I'm afraid I'm missing some step of the reasoning here. Even assuming that the next three primes were only slightly larger than n, a large odd number is expected to at least quadruple with the next step. The following number must be even, and will be halved, yielding a number greater than 2n. Three consecutive halvings will be required to get below n. So it would be required for n plus the next three primes to be divisible by 8 in order to guarantee a decrease. What am I missing? On Fri, Nov 26, 2010 at 9:23 PM, Bill Thurston <wpt4@cornell.edu> wrote:
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, Erdős 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. This implies there is a finite set of loops such that any beginning element ends up in one of them. Bill Thurston
On Nov 26, 2010, at 9:23 AM, Eric Angelini wrote:
Hello math-Fun,
... any taker?
Start from any integer N. If N is even, divide N by 2. If N is odd, add to N the next 3 prime numbers > N. Iterate. Conjecture: all N's will enter in a loop, sooner or later.
http://www.cetteadressecomportecinquantesignes.com/PrimeLatz.htm
Best, É.
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