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A371883
a(n) is the number of divisors d of n such that d^n mod n = d.
4
0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 2, 1, 4, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 3, 1, 1, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 1, 1, 2, 1, 1, 3, 4, 1, 2, 2, 2, 1, 2, 1, 2, 2, 1, 1, 3, 1, 2, 1, 2, 1, 3, 2, 2, 2, 1, 1, 3, 2, 1, 2, 2, 2, 1, 1, 2, 1, 2
OFFSET
1,6
COMMENTS
1 <= a(n) < A000005(n) for n >= 2.
LINKS
EXAMPLE
a(1) = 0: 1 divides 1, but 1^1 mod 1 = 0 (not 1).
a(2) = 1: 1 divides 2, and 1^2 mod 2 = 1;
2 divides 2, but 2^2 mod 2 = 0 (not 2).
a(6) = 2: 1 divides 6, and 1^6 mod 6 = 1;
2 divides 6, but 2^6 mod 6 = 4 (not 2);
3 divides 6, and 3^6 mod 6 = 3;
6 divides 6, but 6^6 mod 6 = 0 (not 6).
MATHEMATICA
a[n_] := DivisorSum[n, 1 &, PowerMod[#, n, n] == # &]; Array[a, 100] (* Amiram Eldar, Apr 11 2024 *)
PROG
(Magma) [#[d: d in Divisors(n) | d^n mod n eq d]: n in [1..100]];
(Python)
from sympy import divisors
def a(n): return sum(1 for d in divisors(n)[:-1] if pow(d, n, n) == d)
print([a(n) for n in range(1, 101)]) # Michael S. Branicky, Apr 10 2024
(PARI) a(n) = sumdiv(n, d, d^n % n == d); \\ Michel Marcus, Apr 20 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved