A342877 - OEIS

%I #38 May 07 2021 16:00:17

%S 1,1,1,0,1,0,1,1,0,1,1,0,1,1,1,0,1,1,0,1,1,1,1,1,0,1,0,1,1,0,1,0,1,0,

%T 1,1,0,1,1,0,1,0,0,0,1,1,0,0,0,1,0,1,1,1,1,0,1,0,0,1,1,0,0,0,1,1,1,0,

%U 0,1,1,1,1,0,1,1,0,1,1,0,1,0,1,0,1,1,0,0,0,1,1,0,1,0,1,1,0,1,1,1,1,1,0,1,1,1

%N 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.

%C 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.

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

%C 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.

%H Andres Cicuttin, <a href="/A342877/a342877_3.png">Log-linear scatter plot of (1/n) Sum_{i=1..n} a(i) for first 2^18 primes</a>.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Poisson_point_process">Poisson point process</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Exponential_distribution">Exponential distribution</a>

%H Yamasaki, Yasuo and Yamasaki, Aiichi, <a href="https://repository.kulib.kyoto-u.ac.jp/dspace/bitstream/2433/84326/1/0887-10.pdf">On the Gap Distribution of Prime Numbers</a>, 数理解析研究所講究録 (1994), 887: 151-168.

%H <a href="/index/Ch#char_fns">Index entries for characteristic functions</a>

%e 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.

%e 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.

%t a={}; nmax=128;

%t Do[If[(Prime[n]-2)/(n-1)>(Prime[n-1]-2)/(n-2),AppendTo[a,1],AppendTo[a,0]],{n,3,nmax}];

%t a

%t (* Uncomment and run next lines to produce the log-linear plot available in Links *)

%t (* a={};

%t nmax=2^18;

%t Do[If[(Prime[n]-2)/(n-1)>(Prime[n-1]-2)/(n-2),AppendTo[a,{n,1}],AppendTo[a,{n,0}]],{n,3,nmax}];

%t 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]],

%t FrameLabel->{Text[Style["n",FontSize->16]],},PlotStyle->{PointSize->Small,Red},GridLines->Automatic] *)

%o (PARI) A342877(n) = (((prime(n)-2)/(n-1)) > ((prime(n-1)-2)/(n-2))); \\ _Antti Karttunen_, Mar 28 2021

%Y Cf. A001223, A079418, A286888.

%K nonn

%O 3,1

%A _Andres Cicuttin_, Mar 28 2021