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Old 2013-07-06, 18:43   #1
Citrix
 
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Default Prime generating series

Question: Are there finite terms in these sequence?

Sierpinski:1, 1, 2, 1, 5, 1, 1, 29, 3, 37, 31, 227, 835, 115
Riesel: 2, 1, 2, 1, 1, 7, 15, 35, 619, 2191, 1267, 187
Checked up to 5000 bits.

How to generate the sequence: (for Sierpinski side)
Start with k=1
Then find lowest n such that k*2^n+1 is prime
n is the first term

Then k=k*2^n+1
Then find lowest n such that k*2^n+1 is prime
n is the second term

and so on....

(Similar for the riesel side).

Any thoughts on what the weights of the k generated will be and if they will ever end in a sierpinski or riesel number.

What is the most efficient way to generate this series. PFGW's script seems to fail after the first few sequences.
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Old 2013-07-08, 04:53   #2
LaurV
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7615 is the next. Small pari script, quite fast. Have no idea about the theoretical question, tho.
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Old 2013-07-08, 10:02   #3
henryzz
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I can add 3755 to the Riesel list.
I can also add 6071 after LaurV's addition on the Sierpinski list.

I have tested upto n=12500 and proved all primes.

I have been using a pfgw script. It runs nice and fast assuming you use -f to turn on trial factoring.
You could do with sieving to go much further than I have done though.

My script will do sequences of the form k*b^n+c. b and c can be changed as can the minimum n to test and the starting k.

Last fiddled with by henryzz on 2013-07-08 at 10:04
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Old 2013-07-08, 10:16   #4
LaurV
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Confirm for 6071. I got it too, but being busy here and forgot to post.
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Old 2013-07-08, 10:23   #5
henryzz
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Other bases don't really work as you end up reaching a k with a trivial factor quickly.
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Old 2013-07-08, 10:37   #6
henryzz
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I don't know how to prove that it never stops at a riesel k for starting k=1 but it is trivial to prove that some prime starting ks terminate.
For k=127301 n=2 is prime and this gives k=509203 for the next iteration. 509203 is a riesel k.
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Old 2013-07-08, 15:48   #7
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Discovered I left riesel k=7 running by accident.
The current status of all sequences is:
k=1, -1: 2, 1, 2, 1, 1, 7, 15, 35, 619, 2191, 1267, 187, 3755 >12500
k=1, +1: 1, 1, 2, 1, 5, 1, 1, 29, 3, 37, 31, 227, 835, 115, 7615, 6071 >12500
k=5, +1: 1, 1, 1, 583, 79, 371, 1439, 27, 1067 >6000
k=7, -1: 1, 3, 3, 11, 35, 3, 19, 23, 11, 59, 315, 2707, 223, 2807, 1627, 2739, 5171, 2243 >26500
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Old 2013-07-08, 16:55   #8
Citrix
 
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Could you share your PFGW script with me. I was getting errors with PFGW when I programmed it. I would like to extend the series higher.

What bases give trivial factors for k=1?

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Old 2013-07-08, 18:29   #9
henryzz
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Quote:
Originally Posted by Citrix View Post
Could you share your PFGW script with me. I was getting errors with PFGW when I programmed it. I would like to extend the series higher.

What bases give trivial factors for k=1?

Code:
SCRIPT

DIM base, 2
DIM k, 5
DIM min_n, 1
DIM n, min_n
DIM type, 1
DIMS type_str,+1
DIMS test_str
OPENFILEAPP prp_file,primeseriesprps.txt

LABEL next_k
SET n, min_n

LABEL next_n
SETS test_str,%d*%d^%d%s;k;base;n;type_str
PRP k * base ^ n + type, test_str

IF (ISPRP) THEN GOTO prp_found
SET n,n+1
GOTO next_n

LABEL prp_found
PRINT test_str
SET k, k * base ^ n + type
WRITE prp_file,test_str
GOTO next_k
I think all bases other than 2 give trivial factors. Give it a try and then put the sequence it gets stuck on into factordb.com . One factor will divide every candidate.
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Old 2013-07-08, 19:36   #10
Citrix
 
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Thanks!
I tried your script for k=1 and base=30, tested upto 40,000 bits but it has not turned into trivial candidates yet.
For base=2 it crashed on me. I am trying it again.

Can we develop some trivial solutions like:
1) If the base is odd... it will turn into trivial solution.
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Old 2013-07-08, 22:40   #11
henryzz
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Well done in finding another that works. I have had very limitted success in finding more. All were multiples of 30 although I am not sure that is a requirement.
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