PARI's commands - Page 108

science_man_88 · 2010-08-30, 16:21

Quote:

Originally Posted by CRGreathouse

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

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:

Originally Posted by science_man_88

well if you take only s sequences that are red you don't hit any that I can find.

Which are red?

Quote:

Originally Posted by science_man_88

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.

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 http://oeis.org/classic/?q=id%3AA180065|id%3AA180066?

Quote:

Originally Posted by 3.14159

@CRG: That any numbers of the supposed format you use form composites via 2^p - 1 ?

It's painfully obvious that for all those numbers, 2^p - 1 is composite.

Wrong, see either of my lists.

3.14159 · 2010-08-30, 16:31

Quote:

Originally Posted by CRGreathouse

Wrong, see either of my lists.

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

Quote:

Originally Posted by CRGreathouse

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

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

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

Captain Obvious's Conjecture:

For any given base b, k * b^b + 1 has an infinite amount of members.

Can you prove or disprove Captain Obvious's conjecture?

CRGreathouse · 2010-08-30, 16:46

Quote:

Originally Posted by 3.14159

Did you factor the b-file numbers on the list of base 2 pseudoprimes? Or did you generate them on your own?

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:

Originally Posted by 3.14159

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

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

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

mdettweiler · 2010-08-30, 16:46

Quote:

Originally Posted by 3.14159

I tested it with WinPFGW. The tests do work, fortunately.

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:

Originally Posted by 3.14159

Captain Obvious's Conjecture:

For any given base b, k * b^b + 1 has an infinite amount of members.

Can you prove or disprove Captain Obvious's conjecture?

given k=-infinity to k = infinity

k * b^b + 1 will therefore have infinite members because infinite k are involved.

CRGreathouse · 2010-08-30, 16:47

Quote:

Originally Posted by 3.14159

Captain Obvious's Conjecture:

For any given base b, k * b^b + 1 has an infinite amount of members.

Can you prove or disprove Captain Obvious's conjecture?

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

3.14159 · 2010-08-30, 16:48

Quote:

Originally Posted by CRGreathouse

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

No! The sequence applies to all b!

Quote:

Originally Posted by 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.

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 * 308³⁰⁸ + 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.

2010-08-30, 16:41	#1183
3.14159 May 2010 Prime hunting commission. 2⁴×3×5×7 Posts	Humor. Captain Obvious's Conjecture: For any given base b, k * b^b + 1 has an infinite amount of members. Can you prove or disprove Captain Obvious's conjecture? Last fiddled with by 3.14159 on 2010-08-30 at 16:44

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