|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,6
|
|
COMMENTS
|
|
|
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)
(PARI) a(n) = sumdiv(n, d, d^n % n == d); \\ Michel Marcus, Apr 20 2024
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|