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

%I A366161 #21 Oct 04 2023 19:04:08
%S A366161 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,
%T A366161 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,
%U A366161 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
%N A366161 The number of ways to express n^n in the form a^b for integers a and b.
%C A366161 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.
%H A366161 Jane Street, <a href="https://www.janestreet.com/puzzles/getting-from-a-to-b-index/">Getting from a to b</a>, September 2023.
%e A366161 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.
%p A366161 a:= n-> add(2-irem(d, 2), d=numtheory[divisors](
%p A366161        igcd(map(i-> i[2], ifactors(n)[2])[])*n)):
%p A366161 seq(a(n), n=2..100);  # _Alois P. Heinz_, Oct 02 2023
%t A366161 intPowCount[n_] := Module[{m, F, i, t},
%t A366161   m = n (GCD @@ FactorInteger[n][[All, 2]]);
%t A366161   t = 0;
%t A366161   While[Mod[m, 2] == 0,
%t A366161    t++;
%t A366161    m = m/2];
%t A366161   t = 2 t + 1;
%t A366161   F = FactorInteger[m][[All, 2]];
%t A366161   If[m > 1,
%t A366161    For[i = 1, i <= Length[F], i++,
%t A366161      t = t (F[[i]] + 1)];
%t A366161    ];
%t A366161   Return[t]]
%o A366161 (Python)
%o A366161 from math import gcd
%o A366161 from sympy import divisor_count, factorint
%o A366161 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
%Y A366161 Cf. A000312, A366196.
%K A366161 nonn
%O A366161 2,1
%A A366161 _Andy Niedermaier_, Oct 02 2023