A116664 Triangle read by rows: T(n,k) is the number of partitions of n into odd parts and having exactly k parts that appear exactly once (n>=0, k>=0).

1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 0, 2, 1, 1, 1, 4, 0, 2, 2, 2, 3, 4, 0, 1, 3, 4, 3, 0, 3, 7, 1, 1, 5, 4, 6, 0, 4, 10, 2, 2, 6, 7, 9, 0, 7, 12, 5, 3, 7, 13, 11, 0, 1, 8, 18, 7, 5, 0, 11, 15, 18, 1, 1, 10, 25, 11, 8, 0, 13, 23, 24, 2, 2, 15, 32, 16, 13, 0, 16, 33, 32, 5, 3, 18, 43, 24, 19, 0, 23, 40

Offset: 0

Author

Emeric Deutsch, Feb 22 2006

Keywords

• nonn
• tabf

Comments

Row n contains 1+floor(sqrt(n)) terms (at the end of certain rows there is an extra 0). Row sums yield A000009. T(n,0)=A097242(n). Sum(k*T(n,k), k>=0)=A116665(n).

Examples

			T(10,2) = 3 because the only partitions of 10 into odd parts and having exactly 2 parts that appear only once are [9,1],[7,3] and [5,3,1,1].
Triangle starts:
1;
0, 1;
1, 0;
1, 1;
1, 0, 1;
1, 2, 0;
2, 1, 1;
1, 4, 0;
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..600, flattened

Crossrefs

Cf. A000009, A097242, A116665.

Programs

Maple

g:=product(1+t*x^(2*j-1)+x^(2*(2*j-1))/(1-x^(2*j-1)),j=1..30): gser:=simplify(series(g,x=0,30)): P[0]:=1: for n from 1 to 25 do P[n]:=coeff(gser,x^n) od: for n from 0 to 25 do seq(coeff(P[n],t,j),j=0..floor(sqrt(n))) od; # yields sequence in triangular form, with one extra 0 in some rows
# second Maple program:
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
       expand(add(b(n-i*j, i-2)*`if`(j=1, x, 1), j=0..n/i))))
    end:
T:= n-> (p-> seq(coeff(p, x, k), k=0..floor(sqrt(n))))
        (b(n, n-irem(n+1, 2))):
seq(T(n), n=0..25);  # Alois P. Heinz, Mar 16 2014

Mathematica

b[n_, i_] :=  b[n, i] = If[n == 0, 1, If[i<1, 0, Expand[Sum[b[n-i*j, i-2]*If[j == 1, x, 1], {j, 0, n/i}]]]]; T[n_] := Function[{p}, Table[Coefficient[p, x, k], {k, 0, Floor[Sqrt[n]]}]][b[n, n-Mod[n+1, 2]]]; Table[T[n], {n, 0, 25}] // Flatten (* Jean-François Alcover, May 13 2015, after Alois P. Heinz *)

Formula

G.f.: product(1+tx^(2j-1)+x^(4j-2)/(1-x^(2j-1)), j=1..infinity).