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

2,1

Links

Table of n, a(n) for n=2..85.

Example

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.

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]]

Prog

(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

Crossrefs

Cf. A000312, A366161.

Sequence in context: A318312 A318474 A326306 * A278525 A357531 A318476

Adjacent sequences: A366193 A366194 A366195 * A366197 A366198 A366199

Keyword

nonn

Author

Andy Niedermaier, Oct 03 2023

Status

approved