A332276 Heinz numbers of widely totally normal integer partitions.

1, 2, 4, 6, 8, 12, 16, 18, 30, 32, 60, 64, 90, 128, 150, 180, 210, 256, 300, 360, 450, 512, 540, 600, 630, 1024, 1050, 1350, 1500, 2048, 2100, 2250, 2310, 2520, 2940, 3150, 3780, 4096, 4200, 4410, 5880, 8192, 8820, 9450, 10500, 11550, 12600, 13230, 14700

Offset: 1

Author

Gus Wiseman, Feb 12 2020

Keywords

• nonn

Comments

First differs from A317246 in having 630.

A sequence of positive integers is widely totally normal if either it is all 1's (wide) or it covers an initial interval of positive integers (normal) and has widely totally normal run-lengths.

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).

Examples

			The sequence of terms together with their prime indices begins:
    1: {}
    2: {1}
    4: {1,1}
    6: {1,2}
    8: {1,1,1}
   12: {1,1,2}
   16: {1,1,1,1}
   18: {1,2,2}
   30: {1,2,3}
   32: {1,1,1,1,1}
   60: {1,1,2,3}
   64: {1,1,1,1,1,1}
   90: {1,2,2,3}
  128: {1,1,1,1,1,1,1}
  150: {1,2,3,3}
  180: {1,1,2,2,3}
  210: {1,2,3,4}
  256: {1,1,1,1,1,1,1,1}
  300: {1,1,2,3,3}
  360: {1,1,1,2,2,3}
For example, starting with (4,3,2,2,1), the partition with Heinz number 630, and repeatedly taking run-lengths gives (4,3,2,2,1) -> (1,1,2,1) -> (2,1,1) -> (1,2) -> (1,1). These are all normal and the last is all 1's, so 630 belongs to the sequence.
		

Crossrefs

Contains all powers of two A000079 and the primorials A002110.

Heinz numbers of normal integer partitions are A055932.

The case of reversed integer partitions is A332276 (this sequence).

The enumeration of these partitions by sum is A332277.

The enumeration of the generalization to compositions is A332279.

The co-strong version is A332290.

The strong version is A332291.

Cf. A007097, A056239, A133808, A181819, A182850, A305732, A317081, A317089, A317090, A317246, A317492, A329747, A332295, A332296.

Programs

Mathematica

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
gnaQ[y_]:=Or[y=={},Union[y]=={1},And[Union[y]==Range[Max[y]],gnaQ[Length/@Split[y]]]];
Select[Range[1000],gnaQ[primeMS[#]]&]