A384237 The number of divisors d of n such that d^d == d (mod n).

1, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 4, 2, 3, 3, 2, 2, 3, 2, 3, 3, 3, 2, 3, 2, 3, 2, 4, 2, 6, 2, 2, 3, 3, 2, 4, 2, 3, 3, 3, 2, 4, 2, 3, 3, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 4, 3, 3, 2, 4, 2, 3, 3, 2, 4, 5, 2, 3, 3, 3, 2, 3, 2, 3, 3, 3, 2, 4, 2, 4, 2, 3, 2, 6, 3, 3, 3, 3, 2, 5, 3, 3, 3, 3, 3, 2, 2, 3, 2, 3

Offset: 1

Author

Juri-Stepan Gerasimov, May 22 2025

Keywords

• nonn
• changed

Examples

			a(2) = 2 because 1^1 = 1 (mod 2) and 2^2 = 2 == 0 (mod 2) where 1 and 2 are divisors of 2.
		

Crossrefs

Cf. A000005, A027750.

Programs

Magma

[1 + #[d: d in Divisors(n) | Modexp(d, d, n) eq d]: n in [1..100]];

Mathematica

a[n_]:=1+Length[Select[Divisors[n],PowerMod[#,#,n]==# &]]; Array[a,100] (* Stefano Spezia, May 25 2025 *)

PARI

a(n) = sumdiv(n, d, Mod(d,n)^d == d); \\ Michel Marcus, May 25 2025

Python

from sympy import divisors
def A384237(n): return 1+sum(1 for d in divisors(n,generator=True) if dChai Wah Wu, May 29 2025

Formula

a(n) = 1 + number of proper divisors d of n such that d^(d-1) == 1 (mod n/d). - Chai Wah Wu, May 29 2025