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%I #11 Apr 17 2014 00:42:49
%S 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,
%T 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,
%U 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
%N 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).
%C 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).
%F 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).
%e T(12,5)=3 because we have [7,3,2],[6,5,1] and [6,3,2,1].
%e Triangle starts:
%e 1;
%e 1;
%e 1,1;
%e 1,0,1;
%e 1,1,0,1;
%e 1,0,2,0,1;
%p 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
%t 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}];
%t TableForm[u] (* A117454 as an array *)
%t Flatten[u] (* A117454 as a sequence *)
%t (* _Clark Kimberling_, Mar 14 2014 *)
%Y Cf. A000009, A117455.
%K nonn,tabf
%O 1,14
%A _Emeric Deutsch_, Mar 18 2006
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