A209407 - OEIS

%I #29 Jun 26 2014 07:44:44

%S 3,7,13,23,31,47,113,139,199,211,293,317,523,1327,1669,1951,2179,2477,

%T 2971,3271,4297,4831,5591,5749,5953,6491,6917,7253,8467,9551,9973,

%U 10799,11743,15683,19609,31397,34061,35617,35677,43331,44293,45893,48679,58831

%N 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.

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

%C 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.

%C 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.

%C 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

%C 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.

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

%C 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

%H Charles R Greathouse IV, <a href="/A209407/b209407.txt">Table of n, a(n) for n = 1..67</a>

%e 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).

%o (PARI) {np=200000;a=vector(44);q=vector(np,k,prime(k+1)/prime(k));m=n=0;

%o 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.

%o (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

%Y Cf. A002386, A144104.

%K nonn

%O 1,1

%A _Thomas Nordhaus_, Mar 08 2012