a(n) exists for every n, since the sum of the inverses of the primes is infinite.
Heuristic: square the first several primes and then add successive primes until the number is abundant.  Fred Schneider, Sep 20 2006
For terms 3 4 and 5, squaring only the first two will be part of the minimal solution: 49061132957714428902152118459264865645885092682687973 = 11^2 * 13^2 * 17 * 19 * 23 * 29 * 31 * 37 * 41 * 43 * 47 * 53 * 59 * 61 * 67 * 71 * 73 * 79 * 83 * 89 * 97 * 101 * 103 * 107 * 109 * 113 * 127 * 131 * 137  Fred Schneider, Sep 20 2006
a(5) = 13^2 * 17^2 * 19 * 23 * ... * 223 * 227  Fred Schneider, Sep 20 2006
a(6) = 17^2 * 19^2 * 23^2 * 29 * 31 * ... * 347 * 349 and
a(7) = 19^2 * 23^2 * 29^2 * 31 * 37 * ... * 491 * 499 (both coming from 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
