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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). 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
Carlos Rivera, Conjecture 30. The Firoozbakht Conjecture PrimePuzzles.net.
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. A249566.
Sequence in context: A003894 A084622 A155147 * A241748 A298591 A225102
KEYWORD
nonn
AUTHOR
Farideh Firoozbakht, Oct 12 2014
EXTENSIONS
a(43)-a(50) from Robert Price, Oct 24 2014
STATUS
approved

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Last modified April 23 12:27 EDT 2024. Contains 371912 sequences. (Running on oeis4.)