A304465 If n is prime, set a(n) = 1. Otherwise, start with the multiset of prime factors of n, and given a multiset take the multiset of its multiplicities. Repeating this until a multiset of size 1 is reached, set a(n) to the unique element of this multiset.
0, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 2, 1, 2, 2, 4, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 2, 1, 3, 1, 5, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 6, 2, 3, 1, 2, 2, 3, 1, 2, 1, 2, 2, 2, 2, 3, 1, 2, 4, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 3
Offset: 1
Keywords
Examples
Starting with the multiset of prime factors of 2520 we have {2,2,2,3,3,5,7} -> {1,1,2,3} -> {1,1,2} -> {1,2} -> {1,1} -> {2}, so a(2520) = 2.
Links
Crossrefs
Programs
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Mathematica
Table[Switch[n,1,0,?PrimeQ,1,,NestWhile[Sort[Length/@Split[#]]&,Sort[Last/@FactorInteger[n]],Length[#]>1&]//First],{n,100}]
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PARI
A181819(n) = factorback(apply(e->prime(e),(factor(n)[,2]))); A304465(n) = if(1==n,0,my(t=isprimepower(n)); if(t,t, t=omega(n); if(bigomega(n)==t),t,A304465(A181819(n)))); \\ Antti Karttunen, Nov 08 2018
Formula
a(p^n) = n where p is any prime number.
a(product of n distinct primes) = n.
a(1) = 0; and for n > 1, if n = prime^k, a(n) = k, otherwise, if n is squarefree [i.e., A001221(n) = A001222(n)], a(n) = A001221(n), otherwise a(n) = a(A181819(n)). - Antti Karttunen, Nov 08 2018
Extensions
More terms from Antti Karttunen, Nov 08 2018
Comments