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 idontknow March 15th, 2019 06:54 AM

Equation with factorial

$\displaystyle a(n!) =[a(n)]! \; \;$ , $\displaystyle n\in \mathbb{N}$ .
$\displaystyle a(n)$=?

By plugging values: n=1 , a(1)=a(1)! , a(1)=1 .
Continuing like this, it gives only the solution $\displaystyle a(n)$ equals constant.
The other solution is $\displaystyle a(n)=n$, how to find it?

 idontknow March 15th, 2019 08:28 AM

Not sure but I arrived to $\displaystyle a((n-1)!)=[a(n)-1]!$ .
$\displaystyle [a(n-1)]!=[a(n)-1]! \; \;$ by removing factorials :
$\displaystyle a(n-1)=a(n)-1 \; \;$ or $\displaystyle 1+a(n)=a(1+n)\;$ where $\displaystyle a(1)=1$.
The equation above is a recurrence relation which has solution $\displaystyle a(n)=n$ .

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