A209407 Primes p(i) such that p(i+1)/p(i) > p(k+1)/p(k) for all k>i, where p(i) is the i-th prime.

3, 7, 13, 23, 31, 47, 113, 139, 199, 211, 293, 317, 523, 1327, 1669, 1951, 2179, 2477, 2971, 3271, 4297, 4831, 5591, 5749, 5953, 6491, 6917, 7253, 8467, 9551, 9973, 10799, 11743, 15683, 19609, 31397, 34061, 35617, 35677, 43331, 44293, 45893, 48679, 58831

Offset: 1

Author

Thomas Nordhaus, Mar 08 2012

Keywords

• nonn

Comments

p(i) belongs to the sequence if p(i+1)/p(i) > p(k+1)/p(k) for all k>i.

It follows from the prime number theorem that p(i+1)/p(i) converges to 1 as i tends to infinity. a(n) is an infinite sequence therefore. The a(n) constitute "record holders" for the relative size of the prime number gaps.

The values a(n) given above were obtained by comparing p(i+1)/p(i) with p(k+1)/p(k) for 1<=i<=5949 and k ranging from i+1 to 200000 for given i.

In order to show that these values are correct one has to analyze the error terms in the formula p(k) ~ k*log(k) and extend the "test range" if needed. Using Dusart's bound: n*(log(n)+loglog(n)-1) < p(n) < n*(log(n)+loglog(n)) for n>=6 one gets

p(k+1)/p(k) < f(k):=(1+1/k)*(log(k+1)+loglog(k+1))/(log(k)+loglog(k)-1) for all k>=6. However this bound tends to 1 like 1+1/log(k) as k->oo. In order to verify, for example, that the term a(9)=199=p(46) is correct one must make sure that p(k+1)/p(k) < p(47)/p(46) = 211/199 =~ 1.0603 for all k>47. However f(10^7)~=1.06666 still, so k <= 10^7 is not sufficient to validate a(9). a(8) however is validated by checking the range k<=10^7.

In order to validate terms up to a(n)=31397 for example one even needs k<=10^20 roughly which needs considerable computational power.

This can be improved with another of Dusart's bounds: there is always a prime in (x, x + x/(25log^2 x)) for x > 396738. Hence it suffices to check up to the higher of exp(1/(25 (prime(i+1)/prime(i)-1))) and 396738. - Charles R Greathouse IV, Mar 06 2013

Examples

			The smallest prime belonging to the sequence is p(2)=3 because p(3)/p(2) = 5/3 > 7/5, 11/7, 13/11, 17/13,... p(1)=2 does not belong to the sequence since p(2)/p(1) = 3/2 <5/3 = p(3)/p(2).
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..67

Crossrefs

Cf. A002386, A144104.

Programs

PARI

{np=200000;a=vector(44);q=vector(np,k,prime(k+1)/prime(k));m=n=0;
while(n<=44,if(q[m++]>vecmax(vector(np-m,j,q[m+j])),a[n++]=prime(m)))} \\ computes the first 44 terms of sequence.

PARI

list(lim)=my(v=List([3]),u=List([2/3]),mn=.04/log(lim)^2,p=7,t);forprime(q=11,nextprime(lim+1),t=(q-p)/p;if(t>mn,if(t>u[#v],v[#v]=p;u[#u]=t,listput(v,p);listput(u,t)));p=q);t=u[#u];forstep(i=#u-1,6,-1,if(u[i]>t,t=u[i],v[i]=3));Set(v) \\ valid for lim > 396738; Charles R Greathouse IV, Jun 25 2014