A237750 Number of partitions of n having depth 2; see Comments.

0, 0, 0, 1, 0, 1, 1, 4, 2, 7, 6, 13, 15, 25, 26, 46, 53, 74, 92, 136, 157, 218, 274, 356, 443, 583, 703, 899, 1125, 1447, 1746, 2182, 2661, 3331, 4077, 4997, 6066, 7432, 8984, 10904, 13212, 15845, 19161, 22932, 27526, 32968, 39351, 46778, 55791, 66272, 78480

Offset: 1

Author

Clark Kimberling, Feb 19 2014

Keywords

• nonn
• easy

Comments

Suppose that P is a partition of n. Let x(1), x(2),...,x(k) be the distinct parts of P, and let m(i) be the multiplicity of x(i) in P. Let f(P) be the partition [m(1)*x(1), m(2)*x(2),...,x(k)*m(k)] of n. Define c(0,P) = P, c(1,P) = f(P),..., c(n,P) = f(c(n-1,P)), and define d(P) = least n such that c(n,P) has no repeated parts; d(P) is introduced here as the depth of P. Clearly d(P) = 0 if and only if P is a strict partition, as in A000009. Conjecture: if d >= 0, then 2^d is the least n that has a partition of depth d.

Examples

			The 11 partitions of 6 are partitioned by depth as follows:
  depth 0: 6, 51, 42, 321;
  depth 1: 411, 33, 222, 2211, 21111, 11111;
  depth 2: 3111.
Thus, a(6) = 6, A000009(6) = 4, A237750(6) = 1, A237978(6) = 0.
		

Crossrefs

Cf. A237685, A237978, A366063, A000009, A000041.

Programs

Mathematica

z = 60; c[n_] := c[n] = Map[Length[FixedPointList[Sort[Map[Total, Split[#]], Greater] &, #]] - 2 &, IntegerPartitions[n]]
Table[Count[c[n], 1], {n, 1, z}] (* A237685 *)
Table[Count[c[n], 2], {n, 1, z}] (* this sequence *)
Table[Count[c[n], 3], {n, 1, z}] (* A237978 *)
(* Peter J. C. Moses, Feb 19 2014 *)