For integers \(n \in \mathbb{Z}\), we know that the partial sum of their reciprocals grows asymptotically as \[ \displaystyle \sum_{n=1}^x \frac{1}{n} \simeq \ln x \]

Likewise, we know that an analogous result holds for prime numbers \(p \in \mathbb{Z}\):

\[ \displaystyle \sum_{p=1}^x \frac{1}{p} \simeq \ln(\ln x) \]

I'm curious to know if we can go any deeper. I'm looking for a subset \(\mathbb{A}\) of \(\mathbb{Z}\) such that, for \(a \in \mathbb{A} \), we have

\[ \displaystyle \sum_{a=?}^x \frac{1}{a} \simeq \ln(\ln(\ln x)) \]

Does a set like this exist? Is it a subset of the prime numbers, or are there elements of this set which are not primes? What special properties do these numbers have?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestMaybe the primes whose index is also prime, so:

\( 3(2), 5(3), 11(5), 17(7), 31(11), 41(13), \ldots \)

There are \( \frac {N}{\ln N} \) primes up to \( N \in \mathbb{Z} \) and \( \frac {\frac {N}{\ln N}} {\ln \left( \frac {N}{\ln N} \right)} = \frac {N} {(\ln N)^2 - (\ln N) \cdot (\ln (\ln N))} \) of them have a prime index, maybe this helps, but I have no idea how to prove something like that.

Edit:

\( \frac 1{\ln N} \) of the integers up to \( N \) are primes and \( \frac 1 {\ln N} \) of them have a prime index, so there are \( \frac 1 {(\ln N)^2} \) primes with prime index up to \( N \).

How does this fit with \( \frac {N} {(\ln N)^2 - (\ln N) \cdot (\ln (\ln N))} \)?

Log in to reply

I think your asymptotic formulas are correct - I found this paper which corroborates what you did. It looks like the \(-ln \: N \cdot ln(ln \: N)) \) term is a higher-order correction.

I also found that the series

\[ \displaystyle \sum_{x=1}^\infty \frac{1}{x\cdot lnx \cdot lnlnx} \]

Will asymptotically grow on the order of \(lnlnlnx\). I'm not sure if the same will hold for the prime-primes, though.

Assuming \(p_n \simeq n \cdot ln \: n\), maybe we can change the sum above to

\[ \displaystyle \sum_{x=1}^\infty \frac{1}{p_x \cdot ln(p_x / x)} \]

This roughly corresponds to the sequence generated by \(p_x \cdot \lceil ln(p_x / x) \rceil \), which I found (using Python) to be

\[ [2, 3, 5, 7, 11, 13, 17, 19, 23, 58, 62, 74, 82, 86, 94, 106, 118, 122, 134, 142, 146, 158, 166, 178, 194, 202, 206, 214, 218, 226, 254, 262, 274, 278, 298, 302, 314, \] \[326, 334, 346, 358, 362, 382, 386, 394, 398, 422, 446, 454, 458, 466, 478, 482, 502, 514, 526, 538, 542, 554, 562, 566, 586, 614, 622, 626, 634, 662, 674, 694,\] \[698, 706, 718, 734, 746, 758, 766, 778, 794, 802, 818, 838, 842, 862, 866, 878, 886, 898, 914, 922, 926, 934, 958, 974, 982, 998 ] \]

Log in to reply

The formula

\( \displaystyle \sum_{x=1}^{\infty} \frac 1 {x \cdot \ln x \cdot \ln(\ln x)} \)

looks like it could be generalized to give an asymptotic growth of

\( \ln (\ln (\ln (\ldots \ln x \ldots))) \).

About the prime-primes, it was just the first thing that came to my mind and a simple application of the prime number theorem, but of course the growth of prime-primes does not directly connect to the growth of the sum of their reciprocals.

Or does it? Maybe there is a formula that "converts" between these two growth rates...

Log in to reply

Should hold for any integer \(k>0\). I think the \(k=0\) case just returns us to the harmonic series.

There might be some way to find the asymptotic density from those growth rates, which will give us information about whether a set is "dense" or not. I'm not entirely sure how to go about that though.

Log in to reply

I think the best thing is that the formula is so simple, just composed logarithms multiplied.

Log in to reply