Problems & Puzzles: Conjectures

Conjecture 91. A conjecture about strong primes

Alain Rochelli sent on October 18, 2021, the following conjecture:

Today, I send you the following conjecture which I hope has not already been published elsewhere.

Strong Primes, p(n) > [ p(n-1) + p(n+1) ] / 2, are explained in sequence OEIS A051634.

Let DS(n) = nextprime[Sp(n)] - Sp(n) as the difference between consecutive primes Sp(n) and the next prime of Sp(n), with Sp(n) the nth Strong prime.

For N to infinite, we conjecture that : SUM_{i=1..N} DS(i) ---> Sp(N) / 4

In practice, we consider the set of N Strong primes between A and B with a sufficiently representative sample [A,B].

For example, with A = 5 000 and B = 15 000 (B-A = 10 000) :

Sp(304) = 5 009  ;  DS(304) = 5 011 - 5 009 = 2

Sp(305) = 5 021 ;  DS(305) = 5 023 – 5 021 = 2

Sp(306) = 5 039 ;  DS(306) = 5 051 – 5 039 = 12

------

Sp(813) = 14 923 ; DS(813) = 14 929 – 14 923 = 6

Sp(814) = 14 939 ; DS(814) = 14 947 – 14 939 = 8

Sp(815) = 14 947 ; DS(815) = 14 951 – 14 947 = 4

Sp(816) = 15 013

and  SUM_{i=304..815} DS(i) = 2 682 ---> 10 000 / 4 = 2 500

For increasingly values of A with B-A = 10^8, using PARI application I computed :

A = 10^10  ;  SUM[DS] = 26 485 910

A = 10^20  ; SUM[DS] =  25 890 234

A = 10^40 ;  SUM[DS] =  25 552 656

A = 10^100 ; SUM[DS] = 25 397 510

A = 10^140 ; SUM[DS] = 25 145 346

A = 10^180 ; SUM[DS] = 24 994 648

A 10^200  ;  SUM[DS] =  25 225 928

Q1. Can you prove this conjecture ?

Q2. Send your own verifications.

During the week that ended the Feb 4, 2022, contributions came from Alain Rochelli,

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Alain wrote:

I am sending you a heuristic proof based on Cramer 's model and the independence of gaps between consecutive primes.

In mathematical literature, it is mentioned that :

The distribution of gaps between prime around their average spacing is expected to be distributed in a Poisson distribution.

In practice, at the large scale, we consider the random variable X defined by X(n) = Dp(n) / log p(n) = [p(n+1) - p(n)] / log p(n).

According to the previous considerations, the mathematical notation is that the probability density for X is f(x) = e^(-x).

Let Int{0..oo} f(x) be the integral from 0 to infinity of the function f(x), we get :

Int{0..oo} f(x) = Int{0..oo} e^(-x) = 1  [ because f(x) : probability density of X ]

Int{0..oo} x*f(x) = Int{0..oo} x*e^(-x) = 1  [ because Prime Number Theorem : average of Dp(n) ~ log p(n) ]

Int{0..oo} x^2*f(x) = Int{0..oo} x^2*e^(-x) = 2  [ cf. Conjecture 89 corrected ]

Int{0..oo} log(x)*f(x) = Int{0..oo} log(x)*e^(-x) = - G  [ cf. Conjecture 82 corrected with G as Euler 's constant ]

Using this mathematical model for Conjecture 91 about Strong primes [with Dp(n) < Dp(n-1)], we get :

The probability that x=Dp(n)/logp < y=Dp(n-1)/logp is equal to Int{x..oo} e^(-x) = e^(-x)  [because logp(n)~logp(n-1)~logp and the independence of x with y]

Then, as e^(-x)*e^(-x) = e^(-2*x), Int{0..oo} x*e^(-2*x) = 1/4  (it 's OK)

Now, if we consider the Weak primes [OEIS A051635 with Dp(n) > Dp(n-1)], we get :

The probability that x=Dp(n)/logp) > y=Dp(n-1)/logp is equal to Int{0..x} e^(-x) = 1 - e^(-x)

Then, Int{0..oo} x*e^(-x)*(1-e^(-x)) = 3/4 (it 's also OK)

In summary, considering OEIS A001223 (gaps between consecutive primes) and denoting :

b(n) = - a(n) = - Dp(n) if Dp(n) < Dp(n-1)

b(n) = 0 if Dp(n) = Dp(n-1)

b(n) = a(n) = Dp(n) if Dp(n) > Dp(n-1)

we get the nice conjecture : Sum{n=2 .. N} b(n) ~ p(N) / 2 for N to infinite

...

At the beginning of the previous article “Distribution of Large Gaps Between Primes”, it is defined :
N(x,H) = Sum{p(n+1)<=x and Dp(n)>=H} 1 as the number of gaps Dp(n) >= H with p(n+1) <= x
and it is mentioned :
Gallagher showed that a uniform version of the prime k-tuples conjecture of Hardy and Littlewood implies that the primes are distributed in a Poisson distribution around their average with :
For fixed h > 0, N(x,h*logx) ~ e^(-h)*(x/logx)
Let PI(x) as the number of primes p such that p <= x, it follows that the ratio R(h) = N(x,h*logx) / PI(x) is equal to e^(-h) because PI(x) ~ x/logx (Prime Number Theorem).

This ratio is valid both on a large scale (when x tends to infinity) and locally (for a sufficiently representative sample of consecutive p(n) around x). This results from the logarithm function having the same behavior as a constant function for large values of x.

Already noted f(x) for the probability density of the continuous random variable X defined by Dp(n) / log p(n) , we get :

R(h) = 1 - Int{0..h} f(x) or else : Int{0..h} f(x) = 1 - e^(-h)

It is easy to show that the function f(x) = e^(-x) is suitable for verifying the previous equality when h > o varies.

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