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 March 15th, 2019, 06:54 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 536 Thanks: 81 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? Last edited by skipjack; March 15th, 2019 at 01:15 PM.
 March 15th, 2019, 08:28 AM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 536 Thanks: 81 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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