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%I #6 May 14 2018 20:23:45
%S 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,
%T 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,
%U 2,3,19,2,2,3,20,2,21,2,2,2,2,3,22,2,4,2,23
%N 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.
%C a(1) = 0 by convention.
%F a(prime(n)) = n.
%F a(p^n) = n where p is any prime number and n > 1.
%F a(product of n > 1 distinct primes) = n.
%e 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.
%t 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}]
%Y Cf. A000005, A001222, A001597, A005117, A007916, A055932, A056239, A112798, A181819, A182850, A182857, A275870, A296150, A303945, A304465.
%K nonn
%O 1,3
%A _Gus Wiseman_, May 13 2018
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