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2020-10-04, 16:43   #2
Dr Sardonicus

Feb 2017
Nowhere

24×271 Posts

Quote:
 Originally Posted by jshort Let $a_{n}$ be the nth Highly Composite Integer* and $b_{n} = lcm(2^{a_{1}} + 1, 2^{a_{2}} + 1, ...,2^{a_{n}} + 1)$. If $c_{n} = \frac{2^{a_{n}} + 1}{gcd(2^{a_{n}} + 1,b_{n-1})}$, will $c_{n}$ always be equal to either 1 or a prime number? *https://en.wikipedia.org/wiki/Highly...n%20the%20OEIS).
No. The first six highly composite numbers are 1, 2, 4, 6, 12, and 24.

We have b5 = lcm([3,5,17,65,4097]) = 3*5*13*17*241, and

224 + 1 = 16777217 = 97*257*673. Thus

c6 = (224 + 1)/gcd(3*5*13*17*241,97*257*673) = (224 + 1)/1 = 97*257*673