(1-x^i)product(1-tx^j, j=1..i-1)], i=1..infinity).