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This is the sequence of prime numbers which are the elementary building blocks for the superior highly composite numbers:

$2, 3, 2, 5, 2, 3, 7, 2, 11, 13, 2, 3, 5, 17, 19, 2, 23, 7, 29, 3, 31, 2, 37, 41, 43, 47, ...$

The $n^{th}$ superior highly composite number is the product of the first $n$ primes in this sequence.

To generate this sequence we need to calculate $$\frac{\log \left(1 + \frac{1}{\epsilon} \right)}{\log p}. $$ for many values of the prime $p$ and the exponent $\epsilon$.

Print these values out and sort them in decreasing order. And we obtain the sequence in question.

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This is the sequence of prime numbers which are the elementary building blocks for the colossally abundant numbers:

$2, 3, 2, 5, 2, 3, 7, 2, 11, 13, 2, 3, 5, 17, 19, 23, 2, 29, 31, 7, 3, 37, 41, 43, 2, ...$

The $n^{th}$ colossally abundant number is the product of the first $n$ primes in this sequence.

Notice that the first $15$ terms in this second sequence are the exact same as the first $15$ terms in the first sequence. But after the first $15$ terms the two sequences are different.

My question is: how can we generate this second sequence of primes?

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  • $\begingroup$ this is in Alaoglu and Erdos renyi.hu/~p_erdos/1944-03.pdf Note that, while there are many problems that can be dealy with by Ramanujan's "superior" method, there are only a few, including colossally abundant numbers, where we can solve in closed form both ways: (A) given $\delta > 0$ and a prime $p,$ find the correct exponent; (B) given a prime $p$ and an exponent $k,$ find the first (largest) $\delta$ that gives that $k$ $\endgroup$ – Will Jagy Apr 14 '16 at 18:01
  • $\begingroup$ Theorem 10 on page 455 $\endgroup$ – Will Jagy Apr 14 '16 at 18:04

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