

A366161


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


2



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



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



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(2irem(d, 2), d=numtheory[divisors](
igcd(map(i> i[2], ifactors(n)[2])[])*n)):


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:=(m1&~m).bit_length()))*(t<<11) # Chai Wah Wu, Oct 04 2023


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



