A246781 -id:A246781 - cp's OEIS frontend

A182134 Number of primes p such that prime(n) < p < prime(n)^(1 + 1/n).

1, 1, 1, 1, 2, 2, 2, 1, 2, 2, 2, 3, 3, 2, 2, 3, 4, 3, 4, 3, 3, 2, 2, 4, 5, 4, 3, 2, 2, 2, 3, 4, 4, 3, 4, 4, 4, 4, 3, 4, 5, 4, 4, 3, 2, 2, 4, 5, 5, 4, 4, 4, 3, 5, 6, 5, 5, 4, 3, 3, 2, 4, 4, 4, 3, 2, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 5, 6, 5, 7, 6, 6, 5, 5, 5, 5, 4, 4, 5, 4, 5, 4, 3, 3, 3, 3, 5, 5, 5, 5, 6, 5

Offset: 1

Thomas Ordowski, Apr 20 2012

nonn

Firoozbakht's conjecture: prime(n+1)^(1/(n+1)) < prime(n)^(1/n), for all n >= 1.
According to Firoozbakht's conjecture, all terms of this sequence are positive. - Jahangeer Kholdi, Jul 30 2014
Conjecture: a(n)=1 only for n = 1, 2, 3, 4, and 8. - Farideh Firoozbakht, Oct 18 2014
See A246782 and A246781 for indices such that a(n)=2 resp. a(n)=3. - M. F. Hasler, Oct 19 2014
Length of n-th row in A244365; a(n) = A001221(A245722(n)). - Reinhard Zumkeller, Nov 18 2014
a(n) = 2 for n = 5, 6, 7, 9, 10, 11, 14, 15, 22, 23, 28, 29, 30, 45, 46, 61, 66, 216, 217, 367, 3793, 1319945, ... = A246782. - Robert G. Wilson v, Feb 20 2015
a(n) = 3 for n = 12, 13, 16, 18, 20, 21, 27, 31, 34, 39, 44, 53, 59, 60, 65, 96, 97, 98, 99, 136, 154, 202, ... = A246781. - Robert G. Wilson v, Feb 20 2015
First occurrence of k: 1, 5, 12, 17, 25, 55, 83, 169, 207, 206, 384, 953, ... = A246810. - Robert G. Wilson v, Feb 20 2015
Conjecture: lim sup n->oo a(n) = oo. - John W. Nicholson, Feb 28 2015
a(n) is unbounded (that is, the above conjecture is true). In particular, there is a constant c > 1 such that a(n) > c log n infinitely often (by Maier's theorem). - Thomas Ordowski and Charles R Greathouse IV, Apr 09 2015

			a(25) = 5, because p(25) = 97 and there are 5 primes p such that 97 < p < 97^(1 + 1/25) = 121.9299290...: 101, 103, 107, 109, 113.

T. D. Noe, Table of n, a(n) for n = 1..10000
Alexei Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015.
Alexei Kourbatov, Upper Bounds for Prime Gaps Related to Firoozbakht’s Conjecture, arXiv:1506.03042 [math.NT], 2015.
Alexei Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, J. Int. Seq. 18 (2015) 15.11.2.
Wikipedia, Firoozbakht's conjecture

Cf. A010051, A000040, A249669, A244365, A246810.

a182134 = length . a244365_row  -- Reinhard Zumkeller, Nov 16 2014

a:= n-> numtheory[pi](ceil(ithprime(n)^(1+1/n))-1)-n:
seq(a(n), n=1..100);  # Alois P. Heinz, Apr 21 2012

Table[i = Prime[n] + 1; j = Floor[Prime[n]^(1 + 1/n)]; Length[Select[Range[i, j], PrimeQ]], {n, 100}] (* T. D. Noe, Apr 21 2012 *)
f[n_] := PrimePi[ Prime[n]^(1 + 1/n)] - n; Array[f, 105] (* Robert G. Wilson v, Feb 20 2015 *)

A182134(n)=primepi(prime(n)^(1+1/n))-n \\ M. F. Hasler, Nov 03 2014

from sympy import primepi, prime
def a(n): return primepi(prime(n)**(1 + 1/n)) - n # Indranil Ghosh, Apr 23 2017

a(n) = Sum_{m=A000040(n+1)..A249669(n)} A010051(m). - Reinhard Zumkeller, Nov 16 2014

a(n) = primepi(prime(n)^(1+1/n)) - n (see PARI program). - John W. Nicholson, Feb 11 2015

More terms from Alois P. Heinz, Apr 21 2012

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).

Original entry on oeis.org

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

A246810 a(n) is the smallest number m such that np(m) = n, where np(m) is number of primes p such that prime(m) < p < prime(m)^(1 + 1/m).

Original entry on oeis.org

1, 5, 12, 17, 25, 55, 83, 169, 207, 206, 384, 953, 1615, 2192, 2197, 3024, 3023, 10709, 10935, 29509, 29508, 62736, 62735, 94333, 94332, 196966, 314940, 608777, 1258688, 1767259, 2448975, 2448973, 7939362, 9373136, 9373134, 16854966, 16854967

Offset: 1

Farideh Firoozbakht and Jahangeer Kholdi, Oct 10 2014

nonn

Firoozbakht's conjecture says that for every n, there exists at least one prime p where, prime(n) < p < prime(n)^(1 + 1/n). Hence if Firoozbakht's conjecture is true, then there is no m such that np(m) = 0.
Conjecture: For every positive integer n, a(n) exists.
a(65) > 10^12. - Robert Price, Nov 12 2014

			a(6) = 55 since the number of primes p such that prime(55) < p < prime(55)^(1 + 1/55) is 6 and 55 is the smallest number with this property.

Robert Price, Table of n, a(n) for n = 1..64
A. Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015
A. Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, J. Int. Seq. 18 (2015) 15.11.2
Nilotpal Kanti Sinha, On a new property of primes that leads to a generalization of Cramer's conjecture, arXiv:1010.1399v2 [math.NT], 2010.
Wikipedia, Firoozbakht's conjecture.

Cf. A000040, A182134, A246781, A246782, A246783, A246787.

np[n_]:=(b=Prime[n]; Length[Select[Range[b+1, b^(1 + 1/n)],PrimeQ]]); a[n_]:=(For[m=1, np[m] !=n, m++]; m);
Do[Print[a[n]], {n, 37}]

A249566 Numbers n such that A182134(n) = 4, i.e., there exist exactly four primes p with prime(n) < p < prime(n)^(1+1/n).

Original entry on oeis.org

17, 19, 24, 26, 32, 33, 35, 36, 37, 38, 40, 42, 43, 47, 50, 51, 52, 58, 62, 63, 64, 76, 77, 78, 79, 90, 91, 93, 95, 121, 123, 124, 125, 126, 134, 135, 137, 150, 153, 185, 186, 187, 188, 189, 201, 203, 213, 218, 219, 238, 239, 259, 263, 278, 279, 289, 293

Offset: 1

Robert Price, Nov 01 2014

nonn

See A246782 for a more complete description of this sequence.
a(1136) > 10^12.
It is interesting that three consecutive integers n = 20004097201301075, n + 1 and n + 2 are in the sequence. Conjecture: The sequence is infinite. - Farideh Firoozbakht, Nov 01 2014

Robert Price, Table of n, a(n) for n = 1..1135
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, A246782.

a249566 n = a249566_list !! (n-1)
a249566_list = filter ((== 4) . 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] == 4,Print[n]], {n, 293}]
np[n_]:=(a = Prime[n]; b = a^(1 + 1/n); Length[Select[Range[a+1,b], PrimeQ]]); Select[Range[293], np[#]==4&] (* Farideh Firoozbakht, Nov 01 2014 *)

for(n=1,9e9,primepi(prime(n)^(1+1/n))-n==4&&print1(n",")) \\ M. F. Hasler, Nov 03 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

cp's OEIS Frontend

A182134 Number of primes p such that prime(n) < p < prime(n)^(1 + 1/n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Python

Formula

Extensions

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

Extensions

A246810 a(n) is the smallest number m such that np(m) = n, where np(m) is number of primes p such that prime(m) < p < prime(m)^(1 + 1/m).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A249566 Numbers n such that A182134(n) = 4, i.e., there exist exactly four primes p with prime(n) < p < prime(n)^(1+1/n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions