The Guy-Selfridge Conjecture

I recently collaborated with Boris Alexeev, Evan Conway, Matthieu RosenfeldAndrew Sutherland, Terence Tao, and Markus Uhr on a problem posed in this blog post by Tao, which has now culminated in a joint paper.

While the paper and posts give a more detailed treatment, I wanted to briefly outline the motivating question here.

The setup is simple: we can always write

\displaystyle N! = 1 * 2 * ... * N

as a product of N factors. However, this is not a “smooth” factorization: the relative sizes of the factors vary substantially. The question is, can this be done more equitably? That is, can we write N! as a product of N positive integers that are all roughly the same size?

Not quite. The Stirling Approximation tells us that

\displaystyle N! \sim \sqrt{2\pi N}\left(\frac{N}{e}\right)^N .

This means the (geometric) average of the factors in an N-fold factorization must be roughly \frac{N}{e}. But we also know from e.g. the prime number theorem that there’s always a prime \sim N and strictly less than N. Thus, some factor must contain that prime, and hence must also be at least \sim N, meaning no factorization can avoid having at least one relatively large factor.

Still, it’s natural to ask whether we can avoid small factors. For instance, is it asymptotically true that there is a factorization with every factor at least \sim\frac{N}{e}? And, if we choose reasonable constants c < \frac{1}{e}, say c = \frac{2}{7} or c = \frac{1}{3}, can we establish explicit lower bounds on N, past which there is always a factorization with all factors > cN?

In the original post, Tao established the first question in the affirmative (as well as an associated upper bound), and estimated what his method should yield in terms of effective bounds for the second question. The post ended with an open challenge to sharpen those effective bounds through improved constructions or algorithms. After much back-and-forth discussion in the comments section and later in a shared GitHub repo, we were able to do just that, ultimately resolving the conjecture:

N > 56 suffices for c = \frac{2}{7},
N > 3*10^5 suffices for c = \frac{1}{3}.

This was a fun project! I encourage interested readers to check out the original post, as well as the follow-up post discussing the improved results.

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