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CRGreathouse 2010-08-28 03:41

[QUOTE=3.14159;227351]So: Anyone else up for the challenge of posting a cofactor prime of at least 500 digits? (Note: The cofactor has to be of one of the 19 types of prime listed.)[/QUOTE]

By your standards and with my usual method, I find a p501 in 203*2^1658+1. :razz:

3.14159 2010-08-28 03:42

[QUOTE=CRGreathouse]By your standards and with my usual method, I find a p501 in 203*2^1658+1.
[/QUOTE]

Let me guess: 3 * p501.

Okay, wiseguy, the smallest factor must be a p40, and the digits must follow a random distribution that is [B]statistically likely[/B]. Good luck.

3.14159 2010-08-28 03:51

Let's see how you loophole your way out of that one!

CRGreathouse 2010-08-28 03:52

[QUOTE=3.14159;227357]Let me guess: 3 * p501.

Okay, wiseguy, the smallest factor must be a p40, and the digits must follow a random distribution that is [B]statistically likely[/B]. Good luck.[/QUOTE]

No thanks, my cores are busy extending several sequences relating to pseudoprimes and large prime gaps. I do enjoy finding trivial examples, though.

CRGreathouse 2010-08-28 03:55

[QUOTE=3.14159;227358]Let's see how you loophole your way out of that one![/QUOTE]

One possibility: go through 20-digit primes, finding k and n values that give a (521 to 540)-digit multiple of that prime, checking if the cofactor is a 2-pseudoprime. This should be faster than actually factoring them out. But I'm not going to do this search because it's just as silly as it sounds.

3.14159 2010-08-28 03:56

[QUOTE=CRGreathouse]No thanks, my cores are busy extending several sequences relating to pseudoprimes and large prime gaps. I do enjoy finding trivial examples, though.
[/QUOTE]

Ah, but you could have loopholed your way out of this one, as I made no digit requirement. It's as easy as:

1230021358933920151693112553959118376943838 * 400[sup]35[/sup] + 1 = 8826736195555687888973540222790863753879 * 1645175382483954577575076564167802963870136473616621363381331706098141318612978687213529959719.

And, voila! You would have won, yet again.

CRGreathouse 2010-08-28 03:58

[QUOTE=3.14159;227354]Never said the divisors had to be small[/QUOTE]

I suppose you could have cofactor (small factor) and cofactor (large factor) categories. It would be interesting to properly classify the difficulty of finding (1) a general and (2) a special-form cofactor based on the size of both the small and large factor. The main difficulty is the latter, of course, but the former plays a part too.

3.14159 2010-08-28 03:58

[QUOTE=CRGreathouse]One possibility: go through 20-digit primes, finding k and n values that give a (521 to 540)-digit multiple of that prime, checking if the cofactor is a 2-pseudoprime. This should be faster than actually factoring them out. But I'm not going to do this search because it's just as silly as it sounds.
[/QUOTE]

See above

CRGreathouse 2010-08-28 04:00

[QUOTE=3.14159;227363]See above[/QUOTE]

I beat you to it. :razz:

3.14159 2010-08-28 04:00

[QUOTE=CRGreathouse]I suppose you could have cofactor (small factor) and cofactor (large factor) categories. It would be interesting to properly classify the difficulty of finding (1) a general and (2) a special-form cofactor based on the size of both the small and large factor. The main difficulty is the latter, of course, but the former plays a part too.
[/QUOTE]

A special-form cofactor in a special-form number? Please, this only happens in Mersenne numbers. If you wanted a special-form factor, you would have to directly rig it to do so.

CRGreathouse 2010-08-28 04:01

[QUOTE=3.14159;227357]the digits must follow a random distribution that is [B]statistically likely[/B][/QUOTE]

How would you define this, I wonder?


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