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
Keywords
Examples
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.
Links
- Jane Street, Getting from a to b, September 2023.
Programs
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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]] -
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
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