What is the maximum possible value of the greatest common divisor of two consecutive terms of the sequence a_n = n! + n, where n is greater than 0?
If you use a couple of numerical examples, it will be much easier to see the GCD of the two:
a(5) = 5! + 5 = 125
a(6) = 6! + 6 = 726
Now, if you factor the above two consecutive terms for their divisors, this is what you get:
125 = 1, 5, 25, 125 (4 divisors)
726 = 1, 2, 3, 6, 11, 22, 33, 66, 121, 242, 363, 726 (12 divisors)
OK, you are right because I didn't consider the first term, or 1, which will give:
1! + 1 and 2! + 2 =2 and 4.So, the maximum GCD would be 2. However, after the 1st term, the GCD of any two consecutive terms will be 1.
