A005669 -id:A005669 - cp's OEIS frontend

A214935 Index of the primes of A205827, A000720(A205827(n)).

1, 2, 4, 9, 30, 189, 217, 2225, 3385, 14357, 30802, 31545, 104071, 149689, 1094421, 1319945, 10655462, 23163298, 112228683, 182837804, 203615628, 486570087, 1094330259, 11992433550, 17883926781, 50070452577, 52302956123, 72178455400

Offset: 1

John W. Nicholson, Oct 28 2012

nonn

A000040(a(n)) = A205827(n).

With pi(x) being the prime counting function, A000720(x), for n from 1 to 3, a(n) = pi(A111870(n)) = A241542(n), for n from 5 to 28, a(n) = pi(A111870(n-1)) = A241542(n-1). - John W. Nicholson, May 10 2014

			a(4) = 9, A000040(9) = 23, and A205827(4) = 23.

John W. Nicholson, Table of n, a(n) for n = 1..38

Cf. A205827.

record=0;i=0;A214935=vectorsmall(38);for(n=1,75,current=(A000101[n]/A002386[n]*1.)^A005669[n];if(current>record,record=current;i++;A214935[i]=A005669[n]))

a(n) = pi(A205827(n)) = A000720(A205827(n)).

a(13)-a(28) from Donovan Johnson, Oct 28 2012

a(29)-a(38) from John W. Nicholson, Dec 01 2013

A246776 a(n) = floor(prime(n)^(1+1/n)) - prime(n+1).

Original entry on oeis.org

1, 0, 1, 0, 4, 2, 6, 4, 3, 9, 5, 8, 11, 9, 7, 8, 13, 9, 12, 14, 10, 13, 11, 10, 15, 17, 15, 17, 15, 5, 17, 15, 20, 11, 20, 16, 16, 19, 17, 17, 22, 13, 22, 20, 22, 12, 13, 22, 24, 22, 20, 24, 16, 21, 21, 21, 25, 21, 23, 25, 17, 14, 25, 27, 24, 14, 23, 20, 28, 26

Offset: 1

Farideh Firoozbakht, Sep 26 2014

nonn

The Firoozbakht Conjecture, "prime(n)^(1/n) is a strictly decreasing function of n" is true if and only if a(n) is nonnegative for all n, n>1.
A246777 is a hard subsequence of this sequence.
18 is not in the sequence. It seems that, 18 is the only nonnegative integer which is not in the sequence.

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

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 4230 terms from Hal M. Switkay)
Carlos Rivera, Conjecture 30
Alexei Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015-2023.
Alexei Kourbatov, Upper Bounds for Prime Gaps Related to Firoozbakht’s Conjecture, arXiv:1506.03042 [math.NT], 2015-2019.
Alexei Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, J. Int. Seq. 18 (2015) 15.11.2
Wikipedia, Prime gap.
Wikipedia, Firoozbakht Conjecture.

Cf. A000040, A001223, A005669, A246777, A246778.

Cf. A249669.

a246776 n = a249669 n - a000040 (n + 1)
-- Reinhard Zumkeller, Nov 16 2014

Table[Floor[Prime[n]^(1+1/n)]-Prime[n+1],{n,70}]

a(n) = A249669(n) - A000040(n+1). - Reinhard Zumkeller, Nov 16 2014

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

A182514 Primes prime(n) such that (prime(n+1)/prime(n))^n > n.

Original entry on oeis.org

2, 3, 7, 113, 1327, 1693182318746371

Offset: 1

Thomas Ordowski, May 04 2012

nonn

The Firoozbakht conjecture: (prime(n+1))^(1/(n+1)) < prime(n)^(1/n), or prime(n+1) < prime(n)^(1+1/n), prime(n+1)/prime(n) < prime(n)^(1/n), (prime(n+1)/prime(n))^n < prime(n).
Using the Mathematica program shown below, I have found no further terms below 2^27. I conjecture that this sequence is finite and that the terms stated are the only members. - Robert G. Wilson v, May 06 2012 [Warning: this conjecture may be false! - N. J. A. Sloane, Apr 25 2014]
I conjecture the contrary: the sequence is infinite. Note that 10^13 < a(6) <= 1693182318746371. - Charles R Greathouse IV, May 14 2012
[Stronger than Firoozbakht] conjecture: All (prime(n+1)/prime(n))^n values, with n >= 5, are less than n*log(n). - John W. Nicholson, Dec 02 2013, Oct 19 2016
The Firoozbakht conjecture can be rewritten as (log(prime(n+1)) / log(prime(n)))^n < (1+1/n)^n. This suggests the [weaker than Firoozbakht] conjecture: (log(prime(n+1))/log(prime(n)))^n < e. - Daniel Forgues, Apr 26 2014
All a(n) <= a(6) are in A002386, A205827, and A111870.
The inequality in the definition is equivalent to the inequality prime(n+1)-prime(n) > log(n)*log(prime(n)) for sufficiently large n. - Thomas Ordowski, Mar 16 2015
Prime indices, A000720(a(n)) = 1, 2, 4, 30, 217, 49749629143526. - John W. Nicholson, Oct 25 2016

			7 is in the list because, being the 4th prime, and 11 the fifth prime, we verify that (11/7)^4 = 6.09787588507... which is greater than 4.
11 is not on the list because (13/11)^5 = 2.30543740804... and that is less than 5.

Farhadian, R. (2017). On a New Inequality Related to Consecutive Primes. OECONOMICA, vol 13, pp. 236-242.

Reza Farhadian, A New Conjecture On the primes, Preprint, 2016.
R. Farhadian, and R. Jakimczuk, On a New Conjecture of Prime Numbers Int. Math. Forum, vol. 12, 2017, pp. 559-564.
Luan Alberto Ferreira, Some consequences of the Firoozbakht's conjecture, arXiv:1604.03496v2 [math.NT], 2016.
Luan Alberto Ferreira, Hugo Luiz Mariano, Prime gaps and the Firoozbakht Conjecture, São Paulo J. Math. Sci. (2018), 1-11.
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.
Carlos Rivera, Conjecture 78. P_n^((P_n+1/P_n)^n) <= n^P_n, 2016.
Nilotpal Kanti Sinha, On a new property of primes that leads to a generalization of Cramer's conjecture, arXiv:1010.1399 [math.NT], 2010.
Matt Visser, Verifying the Firoozbakht, Nicholson, and Farhadian conjectures up to the 81st maximal prime gap, arXiv:1904.00499 [math.NT], 2019.
Wikipedia, Firoozbakht’s conjecture

Cf. A111870.

Prime[Select[Range[1000], (Prime[# + 1]/Prime[#])^# > # &]] (* Alonso del Arte, May 04 2012 *)
firoozQ[n_, p_, q_] := n * Log[q] > Log[n] + n * Log[p]; k = 1; p = 2; q = 3; While[ k < 2^27, If[ firoozQ[k, p, q], Print[{k, p}]]; k++; p = q; q = NextPrime@ q] (* Robert G. Wilson v, May 06 2012 *)

n=1;p=2;forprime(q=3,1e6,if((q/p*1.)^n++>n, print1(p", "));p=q) \\ Charles R Greathouse IV, May 14 2012

for(n=1,75,if((A000101[n]/A002386[n]*1.)^A005669[n]>=A005669[n], print1(A002386[n],", "))) \\ Each sequence is read in as a vector as to overcome PARI's primelimit \\ John W. Nicholson, Dec 01 2013

q=3;n=2; forprime(p=5, 10^9,result=(p/q)^n/(n*log(n));if(result>1, print(q," ",p, " ", n, " ", result));n++;q=p) \\ for stronger than Firoozbakht conjecture \\ John W. Nicholson, Mar 16 2015, Oct 19 2016

a(6) from John W. Nicholson, Dec 01 2013

A085237 Nondecreasing gaps between primes.

Original entry on oeis.org

1, 2, 2, 4, 4, 4, 6, 6, 6, 6, 6, 6, 6, 8, 14, 14, 14, 18, 20, 22, 34, 34, 36, 36, 36, 44, 52, 52, 72, 86, 86, 96, 112, 114, 118, 132, 132, 148, 154, 154, 154, 180, 210, 220, 222, 234, 248, 250, 250, 282, 288, 292, 320, 336, 336, 354, 382, 384, 394, 456, 464, 468, 474, 486, 490, 500, 514, 516, 532, 534, 540, 582, 588, 602, 652, 674, 716, 766, 778

Offset: 1

Farideh Firoozbakht, Aug 11 2003

nonn

All terms of A005250 are in the sequence, but some terms of A005250 appear in this sequence more than once.

a(n) is the gap between the n-th and (n+1)-th sublists of prime numbers defined in A348178. - Ya-Ping Lu, Oct 19 2021

			a(21) = a(22) = 34 because prime(218) - prime(217) = prime(1060) - prime(1059) = 34 and prime(n+1) - prime(n) is less than 34, for n < 1059 and n not equal to 217.

R. K. Guy, Unsolved problems in number theory.

Cf. A005250, A005669, A053686, A002386, A348178.

f[n_] := Prime[n+1]-Prime[n]; v={}; Do[ If[f[n]>=If[n==1, 1, v[[ -1]]], v1=n; v=Append[v, f[v1]]; Print[v]], {n, 105000000}]
DeleteDuplicates[Differences[Prime[Range[10^7]]],Greater] (* Harvey P. Dale, Jan 17 2024 *)

from sympy import nextprime; p, r = 2, 0
while r < 778:
    q = nextprime(p); g = q - p
    if g >= r: print(g, end = ', '); r = g
    p = q # Ya-Ping Lu, Jan 23 2024

a(53)-a(63) from Donovan Johnson, Nov 24 2008
a(64)-a(76) from Charles R Greathouse IV, May 09 2011
a(77)-a(79) from Charles R Greathouse IV, May 19 2011

A107578 Prime index of A000101(n), maximal gap upper end prime index.

Original entry on oeis.org

2, 3, 5, 10, 25, 31, 100, 155, 190, 218, 1184, 1832, 2226, 3386, 14358, 30803, 31546, 40934, 103521, 104072, 149690, 325853, 1094422, 1319946, 2850175, 6957877, 10539433, 10655463, 20684333, 23163299, 64955635, 72507381

Offset: 1

Alex Beveridge, Apr 25 2007

nonn

Conjecture: log a(n) ~ n/2. That is, record prime gaps occur about twice as often as records in an i.i.d. random sequence of comparable length (see arXiv:1709.05508 for a heuristic explanation). - Alexei Kourbatov, Jan 18 2019

			The prime index of a(3) = 5, so prime(a(3)) = prime(5) = 11.

John W. Nicholson, Table of n, a(n) for n = 1..80 (terms 1..75 from Jens Kruse Andersen; further terms coming from Thomas R. Nicely site).
Alex Beveridge, Table giving known values of A000101(n), A005250(n), A107578(n)
Alexei Kourbatov, On the nth record gap between primes in an arithmetic progression, arXiv:1709.05508 [math.NT], 2017; Int. Math. Forum, 13 (2018), 65-78.
Alexei Kourbatov and Marek Wolf, Predicting maximal gaps in sets of primes, arXiv:1901.03785 [math.NT], 2019.
Thomas R. Nicely, First occurrence prime gaps
Thomas R. Nicely, First occurrence prime gaps [Local copy, pdf only]
Thomas R. Nicely, Some Results of Computational Research in Prime Numbers [See local copy in A007053]
Matt Visser, Verifying the Firoozbakht, Nicholson, and Farhadian conjectures up to the 81st maximal prime gap, arXiv:1904.00499 [math.NT], 2019.

Cf. A000101, A005250, A002386, A005669.

a(n) = A005669(n)+1. - Jens Kruse Andersen, Oct 19 2010
From John W. Nicholson, Oct 29 2021: (Start)
a(n) = A000720(A000101(n)).
a(n) = A000720(A002386(n)) + 1. (End)

Name modified by John W. Nicholson, Nov 19 2013

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

A144309 a(n) is the index of the smallest prime such that the gap to the next prime is not less than 2*n.

Original entry on oeis.org

2, 4, 9, 24, 30, 30, 30, 99, 99, 154, 189, 217, 217, 217, 217, 217, 217, 1183, 1831, 1831, 1831, 1831, 2225, 2225, 2225, 2225, 3385, 3385, 3385, 3385, 3385, 3385, 3385, 3385, 3385, 3385, 14357, 14357, 14357, 14357, 14357, 14357, 14357, 30802, 30802, 30802

Offset: 1

Artur Jasinski, Sep 17 2008

nonn

Amiram Eldar, Table of n, a(n) for n = 1..775

For indices of smallest prime such that gap to next prime is exactly equal to 2*n see A000230. For records in this sequence see A005669.

Cf. A000720, A100964.

a = {}; Do[n = 1; While[(Prime[n + 1] - Prime[n]) < 2 k, n++ ]; AppendTo[a, n], {k, 1, 75}]; a

lista(pmax) = {my(k = 1, prv = 2, m = 2, kprv = 2); forprime(p = 3, pmax, k++; if(p - prv >= m, for(i = 1, (p - prv - m)/2 + 1, print1(k-1, ", ")); m = p - prv + 2; kprv = k); prv = p);} \\ Amiram Eldar, Sep 06 2024

a(n) = primepi(A100964(n)) = A000720(A100964(n)). - Michel Marcus, Nov 02 2013

A241540 Indices of primes p in A182514, i.e., a(n) = primepi(p) = A000720(A182514(n)).

Original entry on oeis.org

1, 2, 4, 30, 217, 49749629143526

Offset: 1

M. F. Hasler, Apr 25 2014

a(6) = A214935(33) = A000720(A205827(33)).

Cf. A053302, A053976, A073861, A111870, A111943, A182514.

See also A214935(n)=A000720(A205827(n)), A005669(n) = A000720(A002386(n)).

A131702 Distances between the locations of new prime gaps (A014320).

Original entry on oeis.org

0, 1, 4, 14, 5, 3, 11, 52, 54, 34, 27, 45, 18, 84, 61, 160, 147, 444, 647, 47, 311, 33, 851, 224, 82, 41, 216, 148, 728, 89, 3357, 57, 659, 3853, 1814, 504, 920, 1222, 2019, 4256

Offset: 1

Giovanni Teofilatto, Sep 16 2007

nonn

A014320 lists "new" gaps in the sequence A001223 of prime gaps (not necessarily records as A005669 does).
The locations of these new gaps in A001223 are 1, 2, 4, 9, 24, 30, 34,...
The present sequence lists the first difference of these locations, minus 1: a(1) = 2-1-1. a(2)=4-2-1. a(3)=9-4-1. a(4)=24-9-1.
The sequence therefore argues: need to skip 0 in A001223 to reach a new gap, need to skip 1 to reach a new gap, need to skip 4 to reach a new gap...

Cf. A001223.

A001223 := proc(n) option remember; ithprime(n+1)-ithprime(n) ; end proc:
A014320 := proc(n) option remember; if n = 1 then return 1; else for k from 1 do t := A001223(k) ; isn := true; for i from 1 to n-1 do if procname(i) = t then isn := false; end if; end do: if isn then return t; end if; end do: end if; end proc:
locng := proc(n) option remember; g := A014320(n) ; for k from 1 do if A001223(k) = g then return k; end if; end do: end proc:
A131702 := proc(n) locng(n+1)-locng(n)-1 ; end proc: seq(A131702(n),n=1..40) ;

More terms, program and comment by R. J. Mathar, Aug 23 2010

cp's OEIS Frontend

A214935 Index of the primes of A205827, A000720(A205827(n)).

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A246776 a(n) = floor(prime(n)^(1+1/n)) - prime(n+1).

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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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A182514 Primes prime(n) such that (prime(n+1)/prime(n))^n > n.

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A085237 Nondecreasing gaps between primes.

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A107578 Prime index of A000101(n), maximal gap upper end prime index.

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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).

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