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A116681
Triangle read by rows: T(n,k) is the number of partitions of n into distinct parts, in which the sum of the odd parts is k (n>=0, 0<=k<=n).
2
1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 2, 0, 0, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 1, 2, 0, 0, 0, 1, 0, 1, 0, 2, 0, 2, 0, 2, 0, 1, 0, 1, 0, 2, 3, 0, 0, 0, 2, 0, 1, 0, 2, 0, 2, 0, 3, 0, 2, 0, 2, 0, 1, 0, 2, 0, 2, 4, 0, 0, 0, 2, 0, 2, 0, 2, 0, 2, 0, 3, 0, 4, 0, 3, 0, 2, 0, 2, 0, 2, 0, 2, 0, 3
OFFSET
0,22
COMMENTS
Row sums yield A000009. T(2n,0)=A000009(n), T(2n-1,0)=0. T(2n,1)=0, T(2n+1,1)=A000009(n), T(n,2)=0. T(n,n)=A000700(n). Sum(k*T(n,k), k=0..n)=A116682(n).
FORMULA
G.f.=product((1+tx^(2j-1))(1+x^(2j)), j=1..infinity).
EXAMPLE
T(10,4)=2 because we have [6,3,1] and [4,3,2,1].
Triangle starts:
1;
0,1;
1,0,0;
0,1,0,1;
1,0,0,0,1;
0,1,0,1,0,1;
MAPLE
g:=product((1+(t*x)^(2*j-1))*(1+x^(2*j)), j=1..30): gser:=simplify(series(g, x=0, 20)): P[0]:=1: for n from 1 to 15 do P[n]:=sort(coeff(gser, x^n)) od: for n from 0 to 15 do seq(coeff(P[n], t, j), j=0..n) od; # yields sequence in triangular form
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Emeric Deutsch, Feb 22 2006
STATUS
approved