Timeline for Is there a permutation $\tau\in S_n$ with $\tau(1)^{\tau(2)}+\cdots+\tau(n-1)^{\tau(n)}+\tau(n)^{\tau(1)}$ a square?
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| Jun 2, 2021 at 18:45 | comment | added | Erick Wong | @Zhi-WeiSun Randomly one expects $a(n)$ to be at least as large as $(n-1)!/(n-1)^{n/2}$ (significantly higher as the worst case upper bound occurs only $1/(n-1)$ of cases). This doesn’t look that large for small $n$ but it surely grows faster than exponentially. | |
| Apr 3, 2021 at 23:59 | comment | added | Zhi-Wei Sun | I have verified the conjecture for $n=12$ with the example $$1^2 + 2^5 + 5^6 + 6^8 + 8^4 + 4^{11} + 11^9 + 9^7 + 7^{10} + 10^3 + 3^{12}+ 12^1 = 51494^2$$. Jinyuan Wang has computed $a(n)$ for $n=11,12,13$ (cf. oeis.org/A342965), their values are $$a(11)=105,\ a(12)=245,\ a(13)=525.$$ In view of the values of $a(4),\ldots,a(13)$, it seems that $a(n)$ does not grow quickly. By Stirling's formula, $n!\sim (n/e)^n\sqrt{2\pi n}$. | |
| Apr 2, 2021 at 0:16 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
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| Apr 1, 2021 at 16:25 | comment | added | Zhi-Wei Sun | Let $a(n)$ denote the number of permutations $\tau\in S_n$ with $\tau(n)=n$ satisfying the requirement in the conjecture. I have computed $a(n)$ for $n=4,\ldots,11$, namely, $$a(4)=1,\ a(5)=2,\ a(6)=1,\ a(7)=a(8)=6,\ a(9)=10,\ a(10)=27.$$ | |
| Apr 1, 2021 at 16:08 | comment | added | Wolfgang | (But such a pattern would surely be VERY surprising) | |
| Apr 1, 2021 at 15:34 | comment | added | Wolfgang | Maybe there is a secret hope to find a pattern which yields a solution for each n... | |
| Apr 1, 2021 at 15:23 | comment | added | Yaakov Baruch | Why is this interesting? You have $(n-1)!$ shots in the range $[0,n^n]$ (or so). The chances of hitting a square grow fast towards $1$ as $n$ grows. | |
| Apr 1, 2021 at 11:45 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
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| Apr 1, 2021 at 11:32 | history | asked | Zhi-Wei Sun | CC BY-SA 4.0 |