"Somebody incorrectly remembered Fermat's little theorem as saying that the congruence a^{n+1} = a (mod n) holds for all a if n is prime" (Zagier). The sequence gives the set of integers n for which this property is in fact true.
If i == j (mod n), then m^i == m^j (mod n) for all m. The latter congruence generally holds for any (m, n)=1 with i == j (mod k), k being the order of m modulo n, i.e., the least power k for which m^k == 1 (mod n). -
Lekraj Beedassy, Jul 04 2002
Also, numbers n such that n divides denominator of the n-th Bernoulli number B(n) (cf.
A106741). Also, numbers n such that 1^n + 2^n + 3^n + ... + n^n == 1 (mod n). Equivalently, numbers n such that B(n)*n == 1 (mod n). Equivalently, Sum_{prime p, (p-1) divides n} n/p == -1 (mod n). It is easy to see that for n > 1, n must be an even squarefree number. Moreover, the set P of prime divisors of all such n satisfies the property: if p is in P, then p-1 is the product of distinct elements of P. This set is P = {2, 3, 7, 43}, implying that the sequence is finite and complete. -
Max Alekseyev, Aug 25 2013
In 2005, B. C. Kellner proved E. W. Weisstein's conjecture that denom(B_n) = n only if n = 1806. -
Jonathan Sondow, Oct 14 2013
Squarefree numbers n such that b^n == 1 (mod n^2) for every b coprime to n. Squarefree terms of
A341858. -
Thomas Ordowski, Aug 05 2024
Conjecture: Numbers n such that gcd(d+1, n) > 1 for every proper divisor d of n. Verified up to 10^696. -
David Radcliffe, May 29 2025
Also, numbers m such that k^m + (k+1)^m + ... + (k+m-1)^m is prime for some k (cf.
A242927). -
Max Alekseyev, Nov 04 2025
