I was doing some problem on factorial. Then I was striked by an hypothesis....

The hypothesis was....

Is it possible to show a factorial as a sum of other factorials???


a! = b! + c! + d!.......

All numbers are unique Natural number without repetition.

I have tried to find out some solution but I was unable to find any!

Can anyone will try this as a challenge and give me the solution !!!

  • $\begingroup$ Does this count? $$2!=1!+0!$$ $\endgroup$ – projectilemotion Apr 15 '17 at 17:08
  • $\begingroup$ Sorry forgot to tell all should be natural number $\endgroup$ – Creepy Creature Apr 15 '17 at 17:41
  • $\begingroup$ Some definitions have $0\in \mathbb{N}$ such as the standard ISO 80000-2. $\endgroup$ – projectilemotion Apr 15 '17 at 17:43

Apart from the trivial example $2!=1!+0!$, the answer is "no".

This follows from $$0!+1!+2!+3!+\ldots+(n-1)!\le n\dot (n-1)!=n! $$ with equalitiy iff all summands are equal to $(n-1)!$, i.e., $(n-1)!=1$.

  • $\begingroup$ I'm not sure if I understand how it follows from that. $\endgroup$ – mrnovice Apr 15 '17 at 17:50
  • $\begingroup$ How?? Any examples $\endgroup$ – Creepy Creature Apr 15 '17 at 18:08
  • 1
    $\begingroup$ I believe he's saying that even if you sum all factorials less than a given factorial the sum is too small. In other words the factorial grows too fast. $\endgroup$ – Χpẘ Apr 15 '17 at 19:00

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