A366196 The number of ways to express n^n in the form a^b for positive integers a and b.

2, 2, 4, 2, 4, 2, 8, 6, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 6, 4, 4, 2, 8, 6, 4, 5, 6, 2, 8, 2, 12, 4, 4, 4, 12, 2, 4, 4, 8, 2, 8, 2, 6, 6, 4, 2, 10, 6, 6, 4, 6, 2, 8, 4, 8, 4, 4, 2, 12, 2, 4, 6, 16, 4, 8, 2, 6, 4, 8, 2, 12, 2, 4, 6, 6, 4, 8, 2, 10, 15, 4, 2, 12, 4

Offset: 2

Author

Andy Niedermaier, Oct 03 2023

Keywords

• nonn

Examples

			a(27) = 5, as "27^27 = a^b" has 5 positive integer solutions: 3^81, 27^27, 19683^9, 7625597484987^3, and (3^81)^1.
		

Crossrefs

Cf. A000312, A366161.

Programs

Maple

a:= n-> numtheory[tau](igcd(map(i-> i[2], ifactors(n)[2])[])*n):
seq(a(n), n=2..100);  # Alois P. Heinz, Oct 03 2023

Mathematica

intPowCountPos[n_] := Module[{m, F, i, t},
  m = n (GCD @@ FactorInteger[n][[All, 2]]);
  t = 0;
  While[Mod[m, 2] == 0,
   t++;
   m = m/2];
  t = t + 1;
  F = FactorInteger[m][[All, 2]];
  If[m > 1,
   For[i = 1, i <= Length[F], i++,
     t = t (F[[i]] + 1)];
   ];
  Return[t]]

Python

from math import gcd
from sympy import divisor_count, factorint
def A366196(n): return divisor_count((m:=n*gcd(*factorint(n).values()))>>(t:=(m-1&~m).bit_length()))*(t+1) # Chai Wah Wu, Oct 04 2023