A265255 Triangle read by rows: T(n,k) is the number of partitions of n having k odd singletons (n, k >=0).

1, 0, 1, 2, 1, 2, 4, 0, 1, 2, 5, 8, 1, 2, 4, 11, 14, 3, 5, 9, 20, 0, 1, 24, 8, 10, 16, 37, 1, 2, 41, 15, 21, 28, 65, 3, 5, 66, 30, 39, 49, 108, 9, 10, 104, 57, 69, 0, 1, 80, 178, 19, 20, 163, 99, 120, 1, 2, 128, 286, 39, 37, 248, 170, 201, 3, 5, 203, 448, 73, 68, 372, 284, 327

Offset: 0

Author

Emeric Deutsch, Jan 01 2016

Keywords

• nonn
• tabf
• look

Comments

Sum of entries in row n is A000041(n).

T(n,0) = A265256(n).

Sum_{k>=0} k*T(n,k) = A265257(n).

Examples

			T(6,2) = 2 because each of the partitions [1,2,3], [1,5] of n = 6 has 2 odd singleton, while the other partitions, namely [1,1,1,1,1,1], [1,1,1,1,2], [1,1,2,2], [2,2,2], [1,1,1,3], [3,3], [1,1,4], [2,4], [6], have 0, 0, 0, 0, 1, 0, 0, 0, 0  odd singletons.
Triangle starts:
1;
0, 1;
2;
1, 2;
4, 0, 1;
2, 5;
8, 1, 2.
		

Links

• Alois P. Heinz, Rows n = 0..1000, flattened

Crossrefs

Cf. A000041, A265253, A265256, A265257.

Programs

Maple

g := mul(((1-x^(2*j-1))*(1+t*x^(2*j-1))+x^(4*j-2))/(1-x^j), j = 1 .. 80): gser := simplify(series(g, x = 0, 30)): for n from 0 to 25 do P[n] := sort(coeff(gser, x, n)) end do: for n from 0 to 25 do seq(coeff(P[n], t, q), q = 0 .. degree(P[n])) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(n, i) option remember; expand(
      `if`(n=0, 1, `if`(i<1, 0, add(b(n-i*j, i-1)*
      `if`(j=1 and i::odd, x, 1), j=0..n/i))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2)):
seq(T(n), n=0..30);  # Alois P. Heinz, Jan 01 2016

Mathematica

b[n_, i_] := b[n, i] = Expand[If[n == 0, 1, If[i < 1, 0, Sum[b[n - i*j, i - 1]*If[j == 1 && OddQ[i], x, 1], {j, 0, n/i}]]]]; T[n_] := Function [p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, n]]; Table[T[n], {n, 0, 30}] // Flatten (* Jean-François Alcover, Dec 10 2016, after Alois P. Heinz *)

Formula

G.f.: G(t,x) = Product_{j>=1} ((1 -x^(2j-1))(1+tx^{2j-1}) + x^(4j-2))/ (1-x^j).