A362611 Number of modes in the prime factorization of n.
0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 2, 2, 1, 2, 2, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 3, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 1, 1, 2, 2, 2
Offset: 1
Keywords
Examples
The factorization of 450 is 2*3*3*5*5, modes {3,5}, so a(450) = 2. The factorization of 900 is 2*2*3*3*5*5, modes {2,3,5}, so a(900) = 3. The factorization of 1500 is 2*2*3*5*5*5, modes {5}, so a(1500) = 1. The factorization of 8820 is 2*2*3*3*5*7*7, modes {2,3,7}, so a(8820) = 3.
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
Table[x=Last/@If[n==1,0,FactorInteger[n]];Count[x,Max@@x],{n,100}]
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PARI
a(n) = if(n==1, 0, my(f=factor(n)[,2], m=vecmax(f)); #select(v->v==m, f)) \\ Andrew Howroyd, May 08 2023
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Python
from sympy import factorint def A362611(n): return list(v:=factorint(n).values()).count(max(v,default=0)) # Chai Wah Wu, May 08 2023
Formula
For n > 1, 1 <= a(n) << log n. - Charles R Greathouse IV, May 09 2023
a(n) <= A001221(n), with equality if and only if n is a power of a squarefree number (A072774). - Amiram Eldar, Mar 02 2025
Comments