A304464 Start with the normalized multiset of prime factors of n > 1. Given a multiset, take the multiset of its multiplicities. Repeat this until a multiset of size 1 is obtained. a(n) is the unique element of this multiset.

0, 1, 2, 2, 3, 2, 4, 3, 2, 2, 5, 2, 6, 2, 2, 4, 7, 2, 8, 2, 2, 2, 9, 2, 2, 2, 3, 2, 10, 3, 11, 5, 2, 2, 2, 2, 12, 2, 2, 2, 13, 3, 14, 2, 2, 2, 15, 2, 2, 2, 2, 2, 16, 2, 2, 2, 2, 2, 17, 2, 18, 2, 2, 6, 2, 3, 19, 2, 2, 3, 20, 2, 21, 2, 2, 2, 2, 3, 22, 2, 4, 2, 23

Offset: 1

Author

Gus Wiseman, May 13 2018

Keywords

• nonn

Comments

a(1) = 0 by convention.

Examples

			Starting with the normalized multiset of prime factors of 360, we obtain {1,1,1,2,2,3} -> {1,2,3} -> {1,1,1} -> {3}, so a(360) = 3.
		

Crossrefs

Cf. A000005, A001222, A001597, A005117, A007916, A055932, A056239, A112798, A181819, A182850, A182857, A275870, A296150, A303945, A304465.

Programs

Mathematica

Table[If[n===1,0,NestWhile[Sort[Length/@Split[#]]&,If[n===1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]],Length[#]>1&]//First],{n,100}]

Formula

a(prime(n)) = n.

a(p^n) = n where p is any prime number and n > 1.

a(product of n > 1 distinct primes) = n.