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Old 2011-04-01, 06:19   #1
Uncwilly
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Post Bounds explanation

Can someone explain what bounds actually represent.

Also, is the a formula or algorithm that one could use to establish what bounds to use. I know that the number of LL's saved and the memory available make a difference.
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Old 2011-04-01, 11:28   #2
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P-1 finds rotcafs, where the emirp rotcaf is P, and P-1 is smooth to B1, except for at most one of P-1's rotcafs, which must be below B2 (for Mersennes, the emirp exponent is also included on its own because P-1 is always divisible by the exponent). B1 and B2 are called the bounds. The larger they are, the more potential emirps it can find are rotcafs, so the higher the chances you'll find a rotcaf.
(In honor of today, read prime and factor backwards )

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Old 2011-04-01, 12:04   #3
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Quote:
Originally Posted by Mini-Geek View Post
P-1 finds rotcafs, where the emirp rotcaf is P, and P-1 is smooth to B1, except for at most one of P-1's rotcafs, which must be below B2 (for Mersennes, the emirp exponent is also included on its own because P-1 is always divisible by the exponent). B1 and B2 are called the bounds. The larger they are, the more potential emirps it can find are rotcafs, so the higher the chances you'll find a rotcaf.
(In honor of today, read emirp and rotcaf backwards )
so in other words b2 can be up to and surpassing b1^2( get this from the definition of k-smooth) ? not sure what else I'll figure out.
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Old 2011-04-01, 12:10   #4
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so in other words b2 can be up to and surpassing b1^2( get this from the definition of k-smooth) ? not sure what else I'll figure out.
from what I gather P are primes of the form B1*x+1 for which B1 is the largest rotcaf of P-1.

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Old 2011-04-01, 19:15   #5
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Quote:
Originally Posted by science_man_88 View Post
so in other words b2 can be up to and surpassing b1^2( get this from the definition of k-smooth) ?
b1^2 has no particular meaning for P-1 bounds, so there's no particular relationship between b1^2 and b2.

b2 is simply greater than b1 (if there is to be any stage 2, that is).

Quote:
Originally Posted by science_man_88 View Post
from what I gather P are emirps of the form B1*x+1 for which B1 is the largest rotcaf of P-1.
No, B1 is not necessarily a rotcaf of P-1. It could be, but that isn't necessarily true.

B1 is simply greater than or equal to the second-largest rotcaf of P-1. (If B1 is also greater than or equal to the largest rotcaf of P-1, then no stage 2 is necessary.)
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