A342877 a(n) = 1 if the average distance between consecutive first n primes is greater than that of the first n-1 primes, otherwise a(n) = 0, for n > 2.

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Offset: 3

Author

Andres Cicuttin, Mar 28 2021

Keywords

• nonn

Comments

The average distance between consecutive primes among the first n primes tends to increase with n. This average distance always changes when n is increased to n + 1, but it seems that most of the times this distance decreases. See a log-linear scatter plot of (1/n) Sum_{i=1..n} a(i) in Links.

Conjecture: lim_{n->infinity} (1/n) Sum_{i=1..n} a(i) < 1/2.

In support of the conjecture: If it is assumed, as an approximation, that position of primes follows a Poisson point process then the distance between consecutive primes is a stochastic variable with exponential probability distribution function. The probability that an exponentially distributed stochastic variable takes a value larger than the mean value is about 0.367879.

Examples

			a(3) = 1 because the average distance between consecutive first three primes {2,3,5} is (5 - 2)/2 = 3/2 which is greater than the average distance between consecutive first two primes {2,3} which is (3-2)/1 = 1.
a(6)=0 because the average distance between consecutive first six primes {2,3,5,7,11,13} is (13 - 2)/5 = 11/5 which is smaller than the average distance between consecutive first five primes {2,3,5,7,11} which is (11 - 2)/4 = 9/4.
		

Links

• Andres Cicuttin, Log-linear scatter plot of (1/n) Sum_{i=1..n} a(i) for first 2^18 primes.
• Wikipedia, Poisson point process
• Wikipedia, Exponential distribution
• Yamasaki, Yasuo and Yamasaki, Aiichi, On the Gap Distribution of Prime Numbers, 数理解析研究所講究録 (1994), 887: 151-168.
• Index entries for characteristic functions

Crossrefs

Cf. A001223, A079418, A286888.

Programs

Mathematica

a={}; nmax=128;
Do[If[(Prime[n]-2)/(n-1)>(Prime[n-1]-2)/(n-2),AppendTo[a,1],AppendTo[a,0]],{n,3,nmax}];
a
(* Uncomment and run next lines to produce the log-linear plot available in Links *)
(* a={};
nmax=2^18;
Do[If[(Prime[n]-2)/(n-1)>(Prime[n-1]-2)/(n-2),AppendTo[a,{n,1}],AppendTo[a,{n,0}]],{n,3,nmax}];
ListLogLinearPlot[Transpose[{Range[3,nmax],Accumulate[Transpose[a][[2]]]/Range[3,nmax]}],Frame->True,PlotRange->{All,{0.25,0.75}},PlotLabel->Text[Style["Sum_{i=1..n} a(i)/n",FontSize->16]],
FrameLabel->{Text[Style["n",FontSize->16]],},PlotStyle->{PointSize->Small,Red},GridLines->Automatic] *)

PARI

A342877(n) = (((prime(n)-2)/(n-1)) > ((prime(n-1)-2)/(n-2))); \\ Antti Karttunen, Mar 28 2021