Equation with factorial

Dec 2015
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169
Earth
\(\displaystyle a(n!) =[a(n)]! \; \; \) , \(\displaystyle n\in \mathbb{N}\) .
\(\displaystyle a(n)\)=?

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 a moderator:
Dec 2015
1,082
169
Earth
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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