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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?

 May 2, 2018
 #1
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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.

 May 2, 2018
edited by Guest  May 2, 2018
 #2
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That is incorrect.

Guest May 2, 2018

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