Cyprian’s Last Theorem

I’ve been looking at the value N(n) of N that satisfies the equation

\sum_{i=1}^{n}(N-i)^{n}=N^{n}

Thus turns out to be

N(n)=1.5+\frac{n}{ln2}+O(1/n)

where the O(1/n) term is about 1/400n for n>10.

I’ve verified this by calculation up to about n=1000, using Lenstra’s long integer package LIP.

This result is so beautiful and simple that it must be possible to prove it without brute-force calculation. If anyone has any suggestions as to how to begin then I’d be very grateful.

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2 thoughts on “Cyprian’s Last Theorem

  1. Hi!

    In your equation “Sigma from 1-n (N – i)^n = N^n”

    What is “i”?

    Thanks for your interesting web site,

    Best Wishes,

    Andrew

  2. “i” was the bound variable in the summation. I’ve put it in explicitly now: thank you for pointing out the deficiency!

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