Bases 33-100 reservations/statuses/primes

[robert44444uk]

Some conjectures on the lowest Riesel R for base b:

A. Lowest conjectured Riesels are related according to modular arithmetic on the base value.

The graph below which plots (x-axis) b= base, and y-axis lnln(R) with R = lowest conjectured Riesel, suggests modular patterns and I have discovered the following modular relationships:

The following must be taken in order (eg 428==142mod143 but also 32mod33)
Note: ? are conjectured.

R = 4 b==14mod15
R = 6 b==34,69mod105
R = 8 b==20mod21
R = 10 b==32mod33
R= 12 b==142mod143
R = 14 b==8,38,47,64,77,83,116,122,129 or 155mod195
R = 16 b==50,84,152,203,305?,339?,407?,458?mod765
R = 20 b ==18,37,56,75mod95
R = 22 b== 36?,45? and 68mod69
R = 24 b==114mod115
R = 28 b==8? or 86mod87
R = 32 b==30?,61? or 90mod93
R = 34 b== 20?, 21mod33
R = 36 or 38 b==36,73 or 110 mod111
R = 40 when b==122mod123
etc

The case of R = 36 and 38 is curious.

Odd k relationships seem to be less easy to spot, but that is possibly because of the low number of candidates. But I would conjecture:

R = 13 when b==20,38mod264
R = 21 when b==54mod110

This seems to suggest a much easier way to get to low Riesels!! All that is needed is to check the above, or better still, generate the modular algorithm and run through a modular sieve.

B. I wonder is there is a max lnln value for a lowest Riesel? Possibly not, I would guess as I have encountered some nasty looking b's

C. There is an intriguing hole in the graph, tending to lnln(b)=2

Regards

Robert Smith

[robert44444uk]

Now I understand a bit more about what I am doing, I looked at the Sierpinski side, and found the following modular relationships, which relate to the multiple of the primes in the cover set of the lowest conjectured Sierpinski for a given b. There is at least one anomaly with b=64.

k b
4 14mod15
6 34mod35
8 20mod21: 47,83mod195: 77mod73815: 137mod1551615
10 32mod33
12 142mod143: 562,828,900mod1729: 563mod250705: 597mod1885
901mod19019
13 132,293mod595
14 38mod39: 64mod65 (but not b=64 where k=51!!!)
16 50mod51: 84mod85: 38,47,98,242mod255
18 322mod323: 512mod263683
20 56mod57: 132mod133
21 54mod55
22 68mod69: 160mod161
23 182,878mod795
24 114mod115
25 38mod39
27 90mod91: 538mod2555
28 86mod87
30 898mod899
32 92mod93: 483,747mod2255: 542mod615: 340mod341
34 54mod55: 76mod77
36 159mod233285: 184mod185: 258mod259: 783,993mod1295
38 110mod111: 480mod481: 948mod1105
40 122mod123: 532mod533: 788mod1599

I will go back and redo the Riesel side

[robert44444uk]

Here is the correct list for Riesels:

k b
4 14mod15
6 34mod35
8 20mod21: 83,307mod455:
10 32mod33
12 142mod143: 901mod19019
13 38,47mod255
14 38mod39: 64mod65: 8,47,83,122mod195
16 50mod51: 84mod85
18 3322mod323: 577mod1105
20 56mod57: 132mod133
21 54mod55
22 68mod69: 160mod161: 657mod3395
24 114mod115
27 90mod91: 922mod4745
28 86mod87: 443mod2355
29 908mod91455
30 898mod899
32 92mod93: 340mod341
34 54mod55: 76mod77: 746mod3471
35 50mod51
36 184mod185: 258mod259
38 110mod111: 480mod481:
40 122mod123

It looks very similar to the Sierpinski list

[gd_barnes]

Originally Posted by Siemelink

Yup, that is exactly why I hadn't posted it. I have done some of the doublechecking, but I am no longer clear on what...
The effort is running on a PC with limited access, so no correction possible on the sieve file. I'll wait it out until 25,000 and after that I'll wrap up.

Cheers, Willem.

With only one exception, all the composites that I found on your list turned out to be typos with other primes. Therefore, I tested the one exception, k=109772, up to n=25K. No prime was found. That was the k that you had repeated the n-value prime from the k-value right about it in your list.

I also continued my double-check up to n=6.2K and found no additional problems.

So double-checking to n=6.2K and running primality proofs on your entire list confirms that there are definitely 103 k's remaining for Riesel base 36 at n=22.5K after subtracting off the converted base 6 higher primes that you have already done.


Gary

[gd_barnes]

Originally Posted by mdettweiler

I'll reserve all 31 remaining k's for Riesel base 37--this looks like an easy base to prove, so I think I'll aim to take it to at least n=20K.

Edit: I've moved this post to the "Bases > 32 that are not powers of 2" thread (it was originally in the Reservations/Statuses thread) since it would probably be more appropriate in the former.

Not likely to be proven in your lifetime! (not joking)

I've had this type of discussion with others, especially Kenneth. People well under-estimate the difficulty of finding primes for low weight k's on high bases, especially at high n-values.

Example: I'm testing the ONE final k on Sierp base 12 at n=195K. I'm going to n=250K on a file already fully sieved to P=7T (which took quite a while to get sieved that high), with likely less than a 20% chance of prime by that point. Testing time per candidate on one of my highest-speed machines: Nearly 1 hour! Total CPU time needed to complete it: 80-85 days and that's for just one k on a base 1/3rd as high as yours!! Running on just one core, it is crawling along. Sometime when it passes n=200K; I'll probably put 4 quads on it and knock it out in ~4-5 days just to get it off my plate but that's a LOT of firepower on one simple k for only an n=50K range!

Also, base 37 is not base 31 or base 36. It's not a very prime base at all. Example: There were 41 k's remaining at n=3.2K and 31 k's remaining at n=10K. If you assume a 25% reduction in k's remaining for every tripling of the n-value, you have:

n=10K; 31 k's remain
n=30K; 23 remain
n=90K; 17 remain
n=270K; 13 remain

To get this base to n=100K will be a huge, although very worthwhile effort. As for proving it, even if I'm way off and you only have 4-5 k's remaining at n=270K; this is likely > 100 CPU-year effort to prove it. All that I can say is: Good luck! You'll need it.


Gary

[gd_barnes]

Primes and status in a PM from Karsten on Riesel base 35:

next ones:
269516 6470
81250 6473
65874 6486
166052 6492
32954 6498
163094 6504
236596 6513
231208 6521
272612 6526
281816 6536
232064 6538
227888 6556
48244 6565
110566 6565
225482 6576
148766 6588
128626 6593
166208 6596

at n=6606 with 5.19M candidates left (sieved to 4.3G, will sieving further)

Gary

[kar_bon]

Originally Posted by robert44444uk

(...)
16 50mod51: 84mod85
18 3322mod323: 577mod1105
20 56mod57: 132mod133
(...)

perhaps a misprint?!

should be: 18 322mod323: 577mod1105

[kar_bon]

next PRP-values:

161266 6625
271540 6631
209114 6652
225590 6656
240320 6684
14204 6714
57170 6720
155966 6720
99862 6733

at n=6735

[robert44444uk]

Originally Posted by kar_bon

perhaps a misprint?!

should be: 18 322mod323: 577mod1105

Certainly a misprint from my Excel spreadsheet. There maybe others, hopefully not.

Also I notice that 54mod55 turns up twice, (k=21 and 34) - the second value comes into play if k=21 provides a facile result, both are related to covers [5,11]. And that I now know also explains the base 64 problem.

So the mods I show provide a theoretical low k value and if that produces a facile result then you have to look at other mod combinations. With these low k you don't have to look far.

Dan Krywaruczenko will soon publish his work on determining any k as a Sierpinski/Reisel value excluding k=Mersenne.

[mdettweiler]

Originally Posted by gd_barnes

Not likely to be proven in your lifetime! (not joking)

I've had this type of discussion with others, especially Kenneth. People well under-estimate the difficulty of finding primes for low weight k's on high bases, especially at high n-values.

Example: I'm testing the ONE final k on Sierp base 12 at n=195K. I'm going to n=250K on a file already fully sieved to P=7T (which took quite a while to get sieved that high), with likely less than a 20% chance of prime by that point. Testing time per candidate on one of my highest-speed machines: Nearly 1 hour! Total CPU time needed to complete it: 80-85 days and that's for just one k on a base 1/3rd as high as yours!! Running on just one core, it is crawling along. Sometime when it passes n=200K; I'll probably put 4 quads on it and knock it out in ~4-5 days just to get it off my plate but that's a LOT of firepower on one simple k for only an n=50K range!

Also, base 37 is not base 31 or base 36. It's not a very prime base at all. Example: There were 41 k's remaining at n=3.2K and 31 k's remaining at n=10K. If you assume a 25% reduction in k's remaining for every tripling of the n-value, you have:

n=10K; 31 k's remain
n=30K; 23 remain
n=90K; 17 remain
n=270K; 13 remain

To get this base to n=100K will be a huge, although very worthwhile effort. As for proving it, even if I'm way off and you only have 4-5 k's remaining at n=270K; this is likely > 100 CPU-year effort to prove it. All that I can say is: Good luck! You'll need it.


Gary

Oh, I see. I had assumed that with only 31 k's remaining at n=10K (which, for a lower base, would have meant quite good chances of proving it quickly, even without the advantage of a very prime base), my odds were pretty good--I guess not. Oh well--I'll still take it up to n=20K and see what I can knock out.

[gd_barnes]

Originally Posted by mdettweiler

Oh, I see. I had assumed that with only 31 k's remaining at n=10K (which, for a lower base, would have meant quite good chances of proving it quickly, even without the advantage of a very prime base), my odds were pretty good--I guess not. Oh well--I'll still take it up to n=20K and see what I can knock out.

What difference would it make if it was a very low base; even base 2 or 3? Virtually none at all.

Obviously I still haven't made myself clear on the difficulty in proving these things. Please take a look at the most prime base of all: base 3. On the Sierp side, we'll likely knock out almost exactly half of k's remaining testing k<50M from n=35K to n=100K so for the purposes of estimate, we'll just assume that we halve k's remaining for every 3X increase in n-range; therefore if 31 k's were remaining at n=10K:

10K; 31 k's remaining
30K; 16 k's remaining
90K; 8 k's remaining
270K; 4 k's remaining
810K; 2 k's remaining
1.62M; 1 k remaining
3.24M; 0.5 k remaining

So likely, you'd have to test it to n=3M. Doable as a project over several years but not as an individual.

The point is that 31 k's is a huge # of k's remaining in ANY base at n=10K! For any one person to have a shot at proving a base, there needs to be < ~10 k's remaining at n=10K.

The reason that I want to make this so clear is that people have a tentency to reserve far more than they will ever want to complete. Your reservation to n=20K is quite reasonable for total workload but gives no chance of proving the conjecture in any base for so many remaining k's at n=10K.

Karsten, are you still going to take Riesel base 6 to n=1M? (lol)


Gary