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A182134
Number of primes p such that prime(n) < p < prime(n)^(1 + 1/n).
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
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
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.
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
KEYWORD
nonn
AUTHOR
_Thomas Ordowski_, Apr 20 2012
EXTENSIONS
More terms from _Alois P. Heinz_, Apr 21 2012
STATUS
approved