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Old 2020-06-25, 22:21   #177
R.D. Silverman
 
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Nov 2003

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Quote:
Originally Posted by sweety439 View Post
but 12^319 > 10^323, so 10^323-1 should be factored first

Also Phi_319(12) > Phi_323(10)

12^319-1 has an algebraic factor. We are not factoring 12^319-1, we are factoring (12^319-1)/(12^29-1) ~12^290 ~ C313.

The resulting polynomial is reciprocal, so we can do this number with a quintic.
However, quintic polynomials for numbers this size result in matrices that are
significantly larger than those for numbers of similar
size done with sextics. Greg is LA constrained right now, so he skipped 12^319-1
for the time being. He did C314, C315, C316 C317 and is now working on C318's
via 3^667-1 etc.

Greg may indeed do R323 before he does 12^319-1. I think he will. R323 might well be
done by a reciprocal octic to take advantage of the algebraic factor 10^19-1. Whether
the octic would be easier than the obvious sextic might be an interesting experiment.


It might also be interesting to see if a septic would be any better. I think a septic
will be slightly better in general for numbers of this size.

Let's do a "back of the envelope" look at the norms. Take (10^6, 10^6) == (a,b) as a
'typical lattice point'.

For a sextic, an algebraic norm is ~ a^6 ~ 10^36 and a
linear norm is ~ b * (10^324/6) ~ 10^60. For a septic an anorm is ~a^7 ~ 10^42
and a linear norm is b *(10^322/7) ~ 10^52. The norms are closer for the
septic and their product is slightly smaller. A septic seems slightly superior.
For the reciprocal octic an anorm is a^8 ~ 10^48 and a linear norm is b * (10^38) ~ 10^44 which seems even better still.

Note that one also needs to adjust these estimates by the special-q. The estimates
also ignore the effect of variance on the norms. Since we want smooth numbers
we are more concerned with the tails of the distributions of the norms rather than
the means. However, it does give a quick comparison.

NFS works best when the norms are as nearly equal as possible, other things
being equal.

This very rough estimate is based on the assumption that (10^6, 10^6) is a
typical lattice point. Adjust the analysis if this assumption is not a good enough
estimate. I do now know what sieve areas the lasievef siever uses.

Noone has been calling for him to do 12^319-1.

It is possible that Greg missed the reciprocal octic for R323. He will get to it.

Doing R323 seems to be a compulsion with you.
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