It is sufficient to consider only prime bases: a(n+1) is the least composite number k such that p^(k-1) == 1 (mod k) for every prime p <= lpf(a(n)), with a(1) = 561.
Conjecture: a(n+1) is the smallest Carmichael number k such that lpf(k) > lpf(a(n)), with a(1) = 561. It seems that such Carmichael numbers have exactly three prime factors.
The above conjecture is true if A083876(n) < A285549(n) for all n > 1, but has not been proven; there is no a counterexample up to a(n) < 2^64. - Max Alekseyev and Thomas Ordowski, Mar 13 2018
Carl Pomerance (in a letter to the author) wrote, Mar 13 2018: (Start)
Assuming a strong form of the prime k-tuples conjecture, if there are no small counterexamples, there are likely to be none.
Assuming prime k-tuples, there are infinitely many Carmichael numbers of the form (6k+1)(12k+1)(18k+1), where each factor is prime. And from Bateman--Horn, these are fairly thickly distributed. There are other even better triples such as (60k+41)(90k+61)(150k+101), where "better" means the least prime factor is not so far below the cube root.
So, to get into the sequence, a number needs to be a Fermat pseudoprime for every base up to nearly the cube root.
However, it's a theorem that a sufficiently large number with this property must be a Carmichael number. (end)
Theorem: if lpf(a(n)) < m < a(n), then m is prime if and only if p^(m-1) == 1 (mod m) for every prime p <= lpf(a(n)). - Thomas Ordowski, Mar 13 2018
lpf(a(n)) are listed in A300748. - Max Alekseyev, Mar 13 2018
For m > 1, A135720(m) >= A083876(m-1), with equality iff lpf(a(n)) = prime(m); by this conjecture in the second comment. - Thomas Ordowski, Mar 13 2018