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A182134
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Number of primes p such that prime(n) < p < prime(n)^(1 + 1/n).
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26
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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
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OFFSET
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1,5
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COMMENTS
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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
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
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
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LINKS
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FORMULA
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a(n) = primepi(prime(n)^(1+1/n)) - n (see PARI program). - John W. Nicholson, Feb 11 2015
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EXAMPLE
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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.
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MAPLE
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a:= n-> numtheory[pi](ceil(ithprime(n)^(1+1/n))-1)-n:
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MATHEMATICA
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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 *)
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PROG
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(Haskell)
(Python)
from sympy import primepi, prime
def a(n): return primepi(prime(n)**(1 + 1/n)) - n # Indranil Ghosh, Apr 23 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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