A264405 Triangle read by rows: T(n,k) is the number of integer partitions of n having k repeated parts (each occurrence is counted).

1, 1, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 2, 0, 1, 3, 0, 2, 1, 0, 1, 4, 0, 2, 2, 2, 0, 1, 5, 0, 4, 2, 1, 2, 0, 1, 6, 0, 6, 2, 3, 2, 2, 0, 1, 8, 0, 7, 4, 4, 2, 2, 2, 0, 1, 10, 0, 8, 6, 6, 4, 3, 2, 2, 0, 1, 12, 0, 13, 6, 6, 8, 3, 3, 2, 2, 0, 1, 15, 0, 15, 9, 11, 6, 9, 4, 3, 2, 2, 0, 1, 18, 0, 21, 10, 13, 12, 7, 8, 4, 3, 2, 2, 0, 1

Offset: 0

Author

Emeric Deutsch, Dec 07 2015

Keywords

• nonn
• tabl

Comments

Compare with A264052 where only one occurrence of a repeated part is counted.

Sum of entries in row n = number of partitions of n = A000041(n).

Sum_{k>=0} k*T(n,k) = A194452(n).

Examples

			T(4,2) = 2 because each of the partitions [2,2] and [2,1,1] have 2 repeated parts, while [4], [3,1], [1,1,1,1] have 0 or 4 repeated parts.
Triangle starts:
  1;
  1, 0;
  1, 0, 1;
  2, 0, 0, 1;
  2, 0, 2, 0, 1;
  3, 0, 2, 1, 0, 1;
		

Links

• Alois P. Heinz, Rows n = 0..200, flattened

Crossrefs

Cf. A000009, A000041, A194452, A264052.

Programs

Maple

g := product(1+x^j+t^2*x^(2*j)/(1-t*x^j), j = 1 .. 100): gser := simplify(series(g, x = 0, 30)): for n from 0 to 20 do P[n] := sort(coeff(gser, x, n)) end do: for n from 0 to 20 do seq(coeff(P[n], t, k), k = 0 .. n) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
      add(expand(b(n-i*j, i-1)*`if`(j>1, x^j, 1)), j=0..n/i)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2)):
seq(T(n), n=0..14);  # Alois P. Heinz, Dec 07 2015

Mathematica

b[n_, i_] := b[n, i] = If[n == 0, 1, If[i < 1, 0, Sum[Expand[b[n - i*j, i - 1]*If[j > 1, x^j, 1]], {j, 0, n/i}]]]; T[n_] := Function[p, Table[ Coefficient[p, x, i], {i, 0, n}]][b[n, n]]; Table[T[n], {n, 0, 14}] // Flatten (* Jean-François Alcover, Jan 23 2016, after Alois P. Heinz *)

Formula

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