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Old 2020-11-25, 14:07   #1
sweety439
 
Nov 2016

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Default Excepted number of primes of the form (k*b^n+c)/gcd(k+c,b-1) for n <= given number

According to the page https://oeis.org/wiki/User:Charles_R...special_primes, the excepted number of primes of the polynomial function form a0+a1n+a2n2+a3n3+a4n4+...+arnr (with r>=1 is integer, all ai (0<=i<=r) integers, ar>=1), for 1<=n<=N is (N^(1/r))/(ln(N)), if this is an irreducible polynomial over the integers, and the values of this polynomial at n = 1, 2, 3, ... are relatively prime (Bunyakovsky conjectures is that there are infinitely many n such that this polynomial produces prime values, if this is an irreducible polynomial over the integers, and the values of this polynomial at n = 1, 2, 3, ... are relatively prime)

However, for the exponential function form (k*b^n+c)/gcd(k+c,b-1) (with k>=1 is integer, b>=2 is integer, c is (positive or negative) integer, |c|>=1, gcd(k,c) = 1, gcd(b,c) = 1), what is the excepted number of primes of this form for 1<=n<=N, if this form does not have algebraic covering set (e.g. 1*8^n+1, 8*27^n+1, (1*8^n-1)/7, (1*9^n-1)/8, 4*9^n-1, 2500*16^n+1, etc.) or numerical covering set (e.g. 78557*2^n+1, (11047*3^n+1)/2, (419*4^n+1)/3, 509203*2^n-1, (334*10^n-1)/9, 14*8^n-1, etc.) or partial algebraic/partial numerical covering set (e.g. 4*24^n-1, 4*39^n-1, (4*19^n-1)/3, (343*10^n-1)/9, 1369*30^n-1, 2500*55^n+1, etc.)?
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Old 2020-11-25, 14:09   #2
sweety439
 
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the page only writes "trivial density O(ln(N))", however, I know that the density is different by triple (k,b,c), e.g. 29*2^n-1 has much low density as 17*2^n-1

Last fiddled with by sweety439 on 2020-11-25 at 14:10
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Old 2020-11-30, 00:44   #3
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Finding the expected number of primes of (k*b^n+c)/gcd(k+c,b-1) shouldn't be much different than finding the expected number of primes of the form k*b^n+c. If we treat numbers of the form k*b^n+c as "random" integers, then each candidate has probability of about

{1\over n\ln b + \ln k}

If we care about the expected number of primes from {a} to {n}, then we should expect

L = \sum_{x=a}^{n} {1\over x\ln b + \ln k}

or equivalently

L = {1\over \ln b} \sum_{x=a + \ln k}^{n + \ln k} {1\over x}

(note that for small k, it is practical to ignore the {\ln k} term as it will not make much of a difference in calculating the probability. Therefore, it's acceptable to rewrite as

L = {1\over \ln b} \sum_{x=a}^{n} {1\over x}

So for any sequence of randomly behaving integers, we can assume that there will be

L = C \sum_{x=a}^{n} {1\over x}

primes in the sequence up to nth term, where {C} is a constant. Without any assumptions about the sequence of integers, we can assume that {C = 1\over \ln b} since the growth rate of k*b^n+c (or (k*b^n+c)/gcd(k+c,b-1) in your case), is b.

However, as you mentioned, such sequences are not "completely" random. They are random enough that the prime number theorem can be used to predict their primality, but divisibility by small primes isn't as random and can easily be predicted:

Once one candidate is found to be divisible by a prime p, another predictable candidate will also be divisible by p. This decreases the probability of expected primes.
Sometimes though, the candidates will never be divisible by a prime p, which increases the probability of expected primes.

For example, if (k*b^n+c)/gcd(k+c,b-1) is always odd, we can double our constant: C = {2\over \ln b}, and if (k*b^n+c)/gcd(k+c,b-1) is coprime to 6, we can triple our constant: C = {3\over \ln b}, and this can go on for however many primes we know won't divide any of our candidates.

The most accurate way to predict primes occurring in particular sequences from the a-th term to n-th term, is to sieve or remove candidates that are divisible by primes up to some constant P. Suppose we have r remaining candidates. Then we should expect about 1 in every k = r/(n-1) candidates left to test for primality. The growth rate, in a sense, changes from b, to b^k. The new probability that a random integer N is prime, given that it has no prime factors less than P, is about


{e^{gamma} \ln P \over \ln N}

Putting two and two together, one can obtain a more accurate estimate for the expected number of primes of the form (k*b^n+c)/gcd(k+c,b-1), the new constant C is:

C = {e^{gamma} \ln P \over k\ln b}

Just in case you weren't aware: The Nash value tells you how many candidates remain after sieving a certain number of terms to a certain depth.

Last fiddled with by carpetpool on 2020-11-30 at 00:46
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Old 2020-12-01, 10:42   #4
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Quote:
Originally Posted by sweety439 View Post
Excepted number of primes...
they were excepted from what?
(I had a German colleague who always confused the two terms, haha, he moved to Philippines few years ago, are you that guy?)
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Old 2020-12-01, 16:05   #5
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Quote:
Originally Posted by LaurV View Post
they were excepted from what?
(I had a German colleague who always confused the two terms, haha, he moved to Philippines few years ago, are you that guy?)
No, it is just a typo, should be "expected" ....

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Old 2020-12-02, 15:04   #6
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Previous is a double post from your own thread!
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Old 2020-12-02, 16:26   #7
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Originally Posted by kar_bon View Post
Previous is a double post from your own thread!
Yes, I want to let more people know this conjecture, as Riemann hypothesis and abc conjecture and Schinzel's hypothesis H

This conjecture would imply all Sierpinski conjectures and Riesel conjectures in CRUS to every base b>=2, also the dual Sierpinski conjecture (whether 78557 is the smallest odd number k such that 2^n+k is composite for all n>=1) and the dual Riesel conjecture (whether 509203 is the smallest odd number k such that |2^n-k| is composite for all n>=1), and also imply there are infinitely many such primes:

* Mersenne primes
* Fermat primes
* Generalized repunit primes (b^n-1)/(b-1) to every base b>=2 not of the form m^r with r>1
* Generalized nega-repunit primes (b^n+1)/(b+1) to every base b>=2 not of the form m^r with odd r>1 and not of the form 4*m^4 with integer m
* Generalized Fermat primes b^(2^n)+1 to every even base b>=2 not of the form m^r with odd r>1
* Generalized half Fermat primes (b^(2^n)+1)/2 to every odd base b>=2 not of the form m^r with odd r>1
* Williams primes of the 1st kind (b-1)*b^n-1 to every base b>=2
* Williams primes of the 2nd kind (b-1)*b^n+1 to every base b>=2 (not always true if b-1 is of the form m^r with odd r>1 or of the form 4*m^4 with integer m)
* Williams primes of the 3rd kind (b+1)*b^n-1 to every base b>=2

Last fiddled with by sweety439 on 2020-12-02 at 16:37
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Old 2020-12-02, 21:19   #8
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Quote:
Originally Posted by sweety439 View Post
Yes, I want to let more people know this conjecture, as Riemann hypothesis and abc conjecture and Schinzel's hypothesis H
Then write up a paper explaining your conjectures, and provide more evidence for your claims, like I have given or some of your own research.
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Old 2020-12-03, 06:40   #9
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Quote:
Originally Posted by sweety439 View Post
No, it is just a typo ....
Three or four times in a row?
(also always in the text, not only in the title)
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