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On divisor sums due to Erd\H{o}s and Ramanujan

This paper establishes the asymptotic order of magnitude nx1d(d(n))xloglogx\sum_{n \leq x} \frac{1}{d(d(n))} \asymp \frac{x}{\log \log x} for the hybrid divisor sum combining Erdős's and Ramanujan's problems by employing Golomb's estimate for powerful numbers and Turán's quantitative form of the Hardy-Ramanujan theorem.

Original authors: John M. Campbell

Published 2026-05-04
📖 4 min read🧠 Deep dive

Original authors: John M. Campbell

Original paper licensed under CC BY 4.0 (http://creativecommons.org/licenses/by/4.0/). This is an AI-generated explanation of the paper below. It is not written or endorsed by the authors. For technical accuracy, refer to the original paper. Read full disclaimer

Imagine you are a detective trying to understand the hidden patterns of numbers. In the world of mathematics, every whole number (like 1, 2, 3, 100) has a "family" of divisors. For example, the number 12 has divisors 1, 2, 3, 4, 6, and 12. The count of these divisors is called d(n)d(n).

For over a century, famous mathematicians like Ramanujan and Erdős have been trying to figure out what happens when you add up these divisor counts for millions of numbers. They found some beautiful, predictable patterns.

This paper, written by John Campbell, tackles a new, trickier mystery. Instead of just counting the divisors, the author asks: What happens if we take the reciprocal (the "one over") of a very specific, complicated divisor count?

Here is the breakdown of the paper's story, using simple analogies:

1. The Setup: Two Famous Problems

To understand the new problem, we first need to know the two old ones it mixes together:

  • The Ramanujan Problem: Imagine you have a huge crowd of people (numbers). You ask everyone, "How many friends (divisors) do you have?" Ramanujan figured out that if you add up the reciprocals of these friend counts (1 divided by the number of friends), the total grows in a very specific, predictable way.
  • The Erdős Problem: Now, imagine a game of "telephone." You take a number, count its friends (d(n)d(n)), and then count the friends of that number (d(d(n))d(d(n))). Erdős figured out how the total of these "friends of friends" behaves.

2. The New Mystery: The "Hybrid"

Campbell's paper asks: What if we combine these two?
Instead of just counting the "friends of friends" (d(d(n))d(d(n))), what if we take the reciprocal of that number?
Sum of 1d(d(n)) \text{Sum of } \frac{1}{d(d(n))}

This is like asking: "If I look at the 'friends of friends' for every number up to a million, and I add up the fractions 1/that count1/\text{that count}, what is the total?"

The author notes that this is much harder than the original problems. Taking the reciprocal is like turning up the volume on the "weird" numbers. If a number has a very small "friends of friends" count, its reciprocal becomes huge, throwing off the balance of the sum.

3. The Detective Work: Filtering the Crowd

To solve this, Campbell had to split the crowd of numbers into three groups, like sorting a messy room into piles:

  • Pile A (The "Powerful" Numbers): These are numbers with a very heavy, "powerful" structure (mathematically, they are divisible by squares of primes). Campbell used a rule discovered by Golomb to show that these numbers are rare enough that they don't mess up the total sum too much.
  • Pile B (The "Outliers"): These are numbers that have a weird number of prime factors (either way too many or way too few). Using a classic statistical tool called Turán's inequality (which is like a "variance check" in statistics), Campbell proved that these outliers are also rare enough to be ignored for the main calculation.
  • Pile C (The "Normal" Numbers): This is the vast majority of numbers. For these, the "friends of friends" count behaves nicely. Campbell showed that for this group, the value of 1/d(d(n))1/d(d(n)) is roughly the same size for everyone.

4. The Big Reveal

After filtering out the messy piles (A and B), Campbell looked at the main group (C). He found that the sum behaves in a surprisingly simple way.

The Result:
The total sum grows at a rate of roughly:
xlog(logx) \frac{x}{\log(\log x)}
(Where xx is the size of the crowd you are looking at.)

In plain English: If you double the size of your crowd, the sum doesn't double; it grows slightly slower, following a specific "double-log" curve.

5. The "Guess" (Conjecture)

The paper proves the size (order of magnitude) of this sum. However, the author also makes a bold guess (a conjecture):
Just as Ramanujan found a specific constant number for his problem, Campbell suspects there is a specific constant number for this new hybrid problem too. He writes that proving this specific constant exists is likely very difficult, similar to other famous unsolved problems in math.

Summary

John Campbell took a complex mathematical puzzle involving "divisors of divisors" and their reciprocals. By using a mix of old statistical tools and new filtering techniques, he proved that the sum of these values grows at a predictable rate: proportional to the size of the numbers divided by the logarithm of the logarithm of the size.

It's a story of taking a chaotic, irregular mathematical object, sorting it into "normal" and "abnormal" groups, and showing that the "normal" group dictates the final answer.

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