A172467 T(n,k) = number of summands in the partitions of n into k parts; a triangular array.

1, 1, 2, 1, 2, 3, 1, 4, 3, 4, 1, 4, 6, 4, 5, 1, 6, 9, 8, 5, 6, 1, 6, 12, 12, 10, 6, 7, 1, 8, 15, 20, 15, 12, 7, 8, 1, 8, 21, 24, 25, 18, 14, 8, 9, 1, 10, 24, 36, 35, 30, 21, 16, 9, 10, 1, 10, 30, 44, 50, 42, 35, 24, 18, 10, 11, 1, 12, 36, 60, 65, 66, 49, 40, 27, 20, 11, 12, 1

Offset

1,3

Comments

The triangles A172467 and A066633 have identical row sums, given by A006128.

Links

Alois P. Heinz, Rows n = 1..141, flattened

Example

First six rows:
1
1 2
1 2 3
1 4 3 4
1 4 6 4 5
1 6 9 8 5 6
partition of 5 into 1 part: 5
partitions of 5 into 2 parts: 4+1, 3+2
partitions of 5 into 3 parts: 3+1+1, 2+2+1
partition of 5 into 4 parts: 2+1+1+1
partition of 5 into 5 parts: 1+1+1+1+1;
consequently row 5 of the triangle is 1,4,6,4,5

Maple

b:= proc(n, i, k) option remember; `if`(n=0, [`if`(k=0, 1, 0), 0],
      `if`(i<1 or k=0, [0$2], ((f, g)-> f+g+[0, g[1]])(b(n, i-1, k),
      `if`(i>n, [0$2], b(n-i, i, k-1)))))
    end:
T:= (n, k)-> b(n$2, k)[2]:
seq(seq(T(n, k), k=1..n), n=1..14);  # Alois P. Heinz, Aug 04 2014

Mathematica

p[n_] := IntegerPartitions[n];
l[n_, j_] := Length[p[n][[j]]]
t = Table[l[n, j], {n, 1, 13}, {j, 1, Length[p[n]]}]
f[n_, k_] := k*Count[t[[n]], k]
t = Table[f[n, k], {n, 1, 13}, {k, 1, n}]
TableForm[t] (* A172467 as a triangle *)
Flatten[t]   (* A172467 as a sequence *)
(* second program: *)
b[n_, i_, k_] := b[n, i, k] = If[n==0, {If[k==0, 1, 0], 0}, If[i<1 || k==0, {0, 0}, Function[{f, g}, f + g + {0, g[[1]]}][b[n, i-1, k], If[i>n, {0, 0}, b[n-i, i, k-1]]]]]; T[n_, k_] := b[n, n, k][[2]]; Table[T[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 27 2015, after Alois P. Heinz *)

Crossrefs

Cf. A006128, A066633.

Sequence in context: A355079 A205682 A217762 * A110582 A162507 A091298

Adjacent sequences: A172464 A172465 A172466 * A172468 A172469 A172470

Keyword

nonn,tabl

Author

Clark Kimberling, Mar 03 2012

Status

approved