Problems & Puzzles: Conjectures

Conjecture 82.  Average of log Dn / log(logPn) equal R = 0,877 08...

Alain Rochelle sent the following conjecture, on March 17, 2019:

Denoting D(Pm) = P(n) - P(m) the difference between two consecutive primes Pm and Pn (with n=m+1) and denoting logX the natural logarithm of the number X, we consider the set of N consecutive primes included between X and X+X/(logX)^2.

The mathematic notation is :

X < P1 < P2 < ...... < PN-1 < PN < X+X/(logX)^2 < PN+1

For X to infinite, denoting Dn = Pn+1
- Pn, we conjecture that the average of log Dn / log(logPn) equal R = 0,877 08...

Denoting SIGMA[...] as the sum for n=1 to n=N, we write :

A(X) = (1/N) x SIGMA[ logDn / log(logPn) ] ---> R  (for X to infinite)

In practice, we verify by computer computation for increasingly values of X :MAG[A(X) - R] is of the order of log(X) / SR(X)    (with MAG[ ] the magnitude of [ ] and SR(X) the square root of X)

For example, I compute :

X = 2 x 10^8 ; N = 28 693 ; A(X) = 0,877 20

X = 4 x 10^8 ; N = 51 389 ; A(X) = 0,876 91

X = 6 x 10^8 ; N = 72 791 ; A(X) = 0,876 78

X = 8 x 10^8 ; N = 93 010 ; A(X) = 0,876 69

X = 9,8 x 10^8 ; N = 110 340 ; A(X) = 0,877 51 and logX/SR(X) = 0,000 66

The purpose consists to verify if this conjecture is true for large values of X :

X = 10^9 ; X = 2 x 10^9 ; X = 10^10 ; X = 10^11 ; X = 10^12  and so on ...

Alain Rochelli wrote on Oct 12, 2019

For Conjecture 82, I computed the first fifty million primes.

The result is :

P(1) = 2 ; P(50 000 000) = 982 451 653

Average of log Dn / log(logPn) equal 0,877 095

Who is the Puzzler to compute the next fifty million primes ?

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On June 7, 2021, Alain wrote again:

Using the same mathematical model as for Conjecture 89 (Cramer 's model with the independence of gaps between consecutive primes), we get the value of  R equal to 1 for P(n) to infinite, with P(n) or Pn the nth prime and Dn or Dp(n) = p(n+1) - p(n) the corresponding difference.

More precisely, we have (with AV the average value) the mathematical formula :

AV(Rn) = AV[ logDn / log(logPn) ] = 1 - (G / log(logPn)) where G is Euler 's constant, equal to 0,577215665...

If we consider Dp as the random variable representing the difference between consecutive primes, the mathematical wording is :

Around Pn, Rn is equal to the average exponent of logPn to get Dn ( i.e. : Dn = logPn ^ Rn )

Also AV(Rn) = AV[ log(Dn) / log(logPn) ] = AV{ 1 + [(logDn-log(logPn)) / log(logPn)] } = 1 + {AV[log(Dn/logPn)] / log(logPn)} for Pn to infinite, because log(logPn) is practically a constant.

Now using the main conjecture saying that AV[log(Dn/logPn)] is converging to - G for Pn to infinite, we then obtain AV(Rn) = 1-[G/log(logPn)] as stated previously.

This is equivalent to AV( logDn ) = log(logPn) - G because logDn = Rn * log(logPn)

It follows that the geometric mean GM of Dp is equal to logp * e^(-G) where e^(-G) = 1/e^G = 1 / 1,781072418

noted again by the wonderful formulation GM( Dp ) = logp / e^(G) = AV( Dp ) / 1,78107...

For mathematicians, it 's possible that the proof of this conjecture is of the same level of difficulty as for the Riemann 's hypothesis.

By computer calculation, we need to take into account local variations and go towards very large primes to obtain meaningful results.

For example, it 's necessary to consider primes around 10 ^ 500 to reach R equal to 0,92.

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