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.
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.
