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 A333433 a(n) is the n-th number m that divides n^m - 1 (or 0 if m does not exist). 3
 1, 0, 4, 21, 8, 1555, 9, 6223, 40, 999, 20, 130801, 24, 4484077, 128, 117, 60, 118285781329, 42, 1432001198261, 104, 819, 72, 302508121, 81, 75625, 200, 61731, 78, 14507145975869, 72, 21958351241, 820, 12321, 289, 4375, 144 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS From Jinyuan Wang, Mar 25 2020: (Start) For n > 2, n < a(n) < q^(n-1), where q is a prime factor of n - 1. If p is a prime, then a(p^e+1) is divisible by p. Proof: we can prove that p | m for m > 1 and n = p^e + 1. If n^m == 1 (mod m) and m > 1 is the minimum value that cannot be divisible by p, then gcd(m, eulerphi(m)) = 1. Thus, m must be of the form q*p_2*...*p_k, where q < p_2 < ... < p_k. Note that q | (n^m - 1) = (n^q - 1)*(Sum_{i=0..(m/q)-1} n)^(i*q)) and n^q - 1 can never be divisible by q. Therefore, Sum_{i=0..(m/q)-1} n^(i*q) == n^(m/q) - 1 == 0 (mod q). Because n^(q-1) == 1 (mod q) and gcd(m/q, q - 1) = 1, then n == 1 (mod q), a contradiction! (End) a(38) <= 14948925257859919. - Giovanni Resta, Apr 15 2020 LINKS OEIS Wiki, 2^n mod n FORMULA a(n) = A333432(n,n). PROG (PARI) {a(n) = if(n==2, 0, my(cnt=0, k=0); while(cnt

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Last modified June 22 14:15 EDT 2021. Contains 345380 sequences. (Running on oeis4.)