A116857 Triangle read by rows: T(n,k) is the number of partitions of n into distinct odd parts, the largest of which is k (n>=1, k>=1).

1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 1, 0, 1, 0, 1, 0, 1

Offset

1,137

Comments

Both rows 2n-1 and 2n have 2n-1 terms each. Row sums yield A000700. T(n,2k)=0 Sum(k*T(n,k),k>=1) = A092316(n).

Links

Alois P. Heinz, Rows n = 1..350, flattened

Formula

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

Example

T(20,11)=2 because we have [11,9] and [11,5,3,1].
T(30,17)=3 because we have [17,13],[17,9,3,1] and [17,7,5,1].
Triangle starts:
  1;
  0;
  0,0,1;
  0,0,1;
  0,0,0,0,1;
  0,0,0,0,1;
  0,0,0,0,0,0,1;
  0,0,0,0,1,0,1;
  ...

Maple

g:=sum(t^(2*j-1)*x^(2*j-1)*product(1+x^(2*i-1), i=1..j-1), j=1..30): gser:=simplify(series(g, x=0, 22)): for n from 1 to 20 do P[n]:=sort(coeff(gser, x^n)) od: for n from 1 to 20 do seq(coeff(P[n], t^j), j=1..2*ceil(n/2)-1) od; # yields sequence in triangular form

Crossrefs

Cf. A000700, A092316.

Sequence in context: A370078 A372332 A129182 * A369934 A372505 A322338

Adjacent sequences: A116854 A116855 A116856 * A116858 A116859 A116860

Keyword

nonn,tabf

Author

Emeric Deutsch, Feb 24 2006

Status

approved