20050724, 13:24  #1 
Jun 2005
2×7^{2} Posts 
A New Conjecture
Hi,
Earlier this year when the new largest prime was announced, it restimulated my interest in number theory and prime numbers. In investigating primes in the quest to understand them (silly me), I discovered a new property (that means I have not come across it before and it might be known any way). Let N be an even number and Q an odd number such that P1 = Abs(NQ) and P2 = N+Q are both prime Conjecture 1: For every even number N there are infinite pairs of primes P1, P2 such that either the sum of P1 and P2 divided by two equals N, or the difference of P1 and P2 divided by two equals N . Conjecture 2: GCD(N,Q) = 1 Conjecture 3: The minimum number of distinct factors of N and Q is three. That means if Q = p^y then N has at minimum factors (2 and x) and if N=2^y then Q has at minimum factors (3 and x) Conjecture 4: If N=p# then the smallest value of Q is Prime. or If N=n! then the smallest value of Q is Prime. Conjecture 5: Iff Px and Py are base 2 probable primes, with Qx =(PxPy)/2 and Nx = (Px+Py)/2 and Qx > largest factor of Nx and Qx is the minimum solution to finding Px and Py, and GCD(Qx, Nx)= 1 , then both Px and Py are prime. I have named this the EvenPrimeBalance Conjecture and defined the EPB function EPB(N, Q)= True , Iff N+Q is Prime AND NQ is Prime You can look at the conjecture at (basically the same as above) http://www.geocities.com/al_at_i/epb.html Your comments would be much appreciated (anton@alv.net) and has the EPB conjecture any significance in basic number theory The conjecture is based on research of small and large N. I have tested above for many N, and checked it using Pa = pseudo primes and Pb being the nearest prime (larger and smaller) and found above to hold. best regards Anton 
20050724, 14:50  #2  
Nov 2003
2^{2}·5·373 Posts 
Quote:
immediately from 2. 4 is certainly false. Put N = 101#. It is virtually certain that there exists p1, p2 , each > 101, such that N+p1p2 and Np1p2 are both prime. I do not understand what you are trying to say in 5. Please clarify. 

20050724, 15:45  #3  
Jun 2005
2·7^{2} Posts 
Quote:
Q=523, 739, 1307, 2971, 3709, 3889, 5981, 7393, 7879, 10909, 12757, 14369, 107x139, 16333, 16831,18523, 19273, 19979, 21017, 21937, 23131, 24251, 27751, 107x271, 29759 etc etc The smallest value of Q is prime hence conjecture 4 is true, the conjecture does not say each Q is prime Quote:
Last fiddled with by AntonVrba on 20050724 at 15:50 

20050724, 16:27  #4  
Jun 2005
2×7^{2} Posts 
Quote:
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20050724, 20:48  #5  
Nov 2003
2^{2}·5·373 Posts 
Quote:
is some p such that the smallest Q will be composite. We just haven't look high enough. And I still don't follow what you are trying to say in Conj 5. Please expand upon the conjecture. 

20050725, 02:11  #6 
Aug 2002
Buenos Aires, Argentina
13×107 Posts 
A proof of conjecture 4 is very difficult to obtain.
But if some conjectures about prime gaps are valid, for instance G(n) < (ln n)^2 (see http://mathworld.wolfram.com/PrimeGaps.html), then the conjecture 4 is valid also. This is because ln(n#) < n, so the gap is less than n^2. The difference Q between the first prime after n# and n# must be less than n^2. Notice that Q does not have a divisor m<n because n# is multiple of m also, in which case n#+Q would be also multiple of m, so it would not be a prime. Thus Q must be prime. Last fiddled with by alpertron on 20050725 at 02:13 Reason: Corrected hyperlink 
20050725, 09:54  #7  
Nov 2003
2^{2}·5·373 Posts 
Quote:
There is a requirement that N+Q *and* NQ both be prime. Certainly (by Cramer's conj as you suggest), the first prime gap after N must be less than p^2. Thus Q can't be p1*p2 with p1, p2 > p if we just consider N+Q. But will NQ also be prime? 

20050725, 10:34  #8  
Jun 2005
2·7^{2} Posts 
Quote:
101#+k is prime for k=131, 139, 149, 167, +233, 239, 461, 463, 491, +523, 523 (the negative bias is just coincidental BTW 173#+191 are both Prime and no other primes between these two, whereas 293#+10987 are both prime with 93 primes between these two and all for all 293#+k the k's are prime (for k's negative or positive) Last fiddled with by AntonVrba on 20050725 at 10:38 

20050725, 12:24  #9 
Aug 2002
Buenos Aires, Argentina
13×107 Posts 
Bob, you are right. It appears that at 11 PM I only write gibberish^{TM} and illucid^{TM} statements.
If both n#Q and n#+Q are primes, Q can be possibly greater than (ln n)^2, invalidating my previous argument. So conjecture 4 is stronger than Cramer's conjecture. Notice that still nobody found a prime gap near to (ln n)^2 so it is possible that conjecture 4 is true too. 
20050725, 13:19  #10  
Jun 2005
2·7^{2} Posts 
You got me thinking
Here is conjecture 6: Quote:
Quote:
97#+107x109 101#103x109 103#+107^2 107#109^2 109#+113x191 113#+127x131 127#131x139 131#+149^2 137#139x151 139#149x151 149#157x163 151#163x167 157#163x181 163#173x181 167#179x193 173#193x199 179#181^2 Last fiddled with by AntonVrba on 20050725 at 13:31 

20050725, 13:55  #11  
Nov 2003
2^{2}·5·373 Posts 
Quote:
missed a condition. Here's what makes me think the conjecture is probably false. Put N = p#. The probability that N+Q is prime is about 1/log(N+Q) ~ 1/p The probability that NQ is also prime is ~1/p. We want to search over values of Q, so that both are prime. If we let Q go from 1 to K, then the probability of finding both prime for some Q is sum from Q = 1 to k of 1/p^2 and this is just k/p^2 which is small. for k ~ log^2 N. I expect that for some p's we will have to take k to be bigger than log^2 N, i.e. Q will be p1*p2 for p1,p2 > p. The problem is that 1/p^2 is quite small for large p. 

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