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Also, numbers n such that n divides s(n), where s(1)=1, s(k)=s(k-1)+k*22^(k-1) (cf. A014940).

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

Also, numbers k such that k divides s(k), where s(1)=1, s(j) = s(j-1) + j*25^(j-1).

Equivalently, numbers k that divide ((24*k - 1)*25^k + 1) / 24^2 (cf. A014943).