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Numbers n such that A182134(n)=2, i.e., there exist only two primes p with prime(n) < p < prime(n)^(1+1/n).
+10
7
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
1,1
COMMENTS
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
EXAMPLE
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.
MATHEMATICA
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}]
PROG
(PARI) for(n=1, 9e9, 2==primepi(prime(n)^(1+1/n))-n&&print1(n", ") \\ M. F. Hasler, Nov 03 2014
(Haskell)
a246782 n = a246782_list !! (n-1)
a246782_list = filter ((== 2) . a182134) [1..]
-- Reinhard Zumkeller, Nov 17 2014
KEYWORD
nonn,more
AUTHOR
Farideh Firoozbakht, Oct 12 2014
EXTENSIONS
a(26)-a(27) from Robert Price, Oct 24 2014
a(28)-a(33) from Robert Price, Nov 01 2014
STATUS
approved
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
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
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
KEYWORD
nonn
AUTHOR
Farideh Firoozbakht, Oct 12 2014
EXTENSIONS
a(43)-a(50) from Robert Price, Oct 24 2014
STATUS
approved
Least prime(i) such that prime(i)^(1+1/i) - prime(i) > n.
+10
1
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
1,1
COMMENTS
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.
REFERENCES
Paulo Ribenboim, The little book Of bigger primes, second edition, Springer, 2004, p. 185.
LINKS
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
FORMULA
Log(y) ~= g + x^(1/2) where g = Euler's Gamma.
EXAMPLE
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.
MATHEMATICA
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
PROG
(PARI) a(n) = {i = 0; forprime(p=2, , i++; if (p^(1+1/i) - p > n, return (p)); ); } \\ Michel Marcus, Oct 04 2015
KEYWORD
nonn
AUTHOR
EXTENSIONS
a(2) corrected in b-file by Andrew Howroyd, Feb 22 2018
STATUS
approved

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