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science_man_88 2010-08-30 16:21

[QUOTE=CRGreathouse;227694]110503 = 2^3*(13812+1)-1
132049 = 2^1*(66024+1)-1
216091 = 2^2*(54022+1)-1
756839 = 2^3*(94604+1)-1
859433 = 2^1*(429716+1)-1
1257787 = 2^2*(314446+1)-1
1398269 = 2^1*(699134+1)-1
2976221 = 2^1*(1488110+1)-1
3021377 = 2^1*(1510688+1)-1
6972593 = 2^1*(3486296+1)-1
13466917 = 2^1*(6733458+1)-1[/QUOTE]

figures I suck lol well if you take only s sequences that are red you don't hit any that I can find. and knock out Mersenne exponents if you use Mersenne numbers

can we use your disproof for a way to find Mersenne exponents for example all of these seem to use 2^1 2^2 or 2^3 anyways I guess I need a new idea lol.

CRGreathouse 2010-08-30 16:27

[QUOTE=science_man_88;227697]well if you take only s sequences that are red you don't hit any that I can find.[/QUOTE]

Which are red?

[QUOTE=science_man_88;227697]can we use your disproof for a way to find Mersenne exponents for example all of these seem to use 2^1 2^2 or 2^3 anyways I guess I need a new idea lol.[/QUOTE]

If these are randomly distributed (as Dirichlet tells us for primes in general) then you'd expect about half of them to have 2^1, a quarter to have 2^2, and an eighth to have 2^3. That seems to match my numbers reasonably. Here are more terms, you can check them if you like.

3 = 2^2*(0+1)-1
5 = 2^1*(2+1)-1
7 = 2^3*(0+1)-1
13 = 2^1*(6+1)-1
17 = 2^1*(8+1)-1
19 = 2^2*(4+1)-1
31 = 2^5*(0+1)-1
61 = 2^1*(30+1)-1
89 = 2^1*(44+1)-1
107 = 2^2*(26+1)-1
127 = 2^7*(0+1)-1
521 = 2^1*(260+1)-1
607 = 2^5*(18+1)-1
1279 = 2^8*(4+1)-1
2203 = 2^2*(550+1)-1
2281 = 2^1*(1140+1)-1
3217 = 2^1*(1608+1)-1
4253 = 2^1*(2126+1)-1
4423 = 2^3*(552+1)-1
9689 = 2^1*(4844+1)-1
9941 = 2^1*(4970+1)-1
11213 = 2^1*(5606+1)-1
19937 = 2^1*(9968+1)-1
21701 = 2^1*(10850+1)-1
23209 = 2^1*(11604+1)-1
44497 = 2^1*(22248+1)-1
86243 = 2^2*(21560+1)-1
110503 = 2^3*(13812+1)-1
132049 = 2^1*(66024+1)-1
216091 = 2^2*(54022+1)-1
756839 = 2^3*(94604+1)-1
859433 = 2^1*(429716+1)-1
1257787 = 2^2*(314446+1)-1
1398269 = 2^1*(699134+1)-1
2976221 = 2^1*(1488110+1)-1
3021377 = 2^1*(1510688+1)-1
6972593 = 2^1*(3486296+1)-1
13466917 = 2^1*(6733458+1)-1

CRGreathouse 2010-08-30 16:29

Pi: What did you think of [url]http://oeis.org/classic/?q=id%3AA180065|id%3AA180066[/url]?

[QUOTE=3.14159;227699]@CRG: That any numbers of the supposed format you use form composites via 2[sup]p[/sup] - 1 ?

It's painfully obvious that for all those numbers, 2[sup]p[/sup] - 1 is composite.[/QUOTE]

Wrong, see either of my lists.

3.14159 2010-08-30 16:31

[QUOTE=CRGreathouse]Wrong, see either of my lists.
[/QUOTE]

It was then I realized those were the exponents for the prime Mersenne numbers.


[QUOTE=CRGreathouse]Pi: What did you think of [url]http://oeis.org/classic/?q=id%3AA180065|id%3AA180066?[/url][/QUOTE]

You really did expand on them. Excellent. Also: Did you factor the b-file numbers on the list of base 2 pseudoprimes? Or did you generate them on your own?

I'm up to b = 389. I have an idea. I should write the sequence to the OEIS, starting from 60.

science_man_88 2010-08-30 16:36

[QUOTE]3 = 2^2*(0+1)-1
5 = 2^1*(2+1)-1
7 = 2^3*(0+1)-1
13 = 2^1*(6+1)-1
17 = 2^1*(8+1)-1
19 = 2^2*(4+1)-1
31 = 2^5*(0+1)-1
61 = 2^1*(30+1)-1
89 = 2^1*(44+1)-1
107 = 2^2*(26+1)-1
127 = 2^7*(0+1)-1
521 = 2^1*(260+1)-1
607 = 2^5*(18+1)-1
1279 = 2^8*(4+1)-1
2203 = 2^2*(550+1)-1
2281 = 2^1*(1140+1)-1
3217 = 2^1*(1608+1)-1
4253 = 2^1*(2126+1)-1
4423 = 2^3*(552+1)-1
9689 = 2^1*(4844+1)-1
9941 = 2^1*(4970+1)-1
11213 = 2^1*(5606+1)-1
19937 = 2^1*(9968+1)-1
21701 = 2^1*(10850+1)-1
23209 = 2^1*(11604+1)-1
44497 = 2^1*(22248+1)-1
86243 = 2^2*(21560+1)-1
110503 = 2^3*(13812+1)-1
132049 = 2^1*(66024+1)-1
216091 = 2^2*(54022+1)-1
756839 = 2^3*(94604+1)-1
859433 = 2^1*(429716+1)-1
1257787 = 2^2*(314446+1)-1
1398269 = 2^1*(699134+1)-1
2976221 = 2^1*(1488110+1)-1
3021377 = 2^1*(1510688+1)-1
6972593 = 2^1*(3486296+1)-1
13466917 = 2^1*(6733458+1)-1[/QUOTE]

well technically I know way's to put these all into a form 2^1,2^2 or 2^3, but I doubt anything would come to a pattern.

3.14159 2010-08-30 16:41

Humor.
 
Captain Obvious's Conjecture:

For any given base b, k * b[sup]b[/sup] + 1 has an infinite amount of members.

Can you prove or disprove Captain Obvious's conjecture?

CRGreathouse 2010-08-30 16:46

[QUOTE=3.14159;227704]Did you factor the b-file numbers on the list of base 2 pseudoprimes? Or did you generate them on your own?[/QUOTE]

For both I used Jan Fetisma's list of pseudoprimes, which took several CPU-years to calculate with an advanced algorithm.

The graph of A180066 is intriguing and should be studied further.

[QUOTE=3.14159;227704]I'm up to b = 389. I have an idea. I should write the sequence to the OEIS, starting from 60.[/QUOTE]

One sequence for each b? I hope not...

This is better suited to a project page, like [url]http://www.15k.org/riesellist.html[/url]

mdettweiler 2010-08-30 16:46

[quote=3.14159;227693]I tested it with WinPFGW. The tests do work, fortunately.[/quote]
Hmm...I tried testing the number on my own computer to see if the problem is repeatable (indicating a program bug) but it wasn't--the test finished on the first try, no roundoff error. That points to a possible hardware error on your computer. You may want to try running a Prime95 stress test on all cores for a while (12 hours at least) to make sure that you're not throwing bad results here and there and potentially missing primes.

science_man_88 2010-08-30 16:47

[QUOTE=3.14159;227706]Captain Obvious's Conjecture:

For any given base b, k * b[sup]b[/sup] + 1 has an infinite amount of members.

Can you prove or disprove Captain Obvious's conjecture?[/QUOTE]

given k=-infinity to k = infinity

k * b[sup]b[/sup] + 1 will therefore have infinite members because infinite k are involved.

CRGreathouse 2010-08-30 16:47

[QUOTE=3.14159;227706]Captain Obvious's Conjecture:

For any given base b, k * b[sup]b[/sup] + 1 has an infinite amount of members.

Can you prove or disprove Captain Obvious's conjecture?[/QUOTE]

sm solves that in #1186. Assuming you mean prime members, I addressed it in #1138. :cool:

3.14159 2010-08-30 16:48

[QUOTE=CRGreathouse]One sequence for each b? I hope not...
[/QUOTE]

No! The sequence applies to all b!

[quote=Matt]Hmm...I tried testing the number on my own computer to see if the problem is repeatable (indicating a program bug) but it wasn't--the test finished on the first try, no roundoff error. That points to a possible hardware error on your computer. You may want to try running a Prime95 stress test on all cores for a while (12 hours at least) to make sure that you're not throwing bad results here and there and potentially missing primes.
[/quote]

Well, you stated that this happened to you as well, which makes me doubt that it happens to be only me.

And, it only fucks up when dealing with k * 308[sup]308[/sup] + 1. There have been no other errors before or afterwards. I'm going to test the original Primeform to make sure that this is just a minor problem.


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