A246782 -id:A246782 - cp's OEIS frontend

A246784 Numbers n such that both n and n+1 are in the sequence A246782.

5, 6, 9, 10, 14, 22, 28, 29, 45, 216, 714573709895

Offset: 1

Farideh Firoozbakht, Oct 23 2014

Numbers n such that A182134(n) = A182134(n+1) = 2.
n = 1475067052906945 is a large term in this sequence.
a(12) > 10^12. - Robert Price, Nov 15 2014

Cf. A000040, A182134, A246782.

a(11) from Robert Price, Nov 15 2014

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

Original entry on oeis.org

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

A245396 Largest prime not exceeding prime(n)^(1 + 1/n).

Original entry on oeis.org

3, 5, 7, 11, 17, 19, 23, 23, 31, 37, 41, 47, 53, 53, 59, 67, 73, 73, 83, 83, 89, 89, 97, 107, 113, 113, 113, 113, 127, 131, 139, 151, 157, 157, 167, 173, 179, 181, 181, 193, 199, 199, 211, 211, 211, 223, 233, 241, 251, 251, 257, 263, 263, 277, 283, 283, 293, 293, 293, 307, 307, 317, 331, 337

Offset: 1

M. F. Hasler, Nov 03 2014

nonn

Firoozbakht's conjecture, prime(n+1) < prime(n)^(1 + 1/n), is equivalent to a(n) > prime(n). See also A182134.
Here prime(n) = A000040(n). The conjecture is also equivalent to a(n) - prime(n) >= A001223(n), the n-th gap between primes. See also A246778(n) = floor(prime(n)^(1 + 1/n)) - prime(n).
It is also conjectured that the equality a(n) - prime(n) = A001223(n) holds only for n in the set {1, 2, 3, 4, 8}, see A246782. a(n) is also largest prime less than prime(n)^(1 + 1/n), since prime(n)^(1 + 1/n) is never prime. - Farideh Firoozbakht, Nov 03 2014
a(n) = A007917(A249669(n)) = A244365(n,A182134(n)) = A006530(A245722(n)). - Reinhard Zumkeller, Nov 18 2014

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:1010.1399 [math.NT]
Wikipedia, Firoozbakht's conjecture

Cf. A000040, A001223, A007917, A182134, A246778, A249669.

Cf. A244365.

a245396 n = a244365 n (a182134 n)  -- Reinhard Zumkeller, Nov 16 2014

seq(prevprime(ceil(ithprime(n)^(1+1/n))),n=1..100); # Robert Israel, Nov 03 2014

Table[NextPrime[Prime[n]^(1 + 1/n), -1], {n, 64}] (* Farideh Firoozbakht, Nov 03 2014 *)

PARI
```
a(n)=precprime(prime(n)^(1+1/n))
```

a(n)=precprime(sqrtnint(prime(n)^(n+1),n)) \\ Charles R Greathouse IV, Oct 29 2018

A245396 = A007917 o A249669, i.e., a(n) = A007917(A249669(n)). Although one could say "less than" in the definition of this sequence, one cannot use A151799 in this formula because for n = 2 and n = 4, one has a(n) = A249669(n).

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

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

cp's OEIS Frontend

A246784 Numbers n such that both n and n+1 are in the sequence A246782.

Views

Author

Keywords

Comments

Crossrefs

Extensions

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

A245396 Largest prime not exceeding prime(n)^(1 + 1/n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

PARI

Formula

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

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