What is the maximum possible value of the greatest common divisor of two consecutive terms of the sequence \(a_n = n! + n\), where \(n \ge 0\)?
I'm not sure if it's 2 or not becuase there might be a higher answer...
2 should be the maximum. And that would occur when n =1 and n=2. 1!+1=2 and 2!+2 =4 and the GCD(2, 4)=2. Any two consecutive terms after 2 would have a GCD of 1. So: 3!+3 =9. 4!+4=28 and GCD(9, 28)=1...and so on.
+8 
