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If you take the gamma function of a real positive number n, infinitely recursively (gamma(gamma(gamma..(n))).., for various values it either goes down to 1 or goes to infinity above a certain value. The first few digits of the value where it would just evaluate back to itself seem to be around 3.5623824.

Questions

  1. Does this transcendental number already have a name?

  2. Is it equivalent to any differently rendered analytical solution related to anything else for example the Golden Mean?

  3. Does it have a real-world engineering application?

istaivan
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  • The solution of $x_0 = \Gamma(x_0)$? – A rural reader Mar 31 '21 at 01:34
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    I think that's what' I am after, yes. Out of practice in engineering for decades but was clowning around discovering that the MS Desktop calculator app will now take decimals in Factorial function. And brute forced the value above with a little effort. But wondered if analytical solutions had been reached by real mathematicians that elegantly related it to other mathematical areas. – istaivan Mar 31 '21 at 01:37
  • ...and wouldn't have bothered a no-nonsense forum that is just about getting answers except that i saw under google search that this particular thing didn't seem to have a single search result. – istaivan Mar 31 '21 at 01:39
  • Enjoy @istaivan and keep exploring, that's the thing! – A rural reader Mar 31 '21 at 01:40
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    And if you want to see the company you've stepped into, see A219902 at your OEIS friends. – A rural reader Mar 31 '21 at 03:26
  • See too this thread and this one – Raymond Manzoni Mar 31 '21 at 06:42

1 Answers1

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As @A rural reader commented, you are looking for the non-trivial zero of function $$f(x)=\Gamma(x)-x$$ Using numerical methods does not make any problem. To have something nice, knowing that the solution is close to $x=\frac 72$, we can use one single iteration of high-order iterative methods and have explicit approximations.

For Newton method $$x_{(2)}=\frac 72+\frac{28-15 \sqrt{\pi }}{15 \sqrt{\pi } \psi \left(\frac{7}{2}\right)-8}=3.566255560\cdots$$

We can have this kind of expression for any order but, even for Halley, the formula does not fit on a single line. So, I shall just give the numerical values as a function of the order of the method.

$$\left( \begin{array}{ccc} n & x_{(n)} & \text{method} \\ 2 & 3.5662556 & \text{Newton} \\ 3 & 3.5622765 & \text{Halley} \\ 4 & 3.5623845 & \text{Householder} \\ 5 & 3.5623822 & \text{no name} \\ 6 & 3.5623823 & \text{no name} \end{array} \right)$$

This number has not been identified by inverse symbolic calculators.

Claude Leibovici
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