20061003, 05:49  #1 
Aug 2005
2^{4} Posts 
Arithmetic and Polynomial Progression of Primes?
Interesting paper http://arxiv.org/PS_cache/math/pdf/0610/0610050.pdf .
I barely understand any of it. Can someone give me an example of primes in polynomial progression? Is there also a longest known set of primes in polynomial progression like there are longest known sets of primes in arithmetic progression? 
20061003, 12:31  #2  
Nov 2003
1110100100100_{2} Posts 
Quote:
Then (3,7,11) = (1 + P1(2), 1 + P2(2), 1 + P3(2)) (x = 1) The fact that this is also an AP is a coincidence because of the polynomials that I chose and that I evaluated it at 2. Evaluate it at z = 4 and you get the polynomial progression 7,23,71 with (x = 3) 

20061003, 16:18  #3 
Jan 2005
Transdniestr
503 Posts 
I noticed the next two terms in the sequence are primes as well (263 and 1071)
One small thing: In order to get more than two terms in your sequence (unless z=2 and x=1), the following must be true: z%6 = 4 and x%6 = 3 Last fiddled with by grandpascorpion on 20061003 at 16:33 
20061003, 17:19  #4 
Jan 2005
Transdniestr
767_{8} Posts 
z=4, x=33 yields a sequence of 7 primes: 37, 53, 101, 293, 1061, 4133, 16421
Can anyone find a longer one? 
20061004, 05:39  #5 
Aug 2005
10_{16} Posts 
I think I may have learned something.
How about z=4 x=159 to get: 163, 179, 227, 419, 1187, 4259, 16547, 65699 
20061004, 06:53  #6 
Aug 2005
10000_{2} Posts 
This seems to be the longest polynomial progression of primes for z=4 and x<100000000 for the particular polynomials.
z=4 x=33890799 33890803, 33890819, 33890867, 33891059, 33891827, 33894899, 33907187, 33956339, 34152947, 34939379, 38085107 
20061007, 21:59  #7 
Aug 2005
2^{4} Posts 
Sequence of 12 primes in polynomial progression
z=16 x=244866027 244866043, 244866299, 244870139, 244931579, 245914619, 261643259, 513301499, 4539833339, 68964342779, 1099756493819, 17592430910459, 281475221576699 Can anyone find a sequence of 13? 
20061008, 12:39  #8 
Jan 2005
Transdniestr
503 Posts 
Sequence of 13
z=16
x=836880681 836880697,836880953,836884793,836946233,837929273, 853657913,1105316153,5131847993,69556357433,1100348508473, 17593022925113,281475813591353,4503600464251193 
20061008, 13:12  #9  
Nov 2003
2^{2}·5·373 Posts 
Enough!
Quote:
So what point is there to these posts? They are not even numerical curiosities. I can easily find such progressions of length 50, 100, 1000, etc. Given *any* set of increasing primes, I can find a polynomial to match. These posts are *pointless*. I can find a polynomial and a value of x so that x + (f(1), f(2), f(3), f(4), ....) = 3,5,7,11,13,17,19,23,29,31,.... Or for ANY specified set of primes on the right. Finding a specific instance for one value of x is *meaningless*. Try finding a polynomial and TWO different values of x etc. such that x1 + f(1), x1 + f(2), etc. AND x2 + f(1), x2+ f(2), x2 + f(3) etc. are prime. Or find 3 such x. The Tao result is fascinating. These numerical examples are a total waste of computer and people time. 

20061008, 16:54  #10  
Feb 2006
Denmark
11100110_{2} Posts 
Quote:
Choosing the polynomials first like your z, z^2+z, z^3+z, ... makes it a challenge. Your suggestion to find 2 or 3 values of x with presumably arbitrary polynomials is no challenge. Just compute e.g. the first million prime triplets on form (p, p+2, p+6). Then choose a polynomial of degree one million to run through the p values, and let x1=0, x2=2, x3=6. The solution would not be suited for listing here (or anywhere else). Restrictions are needed to make it computationally interesting. Here is a variant with two x values leading to many consecutive primes: If x is 46863043340173267 or 534402442999154537, then x + 0,2,56,62,80,110,146,150,152,170,230,234,252,264,276,290,294,296,300,302,344 are 21 consecutive primes. Can anybody beat that? Computing a corresponding polynomial would be a simple exercise. 

20061008, 20:52  #11  
Jan 2005
Transdniestr
1F7_{16} Posts 
Quote:
Oh well, that is what I get for not looking at the paper. 

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