- #1
nugae
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I've been looking at the value N(n) of N that satisfies the equation
[tex] \sum_{1}^{n}(N-i)^{n}=N^{n} [/tex]
Thus turns out to be
[tex] N(n)=1.5+\frac{n}{ln2}+O(1/n) [/tex]
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!
[tex] \sum_{1}^{n}(N-i)^{n}=N^{n} [/tex]
Thus turns out to be
[tex] N(n)=1.5+\frac{n}{ln2}+O(1/n) [/tex]
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!