My Math Forum Equation with factorial
 User Name Remember Me? Password

 Real Analysis Real Analysis Math Forum

 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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top