A355522 Triangle read by rows where T(n,k) is the number of reversed integer partitions of n with maximal difference k, if singletons have maximal difference 0.

2, 2, 1, 3, 1, 1, 2, 3, 1, 1, 4, 3, 2, 1, 1, 2, 6, 3, 2, 1, 1, 4, 6, 6, 2, 2, 1, 1, 3, 10, 6, 5, 2, 2, 1, 1, 4, 11, 11, 6, 4, 2, 2, 1, 1, 2, 16, 13, 10, 5, 4, 2, 2, 1, 1, 6, 17, 19, 12, 9, 4, 4, 2, 2, 1, 1, 2, 24, 24, 18, 11, 8, 4, 4, 2, 2, 1, 1

Offset: 2

Author

Gus Wiseman, Jul 08 2022

Keywords

• nonn
• tabl

Comments

The triangle starts with n = 2, and k ranges from 0 to n - 2.

Examples

			Triangle begins:
  2
  2  1
  3  1  1
  2  3  1  1
  4  3  2  1  1
  2  6  3  2  1  1
  4  6  6  2  2  1  1
  3 10  6  5  2  2  1  1
  4 11 11  6  4  2  2  1  1
  2 16 13 10  5  4  2  2  1  1
  6 17 19 12  9  4  4  2  2  1  1
  2 24 24 18 11  8  4  4  2  2  1  1
  4 27 34 22 17 10  7  4  4  2  2  1  1
  4 35 39 33 20 15  9  7  4  4  2  2  1  1
  5 39 56 39 30 19 14  8  7  4  4  2  2  1  1
For example, row n = 8 counts the following reversed partitions:
  (8)         (233)      (35)      (125)    (26)    (116)  (17)
  (44)        (1223)     (134)     (11114)  (1115)
  (2222)      (11123)    (224)
  (11111111)  (11222)    (1124)
              (111122)   (1133)
              (1111112)  (111113)
		

Links

• Gus Wiseman, Sequences counting and ranking integer partitions by the differences of their successive parts.

Crossrefs

Crossrefs found in the link are not repeated here.

Leading terms are A000005.

Row sums are A000041.

Counts m such that A056239(m) = n and A286470(m) = k.

This is a trimmed version of A238353, which extends to k = n.

For minimum instead of maximum we have A238354.

Ignoring singletons entirely gives A238710.

A001522 counts partitions with a fixed point (unproved), ranked by A352827.

A115720 and A115994 count partitions by their Durfee square.

A279945 counts partitions by number of distinct differences.

Cf. A064428, A091602, A179254, A238352, A239455, A286469, A325404, A355524, A355526, A355532.

Programs

Mathematica

Table[Length[Select[Reverse/@IntegerPartitions[n], If[Length[#]==1,0,Max@@Differences[#]]==k&]],{n,2,15},{k,0,n-2}]