k=54459

[igor]

How come the number of total test ( as reported by Project stats) for k=54459 far exceeds that for other k's, while the upper bound is of the same magitude? Was sieving less efficient for this particular k?

Thanks

Igor

[philmoore]

Yes, indeed, sieving is less efficient on this number. The "Proth weight" (or equivalent "Nash weight") measures how efficient sieving is on a particular k. From another thread, the Proth weights of all 17 original k-values are:

4847 .09877
5359 .11761
10223 .11875
19249 .04339
21181 .09649
22699 .04054
24737 .09820
27653 .06394
28433 .05424
33661 .09592
44131 .14273 (eliminated)
46157 .05424 (eliminated)
54767 .12846 (eliminated)
55459 .14102
65567 .04567 (eliminated)
67607 .03540
69109 .08906 (eliminated)

"Typical" k-values have Proth weights closer to 1. The Sierpinski number 78557 has a Proth weight of 0. The fact that these k-values all have small Proth weights is why it has been difficult to eliminate them as Sierpinski candidates. Note that 55459 has the largest Proth weight of the remaining 12 Sierpinski candidates, which also means that it has fewer possible primes eliminated in the sieve.

[Nuri]

Well, I may be wrong on this, but it seems to me like the number of candidates factored out seems to be almost in line with the proth weights.

I looked at the number of eliminated candidates data a couple of days earlier, and compared them with the proth weights just out of curiosity. Just a quick example, proth weight of 55459 is about 4 times that of 67607. So is the number of candidates eliminated.

Well, I deleted that analysis, but will do it with the current data again and post it on this thread today.

What I feel is, (and again, I may be wrong), the percentage of candidates eliminated for each k is proportional to the number of remaining candidates. So, as in the above example, when compared to 67607, 55459 has four times more factors eliminated, four times more remaining candidates (so that four times more prp test needed), and of course, it's proth weight is 4 times higher.

If we're lucky, we'll find the primes for 5359, 10223, and 55459 first. These three k values are only 25% of the remiaining k values, but finding a prime for them will eliminate 37.6% of the remaining prp tests.

On the other hand, if we're unlucky, and if we'll find primes for 27653, 28433, 19249, 22699, and 67607 first, that would mean we will eliminate 23.7% of the remaining prp tests, although we have eliminated 41.7% of the k values (5/12).

And again, anyone please feel free to correct if I'm wrong on my assumptions.

[Moo_the_cow]

1 question:

Why does it seem that the higher k-values (>40,000) are
eliminated first? :confused:

[philmoore]

Good question! One would expect the higher Proth weight k's to go first, but 46157 and 65567 don't fit that expectation. I'm sure that others must have wondered the same thing.

By the way, you could say that we would be lucky to eliminate the low Proth weight candidates first, because we can expect, on average, to test to much higher n before we finally find primes for them.

[Nuri]

Originally posted by Nuri Well, I deleted that analysis, but will do it with the current data again and post it on this thread today.

The data has 256,258 factors (including the duplicates) for p>1G. Factors for n<3,000,000 eliminated.

Well, percentage distribution is not exactly the same, but it looks pretty close to me.

Regards,

Nuri