March 15th, 2019, 06:54 AM  #1 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68  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: iPhone Posts: 436 Thanks: 68 
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 $\displaystyle a(n)=n$ . 

Tags 
equation, factorial, function 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Determining n from the given factorial equation  mared  Algebra  11  October 2nd, 2014 01:46 AM 
An equation with the factorial  Dacu  Number Theory  5  May 25th, 2013 05:54 AM 
Diophantine equation : factorial and powers  momo  Number Theory  8  May 8th, 2009 12:21 PM 
zero factorial (why ??)  sangfroid  Number Theory  8  September 15th, 2008 04:28 AM 
An equation with the factorial  Dacu  Algebra  2  December 31st, 1969 04:00 PM 