Assuming A005382 is infinite, a(n) > 0 for arbitrarily large n, see A348172 for the author's conditional proof that the sequence takes on arbitrarily large values. - Charles R Greathouse IV, Oct 07 2021
a(7) = 10598252544.
Proof:
Let tau(n) be the number of divisors of n (Cf. A000005).
We have q = tau(10598252544)/10598252544 = 1/27599616. So any number k that also got that value tau(k)/k must be <= 4/q^2 = 3046955213389824 < 2^52.
Lemma 1. Any number k such that tau(k)/k = 1/27599616 must be 53-smooth.
Proof by contradiction: If k is not 53-smooth then it has a prime factor p >= 59.
But since the denominator of tau(k)/k does not have such a prime factor we need p | tau(k) as well. However if tau(k) has such a prime factor then there exists a prime factor r such that r^(p-1) | k. Such r is at least 2 so r^(p-1) is at least 2^(59 - 1) > 4/q^2 = 3046955213389824. A contradiction since such k is at most 4/q^2 = 3046955213389824.
Lemma 2. There exist exactly 7 values k (including 10598252544) that have tau(k)/k = 1/27599616.
Proof: We should have 27599616 | k. Else, we cannot have the denominator of tau(k)/k being a multiple of 27599616. Checking all 53-smooth numbers <= 3046955213389824 that are multiples of 27599616 give exactly 7 values k that have tau(k)/k = 1/27599616. Those values are: 10598252544, 17222160384, 17663754240, 18215746560, 18546941952, 28703600640 and 30911569920.
Lemma 3: There are no m < 10598252544 for which there are exactly seven k that have tau(m)/m = tau(k)/k.
Proof: By a similar reasoning as shown in proof for lemma 1 we must have such k is 31-smooth since 10598252544 < 2^34. Checking all 31-smooth numbers < 10598252544 gives no m for which there exist 7 such k. (End)
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