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 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 $\displaystyle a((n-1)!)=[a(n)-1]!$ . $\displaystyle [a(n-1)]!=[a(n)-1]! \; \;$ 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 $\displaystyle a(n)=n$ . Tags equation, factorial, function Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mared Algebra 11 October 2nd, 2014 01:46 AM Dacu Number Theory 5 May 25th, 2013 05:54 AM momo Number Theory 8 May 8th, 2009 12:21 PM sangfroid Number Theory 8 September 15th, 2008 04:28 AM Dacu Algebra 2 December 31st, 1969 04:00 PM

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