A144104 -id:A144104 - cp's OEIS frontend

A144105 Primes at the upper end of the gaps mentioned in A144104.

3, 5, 11, 17, 29, 37, 53, 59, 127, 149, 211, 223, 307, 331, 541, 1361, 1693, 1973, 2203, 2503, 2999, 3299, 4327, 4861, 5623, 5779, 5981, 6521, 6947, 7283, 8501, 9587, 10007, 10831, 11777, 12197, 12889, 15727, 16183, 19661, 31469, 34123, 35671, 35729

Offset: 1

T. D. Noe, Sep 11 2008

nonn

Firoozbakht conjecture: (prime(n+1))^(1/(n+1)) < prime(n)^(1/n), or

prime(n+1) < prime(n)^(1+1/n), which can be rewritten as: (log(prime(n+1))/log(prime(n)))^n < (1+1/n)^n. This suggests a weaker conjecture: (log(prime(n+1))/log(prime(n)))^n < e. - Daniel Forgues, Apr 28 2014

			Examples for (log(prime(n+1))/log(prime(n)))^n < (1+1/n)^n < e:
(log(3)/log(2))^1 = 1.58... < (1+1/1)^1 = 2 < e;
(log(1361)/log(1327))^217 = 2.14... < (1+1/217)^217 = 2.712... < e;
(log(8501)/log(8467))^1059 = 1.59... < (1+1/1059)^1059 = 2.716... < e;
(log(35729)/log(35677))^3795 = 1.69... < (1+1/3795)^3795 = 2.717... < e.
- _Daniel Forgues_, Apr 28 2014

T. D. Noe, Table of n, a(n) for n = 1..176
A. Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015.
Nilotpal Kanti Sinha, On a new property of primes that leads to a generalization of Cramer's conjecture, arXiv:1010.1399 [math.NT], 2010.
Wikipedia, Firoozbakht's conjecture

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.

Original entry on oeis.org

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

Thomas Nordhaus, Mar 08 2012

nonn

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

			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).

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

Cf. A002386, A144104.

{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.

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

A262061 Least prime(i) such that prime(i)^(1+1/i) - prime(i) > n.

Original entry on oeis.org

2, 3, 5, 7, 11, 11, 17, 17, 23, 29, 29, 37, 41, 53, 59, 67, 79, 89, 97, 127, 127, 137, 163, 179, 211, 223, 251, 293, 307, 337, 373, 419, 479, 521, 541, 587, 691, 727, 797, 853, 929, 1009, 1151, 1201, 1277, 1399, 1523, 1693, 1777, 1931, 2053, 2203, 2333, 2521, 2647, 2953, 3119, 3299, 3527, 3847, 4127

Offset: 1

Farideh Firoozbakht and Robert G. Wilson v, Sep 11 2015

nonn

Where A246778(i) first exceeds n, stated by p_i.
Similar to A245396.
Number of terms < 10^n: 4, 19, 41, 75, 120, 176, 242, 319, 407, 506, ..., .
Concerning Firoozbakht's Conjecture (1982): (prime(n+1))^(1/(n+1)) < prime(n)^(1/n), for all n = 1 or prime(n+1) < prime(n)^(1+1/n), which can be rewritten as: (log(prime(n+1))/log(prime(n)))^n < (1+1/n)^n. This suggests a weaker conjecture: (log(prime(n+1))/log(prime(n)))^n < e.
Prime index of a(n): 1, 1, 3, 4, 5, 5, 7, 7, 9, 10, 10, 12, 13, 16, 17, 19, 22, 24, 25, 31, 31, ..., .
All terms are unique for n > 21. Indices not unique: 1 & 2, 5 & 6, 7 & 8, 10 & 11 and 20 & 21.
The distribution of initial digits, 1...9, for a(n), n<508: 140, 91, 60, 50, 44, 36, 32, 27 and 26.

			a(20) = 127 since for all primes less than the 31st prime, 127, p_k^(32/31) - p_k are less than 20.
a(100) = 38113,
a(200) = 2400407,
a(300) = 57189007,
a(400) = 828882731,
a(500) = 8748565643,
a(1000) = 91215796479037,
a(1064) = 246842748060263, limit of Mathematica by direct computation, i.e., the first Mathematica line.

Paulo Ribenboim, The little book Of bigger primes, second edition, Springer, 2004, p. 185.

Farideh Firoozbakht and Robert G. Wilson v, Table of n, a(n) for n = 1..507
Alexei Kourbatov, Upper Bounds for Prime Gaps Related to Firoozbakht's Conjecture, arXiv:1506.03042 [math.NT], 2015.
Alexei Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015

Cf. A144104, A182134, A182514, A245396, A246777, A246778, A246781, A246810, A248855, A249566.

f[n_] := Block[{p = 2, k = 1}, While[n > p^(1 + 1/k) - p, p = NextPrime@ p; k++]; p]; Array[f, 60] (* or  quicker *)
(* or quicker *) p = 2; i = 1; lst = {}; Do[ While[ p^(1 + 1/i) < n + p, p = NextPrime@ p; i++]; AppendTo[lst, p]; Print[{n, p}], {n, 100}]; lst

a(n) = {i = 0; forprime(p=2,, i++; if (p^(1+1/i) - p > n, return (p)););} \\ Michel Marcus, Oct 04 2015

Log(y) ~= g + x^(1/2) where g = Euler's Gamma.

a(2) corrected in b-file by Andrew Howroyd, Feb 22 2018

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A144105 Primes at the upper end of the gaps mentioned in A144104.

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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.

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A262061 Least prime(i) such that prime(i)^(1+1/i) - prime(i) > n.

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