Exponential factorial

The exponential factorial of a positive integer n, denoted by n$, is n raised to the power of n − 1, which in turn is raised to the power of n − 2, and so on and so forth in a right-grouping manner. That is, $n\=n^{(n-1)^{(n-2)\cdots }}$ The exponential factorial can also be defined with the recurrence relation $a_{0}=1,\quad a_{n}=n^{a_{n-1}}$ The first few exponential factorials are 1, 1, 2, 9, 262144, etc. (sequence A049384 in the OEIS). For example, 262144 is an exponential factorial since $262144=4^{3^{2^{1}}}$ Using the recurrence relation, the first exponential factorials are: 0$ = 1
1$= 11 = 1 2$ = 21 = 2
3$= 32 = 9 4$ = 49 = 262144
5$= 5262144 = 6206069878...8212890625 (183231 digits) The exponential factorials grow much more quickly than regular factorials or even hyperfactorials. The number of digits in 6$ is approximately 5×10183230.

The sum of the reciprocals of the exponential factorials from 1 onwards is the following transcendental number:

$\sum _{n=1}^{\infty }{\frac {1}{n\}}={\frac {1}{1}}+{\frac {1}{2^{1}}}+{\frac {1}{3^{2^{1}}}}+{\frac {1}{4^{3^{2^{1}}}}}+{\frac {1}{5^{4^{3^{2^{1}}}}}}+{\frac {1}{6^{5^{4^{3^{2^{1}}}}}}}+\ldots =1.611114925808376736\underbrace {11111111\ldots 11111111} _{183213}272243682859\ldots$ This sum is transcendental because it is a Liouville number.

Like tetration, there is currently no accepted method of extension of the exponential factorial function to real and complex values of its argument, unlike the factorial function, for which such an extension is provided by the gamma function. But it is possible to expand it if it is defined in a strip width of 1.