Bases 33-100 reservations/statuses/primes

[gd_barnes]

I have been working on many of the easier Riesel bases 51-100. Below is a general synopsis of what I have done and found. I will show more details a little later on. All bases were searched to n=10K or until proven unless otherwise stated below.

I am listing any reservations after the k's remaining.

base     k's remaining
 53    [128 k's remaining at n=10K; shown on web pages]
 54    proven
 55    68, 2330, 3060, 3158, 3240, 3578, 3942, 4640, 4878,
       5286, 5354, 5690, & 6098 at n=25K
 56    proven
 57    proven
 59    proven
 60    [81 k's remaining at n=10K; shown on web pages]
 61    198, 1520, 1644, 6168, 9642, 10572, & 10968 at n=25K
 62    proven
 65    proven
 67    242, 1274, 1886, 2228, 2846, & 2906 at n=25K
 68    proven by Flatlander; final prime at n=25395 !
 69    proven
 70    729, 1776, 2202, & 5468 at n=25K
 72    4 at n=54K; sieve file available to n=100K
 73    proven
 74    proven
 76    proven
 77    proven
 80    10, 31, 106, 170, & 214 at n=10K
 81    proven
 83    proven
 84    proven
 86    proven
 87    172, 186, 384, 472, 536, 562, 672, 714, 848, 862, 1004, 1112,
       1132, 1154, 1418, & 1628 at n=20K; sieve file available to n=100K
 89    proven
 90    proven
 92    proven
 93    424 & 452 at n=50K; sieve file available to n=100K
 94    29 at n=51K
 98    proven
 99    proven
100    653 at n=60K; sieve file available to n=150K

Unless shown as reserved, all of the above bases are now unreserved. Anyone can feel free to take them further at this point. The web pages are now fully updated with the details of this effort. I will also continue updating this post for a while until work on these particular bases comes to a near stand still.


Gary

[mdettweiler]

Originally Posted by gd_barnes

I have been working on many of the easier Riesel bases 51-100. Below is a general synopsis of what I have done and found. I will show more details a little later on. All bases were searched to n=10K or until proven.

The only exception to the n=10K search was base 55, which has now been searched to n=15K. I am going to reserve base 55 up to n=25K.

base     k's remaining
 53    [128 k's remaining; to be shown later]
 54    proven
 55    [14 k's remaining; to be shown later]
 56    proven
 57    proven
 59    proven
 62    proven
 65    proven
 68    5 & 7
 69    proven
 72    4, 79, 116, & 291
 73    36
 74    proven
 76    proven
 77    proven
 80    10, 31, 106, 170, & 214
 81    proven
 83    proven
 84    proven
 86    proven
 87    172, 186, 384, 472, 536, 562, 672, 714, 758, 848, 862, 898, 958,
       1004, 1112, 1132, 1154, 1418, & 1628
 89    proven
 90    proven
 92    proven
 93    424 & 452
 94    29
 98    proven
 99    proven
100    74 & 653

All of the above bases are now unreserved except for base 55. Anyone can feel free to take them further at this point. Within a couple of weeks, I'll get the web pages updated with the detailed info.


Gary

Hmm...I think I'll reserve k=36 on Riesel base 73 for at least up to n=20K. With only one k left, that one looks like I might have a pretty decent chance of proving my first conjecture.

[mdettweiler]

Originally Posted by mdettweiler

Hmm...I think I'll reserve k=36 on Riesel base 73 for at least up to n=20K. With only one k left, that one looks like I might have a pretty decent chance of proving my first conjecture.

Hmm...I just started sieving on this a few minutes ago, and oddly enough, after only 6 CPU minutes of sieving having been done, there were only 20 remaining k/n pairs in the sieve! Here's the srsieve output:

srsieve 0.6.9 -- A sieve for integer sequences in n of the form k*b^n+c.
srsieve started: 10000 <= n <= 20000, 3 <= p <= 4000000000003
Split 1 base 73 sequence into 2 base 73^12 subsequences.
WARNING: 36*73^n-1 has algebraic factors.
WARNING: 36*73^n-1 has algebraic factors.
p=409039997, 6801123 p/sec, 9973 terms eliminated, 28 remain
p=846160019, 7272243 p/sec, 9977 terms eliminated, 24 remain
p=1297080067, 7507826 p/sec, 9980 terms eliminated, 21 remain
p=1754920133, 7615436 p/sec, 9980 terms eliminated, 21 remain
p=2217960077, 7713347 p/sec, 9981 terms eliminated, 20 remain
p=2685040051, 7774818 p/sec, 9981 terms eliminated, 20 remain
p=3156040061, 7841635 p/sec, 9981 terms eliminated, 20 remain
Sieving 3 <= p <= 3163531061 eliminated 9981 terms, 20 remain.
Wrote 20 terms for 1 sequence to abc format file `sr_73.pfgw'.
srsieve stopped: at p=3163531061 because SIGINT was received.

Either the sieve somehow got completely messed up, or this is one heck of a low-weight k.

[mdettweiler]

Okay, that was easy. Riesel base 73 is now complete to n=20K. Results are attached.

Heck, if it's this easy, I'll just go on ahead and take it up to n=50K!

[gd_barnes]

Originally Posted by mdettweiler

Hmm...I just started sieving on this a few minutes ago, and oddly enough, after only 6 CPU minutes of sieving having been done, there were only 20 remaining k/n pairs in the sieve! Here's the srsieve output:

srsieve 0.6.9 -- A sieve for integer sequences in n of the form k*b^n+c.
srsieve started: 10000 <= n <= 20000, 3 <= p <= 4000000000003
Split 1 base 73 sequence into 2 base 73^12 subsequences.
WARNING: 36*73^n-1 has algebraic factors.
WARNING: 36*73^n-1 has algebraic factors.
p=409039997, 6801123 p/sec, 9973 terms eliminated, 28 remain
p=846160019, 7272243 p/sec, 9977 terms eliminated, 24 remain
p=1297080067, 7507826 p/sec, 9980 terms eliminated, 21 remain
p=1754920133, 7615436 p/sec, 9980 terms eliminated, 21 remain
p=2217960077, 7713347 p/sec, 9981 terms eliminated, 20 remain
p=2685040051, 7774818 p/sec, 9981 terms eliminated, 20 remain
p=3156040061, 7841635 p/sec, 9981 terms eliminated, 20 remain
Sieving 3 <= p <= 3163531061 eliminated 9981 terms, 20 remain.
Wrote 20 terms for 1 sequence to abc format file `sr_73.pfgw'.
srsieve stopped: at p=3163531061 because SIGINT was received.

Either the sieve somehow got completely messed up, or this is one heck of a low-weight k.

Oops sorry. You can't prove your first conjecture so easily. That one is proven! I goofed. k=36 on Riesel base 73 has algebraic factors as shown by srsieve.

Sheesh, and I just now checked. That is an OBVIOUS one since it's a perfect square:
odd n: factor of 37
even n have algebraic factors: Where n=2q, it factors to (6*73^q-1)*(6*73^q+1)

Note: Not all k's that are perfect squares have algebraic factors to make a full covering set. Only on the Riesel side where all odd n have a trivial "numeric" factor like this one.

I eliminated a lot of algebraic factors on several of the bases but obviously missed this one.

I'll edit my original post. Sorry about that. Any time there are an abnormally low # of candidates remaining after sieving, it almost always means algebraic factors.


Gary

[gd_barnes]

Originally Posted by mdettweiler

Okay, that was easy. Riesel base 73 is now complete to n=20K. Results are attached.

Heck, if it's this easy, I'll just go on ahead and take it up to n=50K!

It looks like you posted this just ahead of my post. Don't waste your time! Like I said, any time there is an abnormally low # of candidates remaining upon sieving, there are likely algebraic factors, hence eliminating the k from consideration. The base is proven.

Edit: It looks like bases 109 and 110 are ripe for a proof.

An interesting note about base 110: k=36 on that base has exactly the same numeric and algebraic factors as does k=36 for base 73. That is a factor of 37 for odd n and where the base = b and n=2q for even n: (6*b^q-1)*(6*b^q+1). That's strange that I found this one and not the one for base 73. Brain fart I guess.


Gary

[mdettweiler]

Originally Posted by gd_barnes

Oops sorry. You can't prove your first conjecture so easily. That one is proven! I goofed. k=36 on Riesel base 73 has algebraic factors as shown by srsieve.

Sheesh, and I just now checked. That is an OBVIOUS one since it's a perfect square:
odd n: factor of 37
even n have algebraic factors: Where n=2q, it factors to (6*73^q-1)*(6*73^q+1)

Note: Not all k's that are perfect squares have algebraic factors to make a full covering set. Only on the Riesel side where all odd n have a trivial "numeric" factor like this one.

I eliminated a lot of algebraic factors on several of the bases but obviously missed this one.

I'll edit my original post. Sorry about that. Any time there are an abnormally low # of candidates remaining after sieving, it almost always means algebraic factors.


Gary

Whoops--okay, I'll be sure to keep on the lookout for those low-#-of-candidates k's in the future.

One quick question though: how do you know that all odd n have a factor of 37? It's probably something blindingly obvious but I'm missing it.

[gd_barnes]

Originally Posted by mdettweiler

Whoops--okay, I'll be sure to keep on the lookout for those low-#-of-candidates k's in the future.

One quick question though: how do you know that all odd n have a factor of 37? It's probably something blindingly obvious but I'm missing it.

The web page that I've expoused many times here:

http://www.alpertron.com.ar/ECM.HTM


Just plug in your form for n=1, 3, 5, 7, 9, etc. You'll see it has a factor of 37 every time. It takes any normal mathematical symbols. I use the page extensively and almost exclusively for determining such things. You just have to see the patterns in the way the factors occur. The factor patterns are one of the most interesting things to me about these conjectures.


Gary

[mdettweiler]

Originally Posted by gd_barnes

The web page that I've expoused many times here:

http://www.alpertron.com.ar/ECM.HTM


Just plug in your form for n=1, 3, 5, 7, 9, etc. You'll see it has a factor of 37 every time. It takes any normal mathematical symbols. I use the page extensively and almost exclusively for determining such things. You just have to see the patterns in the way the factors occur. The factor patterns are one of the most interesting things to me about these conjectures.


Gary

Ah, I see. I assumed that there was some kind of fancy mathematical formula that proved that all odd n had a factor of 37 or something like that.

At this time I haven't been able to get the ECM factoring applet you linked to to work under Linux on my computer (I've used it before under Windows), but I guess the important thing is that now I understand how you came to conclude that all odd n for k=36 have a factor of 37.

[Flatlander]

Originally Posted by mdettweiler

Ah, I see. I assumed that there was some kind of fancy mathematical formula that proved that all odd n had a factor of 37 or something like that.

At this time I haven't been able to get the ECM factoring applet you linked to to work under Linux on my computer (I've used it before under Windows), but I guess the important thing is that now I understand how you came to conclude that all odd n for k=36 have a factor of 37.

I have problems with it too. (Windows.)
But the batch one at the bottom of the page works okay.

I too thought Gary was using fancy stuff. Cheat!

[gd_barnes]

Originally Posted by mdettweiler

Ah, I see. I assumed that there was some kind of fancy mathematical formula that proved that all odd n had a factor of 37 or something like that.

At this time I haven't been able to get the ECM factoring applet you linked to to work under Linux on my computer (I've used it before under Windows), but I guess the important thing is that now I understand how you came to conclude that all odd n for k=36 have a factor of 37.

Oh, that's easy. It's just faster to use Alptertron's site. lol

36*73^1 = 2628 == (1 mod 37)
36*73^2 = 191844 == (36 mod 37)
36*73^3 = 14004612 == (1 mod 37)
36*73^4 = 1022336676 == (36 mod 37)
[etc.]

Now, when you subtract 1 from the above, you get (0 mod 37) for the odd n's. :-)

I don't believe it's a true proof in the classical sense of the word, but it demonstrates conclusively that all odd n are divisible by 37.

BTW, I'm pretty sure I got Alpertron's site to work on one of my Linux machines. I'll check it when I get home. But...I thought you had a Windows machine also.


Gary