A385662 Number of divisors d of n such that d^d == (-d)^d (mod n).
1, 2, 1, 2, 1, 3, 1, 3, 2, 3, 1, 4, 1, 3, 1, 4, 1, 5, 1, 4, 1, 3, 1, 6, 2, 3, 3, 4, 1, 5, 1, 5, 1, 3, 1, 6, 1, 3, 1, 6, 1, 5, 1, 4, 2, 3, 1, 8, 2, 5, 1, 4, 1, 7, 1, 6, 1, 3, 1, 8, 1, 3, 2, 6, 1, 5, 1, 4, 1, 5, 1, 9, 1, 3, 2, 4, 1, 5, 1, 8, 3, 3, 1, 8, 1, 3, 1, 6, 1, 8, 1, 4, 1, 3, 1, 10, 1, 5, 2, 6
Offset: 1
Programs
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Magma
[1 + #[d: d in [1..n-1] | n mod d eq 0 and Modexp(d,d,n) eq Modexp(-d,d,n)]: n in [1..100]];
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Maple
f:= proc(n) if n::odd then nops(select(d -> d &^ d mod n = 0, numtheory:-divisors(n))) elif n mod 4 = 0 then numtheory:-tau(n/2) else numtheory:-tau(n/2) + procname(n/2) fi end proc: map(f, [$1..100]); # Robert Israel, Aug 04 2025
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Mathematica
a[n_] := DivisorSum[n, 1 &, PowerMod[#, #, n] == PowerMod[-#, #, n] &]; Array[a, 100] (* Amiram Eldar, Aug 04 2025 *)
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PARI
a(n) = sumdiv(n, d, Mod(d, n)^d == Mod(-d, n)^d); \\ Michel Marcus, Aug 04 2025
Comments