All the terms are even.
The number a(n)^(2n) + 1 has all divisors d == 1 (mod 2n).
Conjecture: a(n) exists for every n. This is implied by the generalized Bunyakovsky conjecture (Schinzel's hypothesis H).
Theorem: a(n) = 2 if and only if n is a power of 2.
Note: rad(2n) divides rad(a(n)), where rad(m) =
A007947(m).
Even numbers 2n such that a(n) = rad(2n) are powers of two and 6, 10, 12, 14, 26, 36, ... Are there infinitely many such numbers?
We have a(n) = 2n = 2, 6, 10, 14, 20, 26, 28, ...
Problem: are there infinitely many even numbers m <> 2^k such that the number m^m + 1 has all divisors d == 1 (mod m)?
Additional terms: a(46) = 46, a(47) = 94, a(48) = 12, a(49) = 1246, a(50) = 1960, a(52) = 208, a(53) = 636, a(55) = 17600, a(56) = 476.
a(45) > 1000000 (sequence
A298398 likewise has a very large value for n=45).
a(51) >= 16524 (a C241 remains to be factored to verify b=16524).
a(54) >= 6864 (a C201 remains to be factored to verify b=6864).
(End)
