A366161 The number of ways to express n^n in the form a^b for integers a and b.

3, 2, 7, 2, 6, 2, 14, 9, 6, 2, 10, 2, 6, 4, 13, 2, 9, 2, 10, 4, 6, 2, 14, 9, 6, 5, 10, 2, 12, 2, 22, 4, 6, 4, 21, 2, 6, 4, 14, 2, 12, 2, 10, 6, 6, 2, 18, 9, 9, 4, 10, 2, 12, 4, 14, 4, 6, 2, 20, 2, 6, 6, 30, 4, 12, 2, 10, 4, 12, 2, 21, 2, 6, 6, 10, 4, 12, 2, 18, 25, 6, 2, 20, 4, 6, 4, 14

Offset

2,1

Comments

Finding the first appearance of 2023 was the subject of an Internet puzzle in September 2023. (See web link.) The least such n for which a(n) = 2023 is 26273633422851562500 = 2^2 * 3^16 * 5^16.

Links

Table of n, a(n) for n=2..88.

Jane Street, Getting from a to b, September 2023.

Example

a(4) = 7, as "4^4 = a^b" has 7 integer solutions: 2^8, (-2)^8, 4^4, (-4)^4, 16^2, (-16)^2, 256^1.

Maple

a:= n-> add(2-irem(d, 2), d=numtheory[divisors](
       igcd(map(i-> i[2], ifactors(n)[2])[])*n)):
seq(a(n), n=2..100);  # Alois P. Heinz, Oct 02 2023

Mathematica

intPowCount[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 = 2 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 A366161(n): return divisor_count((m:=n*gcd(*factorint(n).values()))>>(t:=(m-1&~m).bit_length()))*(t<<1|1) # Chai Wah Wu, Oct 04 2023

Crossrefs

Cf. A000312, A366196.

Sequence in context: A302714 A193574 A209639 * A174238 A361145 A175920

Adjacent sequences: A366158 A366159 A366160 * A366162 A366163 A366164

Keyword

nonn

Author

Andy Niedermaier, Oct 02 2023

Status

approved