mersenneforum.org > Math Prime progressions...
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2003-08-04, 06:55   #1
Xyzzy

"Mike"
Aug 2002

32·883 Posts
Prime progressions...

On one of the mailing lists lately there was some talk of prime number progressions that were described as equations... I was wondering what method people use to find these and what the largest might be?

Here is an example from the list...

Quote:
 Just in the last week Gary Chaffey reported that the equation 2*x2-88*x+997 generates 51 primes from x=0 to x= 50
I think it might be fun for us to talk about this here... (At least I'm really interested!)

Thanks!

 2003-08-04, 07:27 #2 ET_ Banned     "Luigi" Aug 2002 Team Italia 2×3×17×47 Posts Try a search with the keywords "Ulam spiral": there are plenty of sites that try to implement such an equation, but AFAIK no one has still found "the" equation :-) Luigi
 2003-08-04, 09:29 #3 TTn   22×52×59 Posts Better yet, draw the ulam spiral on graph paper, using only odd numbers. Our beloved Mersenne primes fall directly along the diagonal. :D Hmm here is a wacky equation for that diagonal. __n \ /__ 8(2n+1) n=0 minus one.
 2003-08-04, 23:32 #4 dsouza123     Sep 2002 2·331 Posts Are there any rules, algorithms, heuristics for eliminating numbers or ranges of numbers on the diagonal ? ( Other than doing a TF/LL or under a certain range have all been tested so they can be excluded )
 2003-08-05, 00:20 #5 TTn   3·13·241 Posts I know of none yet drawn, but a real smart mathematician could easily return the probability of n, being prime in the above. You could then mix the current probability of mersenne primes with this, to get a general hybrid prob. Though all Mn, with n odd appear on the diagonal, so I'm not sure any of this will help... well maybe it could then in that case. :?
 2003-08-05, 16:01 #6 koal   Nov 2002 Vienna, Austria 41 Posts Hi there! Somebody proved that a polynomial never will produce just primes. But if I can provide a function, that produces the first 25 primes (those below 100) ... y = ( -1961755 x^24 + 626211420 x^23 - 94460687338 x^22 + 8955156636096 x^21 - 598606037462125 x^20 + 30003592036428780 x^19 - 1170739327278578728 x^18 + 36446058560245549776 x^17 - 920325981234349785205 x^16 + 19064007207309706990260 x^15 - 326336795367642933428338 x^14 + 4636387497021856699615296 x^13 - 54768570875443768152863635 x^12 + 537643906122042660114998340 x^11 - 4373844384977844634682276188 x^10 +29335829725530079227920667696 x^9 - 160950424593447516378672853840 x^8 + 714365040902901226039992880320 x^7 - 2525953621133868833652063283008 x^6 + 6966549186063153961611970043136 x^5 - 14546083395318615094880191933440 x^4 + 22007289038111101518179578490880 x^3 - 22507759972998750778418906726400 x^2 + 13727914564300379016566243328000 x - 3700818363341536479443681280000 ) / 620448401733239439360000 ... what stops me to create a function, that provides the first n primes (or even the first n+1)? Okay, I know, the message is "MACHINE STORAGE EXHAUSTED" :? for x=26, y becomes -1560168 which is unfortunately even, so the prover mentioned before is right, but the function above really produces all primes from 2 to 97. It behaves like a curve, which goes through the 25 points with coordinates (1/2), (2,3), (3,5), ..., (23,83), (24,89), (25,97)! Are there limits except storage? Are there any smaller polynomials for the first 25 (or first n) primes? Koal 8)
 2003-09-20, 02:13 #7 ixfd64 Bemusing Prompter     "Danny" Dec 2002 California 2·32·131 Posts 50 primes? Wow! The best formula I knew before that was p(x) = x2 + x + 41.
 2003-09-22, 11:54 #8 koal   Nov 2002 Vienna, Austria 41 Posts The numbers are 997, 911, 829, 751, 677, 607, 541, 479, 421, 367, 317, 271, 229, 191, 157, 127, 101, 79, 61, 47, 37, 31, 29 , 31, 37, 47, 61, 79, 101, 127, 157, 191, 229, 271, 317, 367, 421, 479, 541, 607, 677, 751, 829, 911, 997 , 1087, 1181, 1279, 1381, 1487, 1597 As you can see, 22 of them are duplicate, which reduces the number of unique primes to 29. x2 + x + 41 => 40 unique primes x2 - 79 + 1601 => 80 primes, 40 of them ar duplicates
 2004-12-02, 14:27 #9 synergy   Aug 2004 2·3·7 Posts Not strictly polynomials but... Though they aren't strictly polynomials (I don't know if you say my previous posts), these will give a lot of early primes without composites... 105 - 2^x 15 - 2^x 45 - 2^x 70 - 3^x 75 - 2^x The "plus" version of these gives alot of them, also. Aaron
2004-12-13, 16:53   #10
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

205210 Posts
Prime progressions.

Quote:
 Originally Posted by koal The numbers are 997, 911, 829, 751, 677, 607, 541, 479, 421, 367, 317, 271, 229, 191, 157, 127, 101, 79, 61, 47, 37, 31, 29 , 31, 37, 47, 61, 79, 101, 127, 157, 191, 229, 271, 317, 367, 421, 479, 541, 607, 677, 751, 829, 911, 997 , 1087, 1181, 1279, 1381, 1487, 1597 As you can see, 22 of them are duplicate, which reduces the number of unique primes to 29. x2 + x + 41 => 40 unique primes x2 - 79 + 1601 => 80 primes, 40 of them ar duplicates

You have made a typographical error in the function in the last line.
Do you mean the function I posted in Thread 'Prime generating polynomials' ? which is f(n) = n^2 -79n +1601

Which function do your duplicate primes refer too ? They certainly dont belong to the function f(n) = n^2 -79n +1601
This function has 80 distinct primes.

Mally

'\\\\\\\\\/l

2004-12-15, 19:46   #11

"Richard B. Woods"
Aug 2002
Wisconsin USA

22×3×641 Posts

Quote:
 Originally Posted by mfgoode You have made a typographical error in the function in the last line. Do you mean the function I posted in Thread 'Prime generating polynomials' ? which is f(n) = n^2 -79n +1601
I'm pretty sure that's what koal meant.

Quote:
 Which function do your duplicate primes refer too ? They certainly dont belong to the function f(n) = n^2 -79n +1601 This function has 80 distinct primes.
Nope. Only 40 distinct prime values for the range n = 0 through 79. (Though who knows how many for n outside that range!) :)

The function is symmetrical around the line n = 39.5

f(0) = 0 - 0 + 1601 = 1601
f(79) = 79^2 - 79^2 + 1601 = 1601 = f(0)

f(39) = 39*39 - 79*39 +1601 = -40*39 +1601 = 41
f(40) = 40*40 -79*40 + 1601 = -39*40 + 1601 = 41 = f(39)

f(79-m) = (79-m)*(79-m) - 79*(79-m) + 1601
= 79^2 - 79m - 79m + m^2 - 79^2 + 79m + 1601
= m^2 - 79m + 1601
= f(m)

Last fiddled with by cheesehead on 2004-12-15 at 19:59

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