From: Dave Rusin Subject: Re: conjecture, T=2Q+R prime implies S=2R+Q prime Date: 19 Nov 1999 16:19:40 GMT Newsgroups: sci.math.research Keywords: Generalization of Goldbach, Twin Prime conjectures In article <3833206C.4CD4BF36@math.kth.se>, Karim Novira wrote: > >CONJECTURE 1a >All primes T>2 can be written on the form T=2*P+Q where P and Q are odd >primes or 1, such that S=2*Q+P is also a prime. It seems unlikely that we could prove such a thing when we cannot even prove that all even T can be written in the form T = P+Q with P and Q odd primes or 1 (with no additional primality requirement as is required of S). Nonetheless the heuristics are in your favor for this conjecture, as for Goldbach's. The number of solutions (P,Q) to the problem, for a particular T, is the number of integers P in the interval (0, T/2) such that P, T-2*P, and 2*T-3*P are all prime (or equal to 1). Well, roughly speaking a given value P makes all three prime with probability 1/ln(P) * 1/ln(T-2P) * 1/ln( 2T-3P ) so that we expect something like integral( dP/( ln(P) ln(T-2P) ln(2T-3P) ) , 3 < P < T/2 ) solutions to the problem; the conjecture is that the correct number of solutions is at least 1. Well, the integral behaves badly near the endpoints, but a tally of the P's near P=T/4 (say, T/6 < P < T/3 ) is likewise estimated by an integral which is easily shown to be a multiple of T/(ln T)^3 , in which positive upper and lower bounds for the multiple can be given for sufficiently large T. In particular, the number of solutions is larger than 1 for T sufficiently large. (As an indication of the weakness of this analysis, let me note that I never used the fact that T is prime, but the conjecture surely fails for some composite T, namely all multiples of 3.) Of course, this is not a proof, as the simultaneous primality of P, T-2P, and 2T-3P is not really a probability (of "independent events", no less!), but the numerical data for small T support these estimates. For the first 100 odd numbers (not multiples of 3) after 10000 we find the number of good P to average 35.18, which is about 2.75*T/log(T)^3; for the first 100 after 100000 the average is 165.03 = 2.52*T/log(T)^3. As you might imagine, people have conjectured similar statements in the past. You have presented k=3 linear expressions in P and asked that all be prime, and I have responded that the expected number of P (in an interval starting near 0) for which this is true is roughly a multiple of L/(log(L))^k (where L is the length of the interval) unless there's an obvious reason where there are no such P. This same reasoning can be used for the Goldbach conjecture (P and N-P both prime) and the twin prime conjecture (P and P+2 both prime) as well as some similar conjectures (P and 2P+1 both prime; P, P+2, P+6 all prime; etc.) Data support the heuristic arguments, so there is every expectation that the supposed number of solutions is roughly correct, but the attempts to prove these things fail because the estimates for the number of primes in an interval have an error term which swamps the number we are trying to estimate! >Other similar expressions of the form W=X*Q+R have been examined where X >have been set 2*2, 2*2*3, 2*5, 2*3*5 but the conjecture does not hold for >these factors. Well, when X=4, you need to exclude values of W which are multiple of 2, 3, and 5 for rather trivial reasons. There are also a few small exceptions (W=1, 107, 557), but beyond a certain point all values of W coprime to 30, prime or otherwise, appear to give not just one but several valid combinations for Q and R. This, too, is consistent with the discussion above -- the number of valid Q should be at least a multiple of W/(log W)^3, and in particular should be greater than 1 for large W. Your other examples should be similar: plenty of pairs (Q,R) should exist for all sufficiently large W not in certain small congruence classes. As I say, this is all heuristic, not a proof. We lack the tools to give proofs of such results at this time. dave