Nearest factorial to a power of 2

Question

Is this possible to find the distance between a big power of 2 lets say 2^77233199 and the nearest factorial before or after this power of 2 ? My guess is to estimate it with Stirling formula.

Answer 1

If I understand you correctly, then you can do the following:

1. Take the $\log$ of $2^M$ and $n!$ to compare the values.

2. Using Stirling approximation, you'd obtain something like: $$ M \approx n \log n - n + \log \sqrt{2\pi n} $$

3. I don't see the way of solving the latter equation, but it's possible to solve approximation of it using Lambert function:

$$ M = x \log x - x \iff x = W(a/e) + 1 $$

Answer 2

An answer in Is there an inverse to Stirling's approximation? suggests that

$$ x = \frac{{\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)}}{{W\!\left( {\frac{1}{\mathrm{e}}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)} \right)}} - \frac{1}{2} $$

is a good approximation to the solution of $y=x!$.

Using R to apply that here

library(pracma)
logy <- 77233199*log(2)
(logy-log(sqrt(2*pi))) / lambertW(exp(-1)*(logy-log(sqrt(2*pi)))) - 1/2
# 3784290.23

and testing either side of this gives

• $2^{77233199} \approx 3.6\times 10^{23249509}$

• $3784290! \approx 1.1\times 10^{23249508}$

• $3784291! \approx 4.2\times 10^{23249514}$

so $3784290!$ is the closest factorial to $2^{77233199}$

How to solve factorial equations with very big numbers is a related question