A345935 Number of divisors d of n for which A002034(d) = A002034(n), where A002034(n) is the smallest positive integer k such that n divides k!.
1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 3, 2, 2, 1, 4, 1, 2, 1, 3, 1, 4, 1, 1, 2, 2, 2, 3, 1, 2, 2, 4, 1, 4, 1, 3, 2, 2, 1, 2, 1, 2, 2, 3, 1, 2, 2, 4, 2, 2, 1, 6, 1, 2, 3, 2, 2, 4, 1, 3, 2, 4, 1, 4, 1, 2, 2, 3, 2, 4, 1, 2, 2, 2, 1, 6, 2, 2, 2, 4, 1, 4, 2, 3, 2, 2, 2, 2, 1, 2, 3, 3, 1, 4, 1, 4, 4
Offset: 1
Keywords
Examples
36 has 9 divisors: 1, 2, 3, 4, 6, 9, 12, 18, 36. When A002034 is applied to them, one obtains values [1, 2, 3, 4, 3, 6, 4, 6, 6], thus there are three divisors that obtain the maximal value 6 obtained at 36 itself, therefore a(36) = 3.
Links
Crossrefs
Programs
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Mathematica
a[n_]:=(m=1;While[Mod[m!,n]!=0,m++];m);Table[Length@Select[Divisors@k,a@#==a@k&],{k,100}] (* Giorgos Kalogeropoulos, Jul 03 2021 *)
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PARI
A002034(n) = if(1==n,n,my(s=factor(n)[, 1], k=s[#s], f=Mod(k!, n)); while(f, f*=k++); (k)); \\ After code in A002034. A345935(n) = { my(x=A002034(n)); sumdiv(n,d,A002034(d)==x); };