A236293 Triangular array T: T(n,1) = number of constant partitions of n; T(n,k) = number of nonconstant partitions of n that have length k, for k = 2..n-1, for n >= 2.

1, 2, 2, 1, 3, 1, 1, 2, 2, 2, 1, 4, 2, 2, 2, 1, 2, 3, 4, 3, 2, 1, 4, 3, 5, 4, 3, 2, 1, 3, 4, 6, 6, 5, 3, 2, 1, 4, 4, 8, 9, 6, 5, 3, 2, 1, 2, 5, 10, 11, 10, 7, 5, 3, 2, 1, 6, 5, 11, 14, 13, 10, 7, 5, 3, 2, 1, 2, 6, 14, 18, 18, 14, 11, 7, 5, 3, 2, 1, 4, 6, 16

Offset

1,2

Comments

This array occurs naturally in a method for counting the distinct cyclic permutations of the partitions of n (A236292). The row sums, limit of column n, and limit of reversed row n are given by A000041, and (column 1) = A000005. Does column 3 give the nonzero terms of A230059?

Links

Alois P. Heinz, Rows n = 1..150, flattened

Formula

Row n: (d(n), f(2), f(3),..., f(n-1)), where d(n) = (number of divisors of n) = (number of constant partitions of n), and f(k) = number of nonconstant partitions of n, for k = 2,3,...,n-1.

Example

First ten rows:
1
2
2 .. 1
3 .. 1 .. 1
2 .. 2 .. 2 .. 1
4 .. 2 .. 2 .. 2 .. 1
2 .. 3 .. 4 .. 3 .. 2 .. 1
4 .. 3 .. 5 .. 4 .. 3 .. 2 .. 1
3 .. 4 .. 6 .. 6 .. 5 .. 3 .. 2 .. 1
4 .. 4 .. 8 .. 9 .. 6 .. 5 .. 3 .. 2 .. 1
(row 6) = (4,2,2,2,1).  The 5 numbers in (4,2,2,2,1) count these partitions:  (6, 33, 222, 1111);  (51, 42); (411, 321); (3111, 2211);  (211111).  ("Constant partition" is exemplified by 6, 33, 222, 1111; i.e., all the parts are the same number.)

Mathematica

t = Map[Flatten[{Length[#[[1]]], Transpose[Tally[Map[Length, #[[2]]]]][[2]]} &[GatherBy[IntegerPartitions[#], Length[Union[#]] > 1 &]]] &, Range[3, 20]] u = Flatten[t]; Flatten[Prepend[u, {1, 2}]]
(* Peter J. C. Moses, Jan 21 2014 *)

Crossrefs

Cf. A000005, A236292, A230059.

Sequence in context: A120967 A116687 A264033 * A056044 A236097 A239319

Adjacent sequences: A236290 A236291 A236292 * A236294 A236295 A236296

Keyword

nonn,tabf,easy

Author

Clark Kimberling, Jan 22 2014

Status

approved