mersenneforum.org - View Single Post - Question about number of the form 2^n + 1

jshort · 2020-10-06, 18:20

Quote:

Originally Posted by Dr Sardonicus

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

Thanks Dr Sardonicus.

I didn't think a counter example could be found for such small $n$ . My mistake was in assuming that the non-primitive part of $2^{n} + 1$ must contain many factors from smaller $2^{m} + 1$ where $m < n$ is another highly composite integer. However just because $m$ is highly composite, doesn't mean that $m | n$ .

In this case, even considering only the primitive part of $2^{24} + 1$ which is $65281 = 97 \cdot 673$ we still get a composite integer, which is kind of surprising because we can also prove that any prime factor of the primitive part of $2^{24} + 1$ must be of the form $1 + 4 \cdot 24k = 1 + 96k$ , which greatly restricts the possible number of "eligible" prime factors to nearly 1% of all primes that the are in the range from $(2, \sqrt{65281})$ . However when $k=1$ , we get $97$ which is a factor lol!