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Old 2020-10-06, 18:20   #3
jshort
 
"James Short"
Mar 2019
Canada

17 Posts
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Quote:
Originally Posted by Dr Sardonicus View Post
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!
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