A263232 Triangle read by rows: T(n,k) is the number of partitions of n having exactly k parts equal to 3 (n >= 0, 0 <= k <= floor(n/3)).

1, 1, 2, 2, 1, 4, 1, 5, 2, 8, 2, 1, 10, 4, 1, 15, 5, 2, 19, 8, 2, 1, 27, 10, 4, 1, 34, 15, 5, 2, 47, 19, 8, 2, 1, 59, 27, 10, 4, 1, 79, 34, 15, 5, 2, 99, 47, 19, 8, 2, 1, 130, 59, 27, 10, 4, 1, 162, 79, 34, 15, 5, 2, 209, 99, 47, 19, 8, 2, 1, 259, 130, 59, 27, 10, 4, 1

Offset: 0

Author

Emeric Deutsch, Nov 01 2015

Keywords

• nonn
• tabf

Comments

Row n has 1+floor(n/3) terms. Row sums are the partition numbers (A000041). T(n,0) = A027337(n). Sum_{k=0..floor(n/3)} k*T(n,k) = A024787(n).

Examples

			T(7,1) = 4 because we have [4,3], [3,2,2], [3,2,1,1], and [3,1,1,1,1].
T(9,2) = 2 because we have [3,3,2,1] and [3,3,1,1,1].
Triangle starts:
  1;
  1;
  2;
  2, 1;
  4, 1;
  5, 2;
  8, 2, 1.
		

Links

• Alois P. Heinz, Rows n = 0..300, flattened

Crossrefs

Cf. A000041, A027337, A024787, A116599.

Programs

Maple

g := 1/((1-x)*(1-x^2)*(1-t*x^3)*(product(1-x^j, j = 4 .. 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, j), j = 0 .. floor((1/3)*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, `if`(i=3, x, 1)*
      `if`(i>n, 0, b(n-i, i)) +b(n, i-1))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2)):
seq(T(n), n=0..25);  # Alois P. Heinz, Nov 01 2015

Mathematica

b[n_, i_] := b[n, i] = Expand[If[n == 0, 1, If[i < 1, 0, If[i == 3, x, 1]* If[i > n, 0, b[n - i, i]] + b[n, i - 1]]]]; T[n_] := Function[p, Table[ Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, n]]; Table[T[n], {n, 0, 25}] // Flatten (* Jean-François Alcover, Jan 21 2016, after Alois P. Heinz *)

Formula

G.f.: (1-x)*(1-x^2)*(1-tx^3)*Product_{j>=4} (1-x^j).