Problems & Puzzles: Conjectures

Conjecture 89.  Limit of R = AVSQr / SQRT[log(AV)] equal 1

Alain Rochelli wrote on April 15, 2020:

Denoting the set of N+1 consecutive primes beginning P(0) and finishing P(N), with N >= 100 000, we consider the average difference between these N+1 consecutive primes.

The mathematic notation is :

AV = (1/N) x SIGMA[P(n)-P(n-1)] = [P(N)-P(0)] / N with SIGMA[...] as the sum for n=1 to n=N

We also consider the average of the square reduced differences AVSQr with the mathematic notation :

AVSQr = (1/N) x SIGMA[(Dn/logPn)^2] with Dn=P(n+1)-P(n)

For N to infinite, we conjecture that the limit of R = AVSQr / SQRT[log(AV)] equal 1 with SQRT() as the square-root of ().

In practice, we verify by computer computation that there are increasingly values P(0) and N such as the value of R converge to 1 :

For example, I computed the first fifty millions primes segmented by one million primes sets. Results are in a jointed EXCEL file. 

Q1. The Puzzlers are invited to verify this conjecture is true for large values of P(0) and N : P(0)>10^9 ; P(0)>10^10 ; P(0)>10^11 and so on ...


During the week ending on June 5, 2021, contributions came from Adam Stinchcombe, Alain Rochelli


Adam wrote:

Using the estimate pn = n*log(n),  AV = [P(N)-P(0)] / N   should be about log(N)

Using the gap size estimate p(n+1)-p(n) = log(n),   AVSQr = (1/N) x SIGMA[(Dn/logPn)^2] with Dn=P(n+1)-P(n) should be "about" 1/N*sigma(1^2)=1  so   R = AVSQr / SQRT[log(AV)] estimates to 1/(root(log(N))

Starting at P0 =next prime after 10^k and adding up 200,000 terms I get the following list of k,AV,AVSQr,R,1/sqrt(log(N)):

15, 34.5523, 1.84024, .3131,.1702
20, 46.1041, 1.87225, .2757,.1474
25, 57.6314, 1.89915, .2502,.1318
30, 69.1783, 1.91524, .2303,.1203
35, 80.6026, 1.91979, .2138,.1114

Perhaps it appears that R is decreasing towards 0 but it does not look like it is heading towards 1.  The estimate 1/(sqrt(logN)) seems to be off by (about) a factor of 2.  Linear regression between my estimate and the actual predicts R will be around 0.1 when P(N) is abour 10^234.  Testing 10^240 with about 27,000 terms I get an R value around .08.


Alain wrote:

Six months ago, based on Cramer 's model and heuristic considerations, I was able to verify using the PARI application that :

for N to infinite, the limit of AVSQr equal 2 (easy to say but very difficult to predict by computer computation !).

Also, with this new conjecture, we can deduce a new formulation :

If we consider Dp as the random variable representing the difference between consecutive primes, Dp(n) = p(n+1) - p(n) , then we can say that the standard deviation of Dp is equal to the logarythm of p.

The mathematic notation is : SQRT[ AV((Dp - logp)^2) ] = AV(Dp) = logp with AV() as the average value of ().

The demonstration of this would be wonderful for mathematicians, isn't it ?

PS : For the same reasons, Conjecture 82 also needs to be corrected : I will send you the correct wording in a few days.



Records   |  Conjectures  |  Problems  |  Puzzles