Re: [math-fun] The PrimeLatz conjecture
From: Bill Thurston
This is true for all **positive** integers N, and false for all integers less than some negative number -C. For large enough n, there are at least 3 primes between n and 2n (from this wikipedia article, Erd?s proved that for any positive integer k, there is a natural number N such that for all n > N, there are at least k primes between n and 2n. An equivalent statement had been proved earlier by Ramanujan (see Ramanujan prime).) Therefore, if the kth member of a PrimeLatz sequence is sufficiently large, either the k+1st element or the k+2nd element is smaller than n.
If the kth member of a PrimeLatz sequence, a_k, is sufficiently large, and prime, then the k+1st element will be less than 7*a_k. The k+2nd element will be less than 7/2*a_k, and may itself be prime, causing subsequent terms to be even bigger. Just from the premises you've quoted, you've not proved the conclusion you claim. Phil
Yes, I retract my assertion, I don't know what I was thinking. Bill On Nov 29, 2010, at 6:50 AM, Phil Carmody wrote:
From: Bill Thurston
This is true for all **positive** integers N, and false for all integers less than some negative number -C. For large enough n, there are at least 3 primes between n and 2n (from this wikipedia article, Erd?s proved that for any positive integer k, there is a natural number N such that for all n > N, there are at least k primes between n and 2n. An equivalent statement had been proved earlier by Ramanujan (see Ramanujan prime).) Therefore, if the kth member of a PrimeLatz sequence is sufficiently large, either the k+1st element or the k+2nd element is smaller than n.
If the kth member of a PrimeLatz sequence, a_k, is sufficiently large, and prime, then the k+1st element will be less than 7*a_k. The k+2nd element will be less than 7/2*a_k, and may itself be prime, causing subsequent terms to be even bigger. Just from the premises you've quoted, you've not proved the conclusion you claim.
Phil
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Phil Carmody