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Search: a246782 -id:a246782
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A246784 Numbers n such that both n and n+1 are in the sequence A246782. +20
0
5, 6, 9, 10, 14, 22, 28, 29, 45, 216, 714573709895 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
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
LINKS
Table of n, a(n) for n=1..11.
CROSSREFS
Cf. A000040, A182134, A246782.
KEYWORD
nonn,more,hard
AUTHOR
Farideh Firoozbakht, Oct 23 2014
EXTENSIONS
a(11) from Robert Price, Nov 15 2014
STATUS
approved
A182134 Number of primes p such that prime(n) < p < prime(n)^(1 + 1/n). +10
26
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,5
COMMENTS
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
LINKS
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
FORMULA
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
EXAMPLE
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.
MAPLE
a:= n-> numtheory[pi](ceil(ithprime(n)^(1+1/n))-1)-n:
seq(a(n), n=1..100); # Alois P. Heinz, Apr 21 2012
MATHEMATICA
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 *)
PROG
(PARI) A182134(n)=primepi(prime(n)^(1+1/n))-n \\ M. F. Hasler, Nov 03 2014
(Haskell)
a182134 = length . a244365_row -- Reinhard Zumkeller, Nov 16 2014
(Python)
from sympy import primepi, prime
def a(n): return primepi(prime(n)**(1 + 1/n)) - n # Indranil Ghosh, Apr 23 2017
CROSSREFS
Cf. A010051, A000040, A249669, A244365, A246810.
KEYWORD
nonn
AUTHOR
Thomas Ordowski, Apr 20 2012
EXTENSIONS
More terms from Alois P. Heinz, Apr 21 2012
STATUS
approved
A245396 Largest prime not exceeding prime(n)^(1 + 1/n). +10
6
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
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
LINKS
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:1010.1399 [math.NT]
Wikipedia, Firoozbakht's conjecture
FORMULA
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).
MAPLE
seq(prevprime(ceil(ithprime(n)^(1+1/n))), n=1..100); # Robert Israel, Nov 03 2014
MATHEMATICA
Table[NextPrime[Prime[n]^(1 + 1/n), -1], {n, 64}] (* Farideh Firoozbakht, Nov 03 2014 *)
PROG
(PARI) a(n)=precprime(prime(n)^(1+1/n))
(PARI) a(n)=precprime(sqrtnint(prime(n)^(n+1), n)) \\ Charles R Greathouse IV, Oct 29 2018
(Haskell)
a245396 n = a244365 n (a182134 n) -- Reinhard Zumkeller, Nov 16 2014
CROSSREFS
Cf. A000040, A001223, A007917, A182134, A246778, A249669.
Cf. A244365.
KEYWORD
nonn
AUTHOR
M. F. Hasler, Nov 03 2014
STATUS
approved
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). +10
6
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
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.
LINKS
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
EXAMPLE
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.
MAPLE
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
MATHEMATICA
np[n_] := (a = Prime[n]; b = a^(1 + 1/n); Length[Select[Range[a + 1, b], PrimeQ]]); Select[Range[10000], np[#] == 3 &]
PROG
(Haskell)
a246781 n = a246781_list !! (n-1)
a246781_list = filter ((== 3) . a182134) [1..]
-- Reinhard Zumkeller, Nov 17 2014
CROSSREFS
Cf. A000040, A182134, A246782.
Cf. A249566.
KEYWORD
nonn
AUTHOR
Farideh Firoozbakht, Oct 12 2014
EXTENSIONS
a(43)-a(50) from Robert Price, Oct 24 2014
STATUS
approved
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). +10
5
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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
LINKS
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.
EXAMPLE
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.
MATHEMATICA
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}]
CROSSREFS
Cf. A000040, A182134, A246781, A246782, A246783, A246787.
KEYWORD
nonn
AUTHOR
Farideh Firoozbakht and Jahangeer Kholdi, Oct 10 2014
STATUS
approved
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). +10
4
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
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
LINKS
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
MATHEMATICA
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 *)
PROG
(PARI) for(n=1, 9e9, primepi(prime(n)^(1+1/n))-n==4&&print1(n", ")) \\ M. F. Hasler, Nov 03 2014
(Haskell)
a249566 n = a249566_list !! (n-1)
a249566_list = filter ((== 4) . a182134) [1..]
-- Reinhard Zumkeller, Nov 17 2014
CROSSREFS
Cf. A000040, A002386, A005669, A182134, A246781, A246782.
KEYWORD
nonn
AUTHOR
Robert Price, Nov 01 2014
STATUS
approved
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Last modified July 25 09:15 EDT 2024. Contains 374587 sequences. (Running on oeis4.)