Bases 33-100 reservations/statuses/primes

[Siemelink]

Hi everyone, here is my base 45 effort. The Riesel conjecture is 22564.
I've taken this to n = 10,000. while ignoring odd k and (k mod 11) = 1.
As 1080 = 24*35 and 16740 = 372*45 I've left hem out.
This leaves the 22 following k:
24
372
1264
1312
2500
2804
4210
4484
5128
6094
6372
7246
10096
10518
12950
13456
13548
15432
17918
19252
20654
21274

There are 2 squares remaining, 2500 and 13456, but that is just coincidence I think. There are many square k's that do have a prime.
Also, 24 = 6*2*2. Several of the conjectures remove k = 6*square, but I don't understand why. How can I check if it can be removed here also?

The top ten of primes:
13546 9069
17734 8019
19102 7368
14324 7281
9938 7240
4628 7209
4622 7116
6554 6462
2230 5892
7750 4586

I've attached and doublechecked all the primes that I found.

Willem.

[gd_barnes]

Originally Posted by Siemelink

There are 2 squares remaining, 2500 and 13456, but that is just coincidence I think. There are many square k's that do have a prime.
Also, 24 = 6*2*2. Several of the conjectures remove k = 6*square, but I don't understand why. How can I check if it can be removed here also?
Willem.

It's not always very easy. The way I do it is to look for patterns in the factors of the various n-values for specific k-values. If there are algebraic factors, it's most common for them to be in a pattern of f*(f+2), i.e.:
11*13
179*181
etc.

In other cases there may be a consistent steady increase in the differences of their factors, which is especially tricky to find but indicates the existence of algebraic factors. That's what I ran into on base 24.

For your 3 cases here, you have:
k=24:
n-value : factors
1 : 13*83
2 : 23*2113
3 : 17*103*1249
4 : 23*163*26251
9 : 2843*6387736694293
Analysis:
For n=3 & 4, multiplying the 2 lower prime factors together does not come close to the higher prime factor so little chance of algebraic factors.
For n=9, the large lowest prime factor that bears no relation to the other prime factor means that there is unlikely to be a pattern to the occurrences of large prime factors so there must be a prime at some point.

k=2500:
n-value : factors
1 : 19*31*191
2 : 13*173*2251
3 : 89*2559691
4 : 19*73^2*103*983
9 : 9439*4280051*46824991
For n=9 same explanation as k=24.

k=13456:
n-value : factors
1 : 269*2251
2 : 17*23*227*307
3 : 31*39554129
4 : 7*23*467*503*1459
7 : 3319*1514943103721
For n=7 same explanation as k=24.


The prime factors for n=9, n=9, and n=7 respectively make it clear to me that these k-values should all yield primes at some point so you are correct to include them as remaining.

The higher-math folks may be able to chime in and answer why there are an abnormally large # of k's that are perfect squares that end up remaining even though they don't have known algebraic factors for most bases. IMHO, it's because there ARE algebraic factors for a subset of the universe of n-values on them but not for all of the n-values. Hence they are frequently lower weight than the other k's but NOT zero weight and so should eventually yield a prime.


Gary

[Siemelink]

Hi everyone,

here is my work on the base 42 Riesel conjecture 15137. The covering set is {5, 42, 353}.
After taking n to 10,000 there were 72 k left. I removed 2058 = 49 * 42 as 49 is still in the list of k.
There are also 5 squares in the list (49, 1369, 2304, 3721 and 10201), but they give no obvious deductions.

This leaves:

49
386
603
1049
1160
1426
1633
1678
2304
2464
2538
2753
3428
3734
4299
4903
5118
5417
5677
5820
5899
5978
6333
6623
6664
6836
6838
6964
7016
7051
7309
7489
7614
7658
7698
7913
8297
8341
8384
8453
8524
9029
9201
9418
9633
9848
10026
10114
10276
10663
10923
11052
11267
11781
11911
11996
12039
12125
12127
12151
12213
12598
13288
13329
13347
13425
13632
13757
13898
14576
15024

Enjoy, Willem.

[gd_barnes]

Originally Posted by Siemelink

Hi everyone,

here is my work on the base 42 Riesel conjecture 15137. The covering set is {5, 42, 353}.
After taking n to 10,000 there were 72 k left. I removed 2058 = 49 * 42 as 49 is still in the list of k.
There are also 5 squares in the list (49, 1369, 2304, 3721 and 10201), but they give no obvious deductions.


Enjoy, Willem.

Looks good. Nice work. Actually only 2 of your squares are remaining: k=49 and k=2304. As you showed in your list, k=1369 has a prime at n=7577, k=3721 has a prime at n=4611, and k=10201 has a prime at n=2129.


Gary

[Siemelink]

Hi there Gary, thanks for clarifying my muddled statement.

here is the next one, base 37. The conjecture is 7772, with set = {5, 19, 137}. At n = 10,000 there are 30 k remaining:

284
498
522
590
672
816
1008
1578
1614
1842
1958
2148
2606
2640
3336
3480
3972
4356
4428
4542
4806
5262
5376
5910
5946
6288
6752
6792
7088
7352
7466

I've attached the primes found. Here is the list of the highest primes:
7058 8314
1334 7883
5156 7797
6480 7763
554 7472
7124 6396
474 3952
998 3572
912 3394
1956 3250

Willem.

[Siemelink]

Hi everyone,

I've double checked my work on the Riesel 35 Conjecture. The value is 287860. I've taken the k's to n=200 with PFGw. Then I sorted out the multiples of 35 and the squares. ThenI took the remainder to n=5000
I've attached the 1559 k's that I had left.
The top ten primes list is this:
65216 4986
248264 4980
104690 4978
126050 4978
286652 4976
129052 4975
229454 4974
48772 4965
169448 4964
7874 4962

All the primes are in a file that zips to 350k, that doesn't fit on the forum. The can be sent if so desired.

Willem.

[gd_barnes]

Originally Posted by Siemelink

Hi everyone,

I've double checked my work on the Riesel 35 Conjecture. The value is 287860. I've taken the k's to n=200 with PFGw. Then I sorted out the multiples of 35 and the squares. ThenI took the remainder to n=5000
I've attached the 1559 k's that I had left.
The top ten primes list is this:
65216 4986
248264 4980
104690 4978
126050 4978
286652 4976
129052 4975
229454 4974
48772 4965
169448 4964
7874 4962

All the primes are in a file that zips to 350k, that doesn't fit on the forum. The can be sent if so desired.

Willem.

Can you please send the primes to me at: gbarnes017 at gmail dot com


Thanks,
Gary

[kar_bon]

i give base 35 a few days shot!

so far:
sieved all 1559 (minus k with primes) upto p=402M for n=5k-100k
checked upto n=5249
38 more primes found
6.5M candidates left

sieving further!

will mail primes when more available.

[kar_bon]

reserved base 35 from n=5k-100k

so Siemelink can check another base :-)

[michaf]

Rest assured that it'll take you quite some time to take it to 100k...

(Base 31, also not very prime (compared to base 3), now runs erm... 4 or 5 months or so :) )

[gd_barnes]

Originally Posted by michaf

Rest assured that it'll take you quite some time to take it to 100k...

(Base 31, also not very prime (compared to base 3), now runs erm... 4 or 5 months or so :) )

To clarify: Actually base 31 is very prime compared to most bases. It is much more prime than base 35 is. But of course nothing compares to base 3. So far, only base 7 comes close. I also suspect base 15 will be quite prime.

It seems that all bases where b=2^q-1 are very prime as compared to their neighbors.

I would expect many CPU years to get base 35 up to n=100K.


Gary