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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.
**
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