It is sufficient to consider only prime bases: a(n+1) is the least composite number k such that p^(k1) == 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 ktuples conjecture, if there are no small counterexamples, there are likely to be none.
Here's why.
Assuming prime ktuples, there are infinitely many Carmichael numbers of the form (6k+1)(12k+1)(18k+1), where each factor is prime. And from BatemanHorn, 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^(m1) == 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(m1), with equality iff lpf(a(n)) = prime(m); by this conjecture in the second comment.  Thomas Ordowski, Mar 13 2018
