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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)
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