a(n) exists for every n, since the sum of the inverses of the primes is infinite.
From Fred Schneider, Sep 20 2006; edited by Danny Rorabaugh, Nov 26 2018: (Start)
Heuristic: Add the squares of several successive primes and then add successive primes until the number is abundant.
a(2) = 5^2 * 7 * 11 * 13 * 17 * 19 * 23 * 29;
a(3) = 7^2 * 11^2 * 13 * 17 * ... * 61 * 67;
a(4) = 11^2 * 13^2 * 17 * 19 * ... * 131 * 137;
a(5) = 13^2 * 17^2 * 19 * 23 * ... * 223 * 227. (End)
a(6) = 17^2 * 19^2 * 23^2 * 29 * 31 * ... * 347 * 349;
a(7) = 19^2 * 23^2 * 29^2 * 31 * 37 * ... * 491 * 499 (both coming from the D. Iannucci paper).  Michel Marcus, May 01 2013
The known terms of this sequence provide Egyptian decompositions of unity in which all the denominators lack the first n primes, as follows: Every term listed in this sequence is a semiperfect number, which means that a subset of its divisors add up to the number itself. The decomposition 1 = 1/a + 1/b + ... + 1/m, where the denominators are a(n) divided by those divisors, is the desired decomposition.  Javier Múgica, Nov 15 2017
a(n) is the product of consecutive primes starting from prime(n+1) raised to nonincreasing powers.  Jianing Song, Apr 10 2021
From Jianing Song, Apr 14 2021: (Start)
By definition, Omega(a(n)) >= A108227(n+1) for all n, where Omega = A001222. For 0 <= n <= 12 we have Omega(a(n)) = A108227(n+1), but this is not true for n = 13, where Omega(a(13)) = 335 > A108227(14) = 334.
We also have omega(a(n)) >= A001276(n+1) for all n, where omega = A001221. The differences for known terms are 0, 0, 1, 1, 2, 3, 2, 3, 4, 4, 5, 6, 6, 6 respectively.
Conjecture: other than a(1) = 945, all terms are cubefree. (End)
