A249566 -id:A249566 - cp's OEIS frontend

A246782 Numbers k such that A182134(k)=2, i.e., there exist only two primes p with prime(k) < p < prime(k)^(1+1/k).

5, 6, 7, 9, 10, 11, 14, 15, 22, 23, 28, 29, 30, 45, 46, 61, 66, 216, 217, 367, 3793, 1319945, 1576499, 8040877, 17567976, 44405858, 445538764, 1478061204, 3643075047, 17440041685, 190836014732, 714573709895, 714573709896

Offset: 1

Farideh Firoozbakht, Oct 12 2014

Firoozbakht's conjecture says that for every n, there exists at least one prime p such that prime(n) < p < prime(n)^(1+1/n).
Let A(m) = {n | A182134(n) = m} where A182134(n) = #{p | p is prime and prime(n) < p < prime(n)^(1+1/n)}. This sequence gives the terms of A(2) and the sequence A246781 gives the terms of A(3).
The only known indices n for which A182134(n) = 1 are {1, 2, 3, 4, 8}.  It is conjectured that this is the complete set A(1).
Conjecture: For all m, where m is greater than one, A(m) is an infinite set.
a1 = 49749629143524, a2 = 1475067052906944 and a3 = 1475067052906945 are three large terms of the sequence. It is interesting that a3 - a2 = 1.
Conjecture: The sequence is infinite.
Next term is greater than 25000000.
a(34) > 10^12. - Robert Price, Nov 01 2014
The conjecture that A(1)={1, 2, 3, 4, 8} holds through 10^12. - Robert Price, Nov 01 2014

			5 is in the sequence since there exists only two primes p, prime(5) < p < prime(5)^(1+1/5). Note that prime(5) = 11, 11^(1+1/5) ~ 17.77 and 11 < 13 < 17 < 17.77.

A. Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015
Wikipedia, Firoozbakht's conjecture

Cf. A000040, A002386, A005669, A182134, A246781, A249566.

a246782 n = a246782_list !! (n-1)
a246782_list = filter ((== 2) . a182134) [1..]
-- Reinhard Zumkeller, Nov 17 2014

np[n_]:=(a = Prime[n]; b = a^(1 + 1/n); Length[Select[Range[a+1,b], PrimeQ]]); Do[If[np[n] == 2,Print[n]], {n, 25000000}]

for(n=1,oo,2==primepi(prime(n)^(1+1/n))-n&&print1(n", ")) \\ M. F. Hasler, Nov 03 2014

a(26)-a(27) from Robert Price, Oct 24 2014

a(28)-a(33) from Robert Price, Nov 01 2014

A246781 Numbers n such that A182134(n) = 3, i.e., there exist only three primes p with prime(n) < p < prime(n)^(1 + 1/n).

Original entry on oeis.org

12, 13, 16, 18, 20, 21, 27, 31, 34, 39, 44, 53, 59, 60, 65, 96, 97, 98, 99, 136, 154, 202, 214, 215, 220, 221, 280, 324, 325, 326, 365, 366, 736, 780, 2146, 2225, 3792, 5946, 5947, 5948, 6902, 6903, 18524, 22078, 23510, 23511, 23512, 31542, 31544, 33606

Offset: 1

Farideh Firoozbakht, Oct 12 2014

nonn

Firoozbakht's conjecture states that for every n, there exists at least one prime p with prime(n) < p < prime(n)^(1+1/n).
The only known indices n for which A182134(n) = 1 are {1, 2, 3, 4, 8}.
See A246782 for indices n such that A182134(n) = 2.
This sequence lists numbers n such that A182134(n) = 3.

			12 is in the sequence since there exists only three primes p where, prime(12) < p < prime(12)^(1 + 1/12). Note that prime(12) = 37, 37^(1 + 1/12) ~ 49.99 and 37 < 41 < 43 < 47 < 49.99.

Robert Price, Table of n, a(n) for n = 1..170
Carlos Rivera, Conjecture 30. The Firoozbakht Conjecture PrimePuzzles.net.
A. Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015
Wikipedia, Firoozbakht's conjecture

Cf. A000040, A182134, A246782.

Cf. A249566.

a246781 n = a246781_list !! (n-1)
a246781_list = filter ((== 3) . a182134) [1..]
-- Reinhard Zumkeller, Nov 17 2014

N:= 10^5: # to get all terms where prime(n)^(1+1/n) < N
Primes:= select(isprime,[2,seq(2*i+1,i=1..floor((N+1)/2))]):
filter:= proc(n) local t; t:= Primes[n]^(n+1); Primes[n+3]^n <= t and Primes[n+4]^n > t end proc:
select(filter, [$1..nops(Primes)-4]); # Robert Israel, Mar 23 2015

np[n_] := (a = Prime[n]; b = a^(1 + 1/n); Length[Select[Range[a + 1, b], PrimeQ]]); Select[Range[10000], np[#] == 3 &]

a(43)-a(50) from Robert Price, Oct 24 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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A246782 Numbers k such that A182134(k)=2, i.e., there exist only two primes p with prime(k) < p < prime(k)^(1+1/k).

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A246781 Numbers n such that A182134(n) = 3, i.e., there exist only three primes p with prime(n) < p < prime(n)^(1 + 1/n).

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

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