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19 divides a(n) for n > 1. All powers of 19 are terms. a(n) = 19^(n-1) for all to n < 8, while a(8) = A128356(8) = 148305659 = 410819*19^2.

Prime divisors of a(n) in the order of appearance are {19, 410819, 617311, 1508981, ...}. - Alexander Adamchuk, May 16 2010

All listed terms have 2 distinct prime divisors. Most listed terms are semiprimes, except a(7) = 20231*17^2 and a(8) = 410819*19^2. p = prime(n) divides a(n). Quotients a(n)/prime(n) are listed in A128357 = {10, 7, 311, 127, 23, 157, 343927, ...}. a(15) = 583223 = 47*12409. a(16) = 3775349 = 53*71233.

A128356 = {20, 21, 1555, 889, 253, 2041, 5846759, ...} = Least number k>1 (that is not the power of prime p) such that k divides (p+1)^k-1, where p = prime(n). Most listed terms are primes, except a(7) = 20231*17 and a(8) = 410819*19. a(15) = 12409. a(16) = 71233.

Note that all prime listed terms of {a(n)} coincide with terms of A128456 = {2, 7, 311, 127, 23, 157, 7563707819165039903, 75368484119, 47, 9629, 311, 25679, 821, ...} = least prime factor of ((p+1)^p - 1)/p^2, where p = prime(n).

A000420 are the only odd terms of the sequence. - Robert Israel, Feb 25 2020

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

Primes that divide at least one term of A014960.

Prime p is in this sequence iff the multiplicative order of 24 modulo p is the product of smaller terms of this sequence. - Max Alekseyev, May 26 2010

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).