Code to explore a! x b! = c! where 1 < a < b < c-1
I became interested in the solutions to this factorial equation. It turns out that there is a countable infinity of solutions of the form: n! x (n!-1)! = (n!)! which comes from the definition k x (k-1)! = k! if k = n!.
However it is also known that 6! x 7! = 10! which doe not fit that general pattern. The question is why? and are there any other solutions that don't fit? Can we prove that there are not?
The program Factorial_Statistics.py is a command line python program that searches the integer solution space for triplets that solve the equation, but don't follow the usual pattern where b! = (c-1)! resulting in a! = c. It does this by succesive tests of integers for c, and then looping back from b = c-2 to c/2, checks to see if c!/b! equals any a! for a < b.
Run "python Factorial_Statistics.py" to do a default search up to 2500! and generate a .csv statistics file on the way. The statistics are useful in a "vanishing possibility" proof that (6, 7, 10) is the only special solution.
Run "python Factorial_Statistics.py --limit 10000000 --no-stats" to search for solutions up to c = 1,000,000 with no statistics file generated (because it would be huge) or even higher if your machine will do it.
If you want to watch it make progress while it runs, use --progress-bar to see how fast it is generating lists of primes and how fast it is searching those lists.
The code uses the python tqdm module for the console parser. You may need to: pip install tqdm
Using this program I was able to find a special number, where c = 2080, such that for all c > 2080, no candidate for b < c-1 and > c/2 can generate a c!/b! that could possibly be equal to a!
-Ken Clements