# Is there a formula that can predict the primes in the sequence of ratios of consecutive superior highly composite numbers? : $2, 3, 2, 5, 2, 3, 7,…$

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

$2, 3, 2, 5, 2, 3, 7, 2, 11, 13, 2, 3, 5, 17, 19, 2, 23, ...$

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

What I would like to know is where do these primes come from?
Is there a formula which directly gives what is the $n^{th}$ term in this sequence? Or some other kind of relation?

At first I thought that the $n^{th}$ term in this sequence was the smallest prime $p$ such that $p^{a+1}$ is the smallest as possible, where $a$ is the number of times $p$ has already occurred in the sequence.
This correctly predicts the first $4$ terms: $2, 3, 2, 5$
But then it predicts that the next few terms are $7, 2, 3$ instead of $2, 3, 7$
So that's clearly not the answer.

• For those that do not know what a superior highly composite number is: en.wikipedia.org/wiki/Superior_highly_composite_number – Per Alexandersson Oct 27 '14 at 11:45
• And the sequence in question: oeis.org/A000705 – Per Alexandersson Oct 27 '14 at 11:45
• start off with this answer mathoverflow.net/questions/43103/… – Will Jagy Oct 27 '14 at 18:04
• What does: a natural number which has more divisors than any other number scaled relative to the number itself mean? Whatever this would mean, it sounds that there is at most one. – Włodzimierz Holsztyński Oct 27 '14 at 19:45
• @WłodzimierzHolsztyński, grammar notwithstanding, it means the number that gives the maximum value of $d(n)/ n^\delta$ for a fixed real $\delta > 0.$ Proof that there is a maximum is in Hardy and Wright, for example. – Will Jagy Oct 27 '14 at 20:10

## 1 Answer

No, no formula. The correct way to get the SHC numbers in order is to use Ramanujan's original recipe. For any prime $p$ and any integer exponent $k \geq 1$ calculate the real number $$\delta = \frac{\log \left(1 + \frac{1}{k} \right)}{\log p}.$$ As this is smaller than $1/ (k \log p)$ there are only finitely many pairs $(p,k)$ such that $\delta$ is above a given lower bound. Print those out, then print them out with the $\delta$ column in decreasing order. The ordered list tells you which prime comes next. I used lower bound $0.145.$ First, raw output:

1.0000000000000000           2           1
0.5849625007211562           2           2
0.4150374992788437           2           3
0.3219280948873623           2           4
0.2630344058337938           2           5
0.2223924213364480           2           6
0.1926450779423958           2           7
0.1699250014423124           2           8
0.1520030934450501           2           9
0.6309297535714574           3           1
0.3690702464285426           3           2
0.2618595071429148           3           3
0.2031140135750123           3           4
0.1659562328535302           3           5
0.4306765580733931           5           1
0.2519296364125923           5           2
0.1787469216608008           5           3
0.3562071871080222           7           1
0.2083678469455574           7           2
0.1478393401624647           7           3
0.2890648263178878          11           1
0.1690920836734384          11           2
0.2702381544273197          13           1
0.1580791866040749          13           2
0.2446505421182260          17           1
0.2354089133666382          19           1
0.2210647294575037          23           1
0.2058468324604344          29           1
0.2018490865820999          31           1
0.1919587200065601          37           1
0.1866524112389434          41           1
0.1842888331487062          43           1
0.1800313266566926          47           1
0.1745834300480449          53           1
0.1699916162869140          59           1
0.1686130986895011          61           1
0.1648508567221603          67           1
0.1626083122716342          71           1
0.1615554674429964          73           1
0.1586349589155960          79           1
0.1568617748594410          83           1
0.1544226628011101          89           1
0.1515171524096389          97           1


Next, ordered by decreasing $\delta.$ You can read off the next prime in the second column, and the exponent of that prime in the resulting SHC number in the final column:

1.0000000000000000           2           1
0.6309297535714574           3           1
0.5849625007211562           2           2
0.4306765580733931           5           1
0.4150374992788437           2           3
0.3690702464285426           3           2
0.3562071871080222           7           1
0.3219280948873623           2           4
0.2890648263178878          11           1
0.2702381544273197          13           1
0.2630344058337938           2           5
0.2618595071429148           3           3
0.2519296364125923           5           2
0.2446505421182260          17           1
0.2354089133666382          19           1
0.2223924213364480           2           6
0.2210647294575037          23           1
0.2083678469455574           7           2
0.2058468324604344          29           1
0.2031140135750123           3           4
0.2018490865820999          31           1
0.1926450779423958           2           7
0.1919587200065601          37           1
0.1866524112389434          41           1
0.1842888331487062          43           1
0.1800313266566926          47           1
0.1787469216608008           5           3
0.1745834300480449          53           1
0.1699916162869140          59           1
0.1699250014423124           2           8
0.1690920836734384          11           2
0.1686130986895011          61           1
0.1659562328535302           3           5
0.1648508567221603          67           1
0.1626083122716342          71           1
0.1615554674429964          73           1
0.1586349589155960          79           1
0.1580791866040749          13           2
0.1568617748594410          83           1
0.1544226628011101          89           1
0.1520030934450501           2           9
0.1515171524096389          97           1
0.1478393401624647           7           3


It is a theorem of Siegel that there cannot be three different primes $p,q,r$ and three different positive integer exponents $i,j,k$ such that $(p,k); (q,i); (r,j)$ all give the same $\delta$ value. This still leaves the distressing possibility of two primes matching, although no-one expects this, and of course it has never been seen regardless of how far computations have been done.