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Old 2009-08-20, 02:16   #1
Dougy
 
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Default On prime chains

I've updated the paper I submitted to the arXiv here. It is now entitled "On prime chains." It gives some interesting, but fairly minor results about sequences of primes (p_k)_{k=0}^{\lambda-1} such that p_k=ap_{k-1}+1 for all 1 \leq k \leq \lambda-1.

The second version expands the results of first version, improves the literature review and corrects some typos I made (which would have been very confusing for whomever read the first version).

I'm somewhat tempted to submit this to some mediocre journal - but I think i'd prefer it if someone came up with some good ideas, helped make it into a better paper and they could become co-author. But, in any case, I'd appreciate any feedback.

Last fiddled with by Dougy on 2009-08-20 at 02:26
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Old 2009-08-20, 02:59   #2
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"Lehmer [7] remarked that Dickson’s Conjecture [3], should it be true, would imply that there are infinitely many prime chains of length \lambda based on the pair (a, b), with the exception of some inappropriate pairs (a, b)."

Out of curiosity, what are the inappropriate pairs? Anything more interesting than just a and b sharing a common factor?
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Old 2009-08-20, 06:11   #3
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There'd also be some others. For example if (a,b)=(3,1) and p(k) is odd, then p(k+1)=3*p(k)+1 is even. Lehmer didn't explain this very well... hmm...

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Old 2009-08-20, 10:54   #4
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Quote:
Originally Posted by Dougy View Post
There'd also be some others. For example if (a,b)=(3,1) and p(k) is odd, then p(k+1)=3*p(k)+1 is even. Lehmer didn't explain this very well... hmm...
I suppose something similar happens anytime there exists an N for which a=1 mod N and b is coprime to N. Working modulo N, p(k)=p(0)+kb mod N, so you'll always get something divisible by N when k=p(0)*b^-1. I feel like there should be a few more ways you can "trivially" guarantee a factor of N in a bounded number of steps if a,b, and N satisfy certain relations, but I'm not prepared to take that on or look up the reference since it's close to 6am local time.
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Old 2009-08-24, 12:15   #5
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I'm trying to track down some references from the Loh paper:

Takao Sumiyama, "Cunningham chains of length 8 and 9," Abstracts Amer. Math. Soc., 4 (1983) p. 192.
Takao Sumiyama, "The distribution of Cunningham chains," Abstracts Amer. Math. Soc., 4 (1983) p. 489.

Has anyone heard of "Abstracts Amer. Math. Soc."? I'm not sure what this means, it could just be a list of talk abstracts or something. Any help would be appreciated.
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Old 2009-08-24, 12:20   #6
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Quote:
Originally Posted by Dougy View Post
I'm trying to track down some references from the Loh paper:

Takao Sumiyama, "Cunningham chains of length 8 and 9," Abstracts Amer. Math. Soc., 4 (1983) p. 192.
Takao Sumiyama, "The distribution of Cunningham chains," Abstracts Amer. Math. Soc., 4 (1983) p. 489.

Has anyone heard of "Abstracts Amer. Math. Soc."? I'm not sure what this means, it could just be a list of talk abstracts or something. Any help would be appreciated.
AFAIK, this particular journal was discontinued some time ago. I did receive
it as an AMS member back in the 80's.
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Old 2009-08-24, 13:44   #7
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Thanks. It looks like they'll be tricky to track down.
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Old 2009-08-24, 13:47   #8
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Quote:
Originally Posted by Dougy View Post
Thanks. It looks like they'll be tricky to track down.
Be aware that it was not peer reviewed in any way. Any member
could submit an abstract at any time. Said abstract did not need to make
sense.
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Old 2009-08-24, 22:29   #9
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Quote:
Originally Posted by R.D. Silverman View Post
Be aware that it was not peer reviewed in any way. Any member
could submit an abstract at any time. Said abstract did not need to make
sense.
If that's the case, it's probably not worth my time. I'll just say something like "Loh said Sumiyama said..."
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Old 2009-09-05, 22:37   #10
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Here's another "trivial" one that I spotted... if you choose a=-1 and b=p_0+p_1. Then the sequence is p_0,p_1,p_0,p_1,... and so on.
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Old 2009-09-11, 21:05   #11
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So anyway, I ended up submitting an expanded version of what's on the arXiv. Every paper counts when you're looking for a postdoc.
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