20210927, 14:26  #1 
Feb 2017
Nowhere
5^{3}·41 Posts 
Proper factorization known, no prime factors known
This is merely recreational mathematics. If this more properly fits in an area other than "Puzzles," feel free to move it there.
My "jumping off place" is the fact that (currently, AFAIK) the character (prime or composite) of F_{33} is unknown. The Prime factors k*2^{n} + 1 of Fermat numbers F_{m} and complete factoring status page lists F_{m} for the values m = 33, 34, 35, 41, 44, 45, 46, 47, 49, 50, 51, . . . as "Character unknown." So it occurred to me that there are numbers with known proper factorizations, for which the "minimal" proper factors (those which are not the product of smaller known proper factors) are all of unknown character. A trivial example would be N = F_{33}*F_{33}. Of course, one can cobble together other such trivial examples by tracking down numbers of unknown character and multiplying them together. It occurred to me that one could probably construct "more natural" examples using algebraic factorizations. As a potential example, I give the number N = googolplex + 1 = 10^{10^100} + 1. We have According to Prime factors of generalized Fermat numbers F_{m}(10) and complete factoring status, F_{100}(10) appears to be of unknown character [though two prime factors are known for F_{99}(10)], so the first factor at least is of unknown character. I'm guessing nobody has tried to find factors of the other algebraic factors, but I don't know. I am not suggesting that anybody waste their time looking for them. I imagine much smaller examples are to be found; a possible candidate is 2^{3*2^33} + 1. But for all I know, someone may have found a divisor of the larger algebraic factor. I don't know whether there's a number of unknown character smaller than F_{33}. Anybody? Last fiddled with by Dr Sardonicus on 20210927 at 14:31 Reason: insert missing multiplication symbol 
20210927, 16:17  #2 
"Daniel Jackson"
May 2011
14285714285714285714
677 Posts 
In terms of smaller numbers, there's M1347826279*M1552958761 (I TF'd both of them to make sure there were no small factors <10^21), but there's also some larger examples, such as MM61*MM89*MM107*MM127*MM521*MM607*...*MM82589933*..., going on as far as there are Mersenne Prime exponents to use, assuming that there are finitely many Mersenne primes (otherwise the resulting number is infinite). (2^2^((Loader's Number)↑↑↑...↑↑↑Loader's Number)+1)^2 (where the number of up arrows is Loader's number) is about as big as I can think of, and it's the square of a Fermat number, albeit a gargantuan one.
Last fiddled with by Stargate38 on 20210927 at 16:25 
20210927, 16:56  #3 
Feb 2017
Nowhere
1010000000101_{2} Posts 
D'oh! Of course! 2^{33} is 8589934592, or almost 8.6 billion! There are plenty of M_{p} whose character (prime or composite) is unknown, with p < 2^{33}. This is likely to remain true for some time to come.

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