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
