Neat! How large is the interval that c must be in here? That is, what is the intersection of all the intervals [ ({c^n}-1/2)^1/n, ({c^n}+1/2)^1/n ] ? For your length-7 example, looks like it's [ (555/2)^1/6, (1417/2)^1/7 ], i.e. [2.55383..., 2.55394...]. So saying c=2.5539 is sufficiently precise. Hmm, I guess if I were a good person, I'd've gone with 166/65 = 2.5(538461)*, the simplest fraction in that interval, rather than the shortest decimal. --Michael On Mon, Jun 1, 2020 at 10:56 AM Simon Plouffe <simon.plouffe@gmail.com> wrote:
Hello,
sometime ago, someone came up with 26 primes in arith. progression. The calculation was hard and took 6 months of intense CPU time.
https://en.wikipedia.org/wiki/Primes_in_arithmetic_progression
But what if the primes are in geometric progression instead like a(n) = { c^n } , where { } is the nearest integer. For example if c = 2.553854696... then the primes are : 3, 7, 17, 43, 109, 277, 709. Only 7 of them. Can we find better ?
By using a Monte-Carlo + Simulated Annealing method, one can find much more. Here is a small table :
Value of c : range : number of primes generated 2027.167168476491219434395... n = 1..97 : 97 primes 31622.7767185595693… n = 2..388 : 387 primes (see A332308) 55237.07504296764715433124 n = 2 ..633 : 632 primes (see A333127)
A conjecture is then : if c is large enough, arbitrary long chain of primes can be generated. The data about this is here : http://plouffe.fr/NEW/
I could not find any mention of <primes in geometric progression> in literrature. Best regards, Simon Plouffe _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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