
#1
Feb812, 07:57 PM

P: 891

Ramsey Primes are those generated from a simple criteria that is easy to check. I checked all odd numbers from 1 to 1 million and 29455 numbers met the criteria. None were composite.
The check is to do the following sequence mod P and check to see that the (P1)/2 term is zero and no term prior to that is zero. The test sequence is S(0) = 2, S(1) = 3, S(n) = 6*S(n1)  S(n2)  6. If P is prime, then S((P1)/2) is divisible by P, but I am interested in a test that is valid only for primes. S((351)/2) is divisible by 35 but that is not the first term divisible by 35. Only primes seem to meet the more restricted criteria, i.e. have no term divisible by P prior to S((P1)/2). The first two Ramsey primes are 11 and 13 which I call a Ramsey Twin. The largest Ramsey Twin under 1 million is (998651, 998653). Any thoughts are welcome. Edit So far about 5 in 13 primes are Ramsey Primes so my test has limited value unless it can be proven that only primes can meet the test. Any way to prove this? 



#2
Feb912, 07:25 AM

P: 891

Still about 5 in 13 primes are Ramsey Primes. Also, Primes 3 and 5 are Ramsey Primes. 



#3
Feb1012, 05:14 PM

P: 891

My code for Mathematical is short so I will include it below for reference. There are some flaws. For instance the Reap/Sow function generates a list of the form {Null,{{prime3,prime5,prime6,prime8,...}}}. I want to generate a simple list {prime3,prime5,prime6,prime8...}. Also, the code took about an hour to pick out 31 primes between 1049000 and 1050000. Overall, about 2/3 of the prime numbers are not picked out, but that is what I am looking for code that will generate only primes, not code that will generate all primes but also include some composites. So far some 30,000 prime numbers between 5 and 1050000 were generated with no false hits. Any help improving my code will be appreciated.
Caa = 5 Reap[While[Caa < 1050001, Co = 1; S0 = 2; S1 = 3; While[Co < ((Caa1)/2),Temp = Mod[6 S1  S0  6,Caa];If[Temp <= 1,Break[]]; S0=S1;S1=Temp;Co++]; n = ((Caa1)/2)  Co; If[n<2,If[n>0, Sow[Caa]]]; Caa++;Caa++]] 



#4
Feb1512, 09:20 AM

P: 891

Ramsey Primes2,3,10,51 is of the Form S(n) = 6S(n1) S(n2)  K K = 6 and Test is S((P1)/2) = 0, Mod P etc 1,2,9,50 is of the Form S(n) = 6S(n1)  S(n2)  K K = 2 and Test is S((P1)/2) = 1, Mod P etc Subtracting 1 from each term reduced K by 4 Doubling each term multiplies K by 2, i.e.: 2,4,18,100 is of the Form 6S(n1)  S(n2)  K K = 4 and Test is S((p1)/2 = 2, Mod P, etc Subtracting 1 from each term again reduces K by 4 i.e. to 0,: 1,3,17,99 is of the Form 6S(n1)  S(n2) K = 0 Test is S((P1)/2) = 3, Mod P, etc. Thus a direct formula exists for S((P1)/2) 



#5
Feb1612, 12:03 AM

P: 891

Bill Simpson help me with the following Mathematica code which runs more than 20 times faster.
In: Timing[f = Compile[{{Co, _Integer}, {S0, _Integer}, {S1, _Integer}, {Caa, \ _Integer}}, Module[{xCo = Co, xS0 = S0, xS1 = S1, Temp}, While[Temp = Mod[6 xS1  xS0  6, Caa]; xCo > 0 && Temp >= 1, xS0 = xS1; xS1 = Temp; xCo]; xCo]]; Caa = 349049001; Reap[While[Caa < 349049501, Co = (Caa  3)/2; S0 = 2; S1 = 3; If[f[Co, S0, S1, Caa] == 1, Sow[Caa]]; Caa += 2]]] Out: {11001.066`, {Null, {{349049059, 349049069, 349049179, 349049291, 349049293, 349049317, 349049333, 349049347, 349049411, 349049443}}}} According to Mathematica there are 21 primes in the range and the code found 10 of them. Again no composites were picked out. The code appears to have a better chance of picking a given prime as the size of the prime increases. I still believe that the code can be made to speed up by first checking whether the (P1)/2 term in the sequence equals 0 mod P before doing the recursive algorithim to see if any prior term equals 0 mod P. Also inspection of the location of a prior zero may be a step in factoring composite numbers P whose (P1)/2 term = 0 mod P. 



#6
Feb1612, 01:21 AM

P: 144

Hi,
I find this interesting, and want to look into what is going on here. Myfirst thougt was about the name Ramsey, have you just named the primes after yourself, or is there some other connection? Then where did the test sequence come from? Did you try all sequences, or did you see that this particular one had some good properties. To prove your claim, if true, one could try to use the closed form expression for S(n) 4S(n) = r^{n}+s^{n}+6, with r=3+2√2 and s=32√2, (hope i got that right) and see what can be done over Q(√2). 



#7
Feb1612, 03:37 AM

P: 688

[tex]\begin{align*} S(n) &= \frac {(3+2\sqrt 2)^n + (32\sqrt 2)^n} 4 + \frac 3 2 \\ &= \frac {\left( \sum_{k=0}^{\lfloor\frac n 2\rfloor} \binom n {2k} 3^{n2k} 8^k \right) + 3} 2 \\ &= \frac {3^n + 8m + 3} 2 \quad,\quad {\scriptsize \mbox{ for some integer }m}\end{align*}[/tex] From "empirical evidence" I was hoping to find something related to the quadratic nature of 2 mod p, as S((p1)/2) appears to be divisible by p (I think) for primes p of the form 8ką3. But to no avail yet. 



#8
Feb1612, 10:58 AM

P: 891

Your direct formula agrees with my data and the formula posted for OEIS A001541 and A003499 so I believe you are right. I am not very well versed in number theory and I often make careless errors, so I appreciate your post. Followup, in the range 349049001349049500, primes 349049123 and 349049227 which are of the form 8n +/ 3 (two such primes out of 12) do not pass my prime test, so I guess we may as well name the group of primes that do pass the test "Ramsey Primes" 



#9
Feb1612, 12:55 PM

P: 688




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