I know about Kneser, but if we take a simple recursion

$$^{1/d}b=c, a(0)=b, a(n)=b^{\frac{1}{^{d-1}(a(n-1))}}$$



and we can quickly find $c$ for positive integer $d>1$ and positive $b$. But $b$ has bounds, ex.

$$d=2, \left[(\frac{1}{e})^{\frac{1}{e}}; e^e\right]$$ $$d=3, [0; \approx2,421]$$ $$d=4, [\approx0,593; \approx1,904]$$ $$d=5, [0; \approx1,762]$$ $$d=6, [\approx0,543; \approx1,689]$$ $$d=7, [0; \approx1,632]$$ $$d=8, [\approx0,511; \approx1,592]$$

What is the law of those bounds? Why is lower bound for odd $d$ is $0$?

  • 1
    $\begingroup$ How do you define $^xb$ for non-integer $x$? $\endgroup$ – Simply Beautiful Art Jan 13 at 23:19
  • $\begingroup$ @SimplyBeautifulArt, thank you for comment! If $^xb=c$, so $^{1/x}c=b$. $\endgroup$ – user514787 Jan 14 at 0:17
  • $\begingroup$ That doesn't define it though. And what's to guarantee it's uniqueness? $\endgroup$ – Simply Beautiful Art Jan 14 at 0:19

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