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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).
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, oo, 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
CROSSREFS
KEYWORD
nonn,more,changed
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