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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)

If we replace "abundant" in the definition with "non-deficient", we get the same sequence with an initial 2 instead of 3, barring an astronomically unlikely coincidence with some as-yet-undiscovered odd perfect number. [This is sequence A107705. - M. F. Hasler, Jun 14 2017]

It appears that all terms >= 5 correspond to the odd primitive abundant numbers (A006038) which are products of consecutive primes (cf. A285993), i.e., of the form N = Product_{0<=iM. F. Hasler, May 08 2017

From Jianing Song, Apr 21 2021: (Start)

Let x_1 < x_2 < ... < x_k < ... be the numbers of the form p of p^2 + p, where p is a prime >= prime(n). Then a(n) is the smallest N such that Product_{i=1..N} (1 + 1/x_i) > 2. See my link below for a proof.

For example, for n = 3, we have {x_1, x_2, ..., x_k, ...} = {5, 7, 11, 13, 17, 19, 23, 29, 5^2 + 5, ...}, we have Product_{i=1..8} (1 + 1/x_i) < 2 and Product_{i=1..9} (1 + 1/x_i) > 2, so a(3) = 9. (End)

A perfect (or abundant) number with prime(n) as its lowest prime factor must be divisible by a prime greater than or equal to a(n).

First negative term is a(510510) = -686785. See A001276.

Each term has at least A001276(4) = 15 distinct prime factors and A108227(4) = 18 prime factors counted with multiplicity. - Jianing Song, Apr 13 2021

The smallest term with exactly 15 distinct prime factors is a(830) = 465709156638373299218537971 = 7^3 * 11^2 * 13^2 * 17^2 * 19 * 23 * ... * 61. - Jianing Song, Apr 14 2021