March 15th, 2019, 06:54 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 636 Thanks: 91  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: 636 Thanks: 91 
Not sure but I arrived to $\displaystyle a((n1)!)=[a(n)1]!$ . $\displaystyle [a(n1)]!=[a(n)1]! \; \; $ by removing factorials : $\displaystyle a(n1)=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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equation, factorial, function 
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