I notice that your long chain starts at {c^2} though, not {c}. Is that important? Seems like there are three possible questions, of increasing generality: 1. {c^k} all prime for k=1..n 2. {c^k} all prime for k=a..b 3. {m c^k} all prime for k=1..n --Michael On Mon, Jun 1, 2020 at 2:56 PM Simon Plouffe <simon.plouffe@gmail.com> wrote:
Hello,
I am begining to compute an empirical law that would give the number of primes for any given c.
if c = 2.5 approx : 7 primes. if c = 180 about 25 primes. here is one output of a program I am running that finds good candidates for c and the length of the chain :
The big one when c = 55237.07... is the biggest that I could compute so far. It has a length of 632 primes. ; When the prime is 3000 digits long , it takes some time to validate that the prime is a probable prime. All of this of course is empirical.
list : value of c : length of the prime chain. 306.68699704 27, 311.02942978 27, 330.70121184 27, 281.31688939 28, 306.84441080 28, 307.21127405 29, 331.16927945 29, 280.67638838 30, 310.63328478 30, 313.45007923 30, 330.83340890 30, 316.79874303 31, 336.70642355 33, 331.47084887 34,
For technical reasons here the length is length -1 that is for 331.4708488... the longest valid chain of primes is of length 33.
I don't know what could be the empirical relation between : size of c : length of prime chain. I was thinking maybe , log(n) with a square root term, nothing substantial so far.
best regards, Simon Plouffe
Le lun. 1 juin 2020 à 19:24, Michael Kleber <michael.kleber@gmail.com> a écrit :
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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