OFFSET
1,1
COMMENTS
The Firoozbakht Conjecture, "prime(n)^(1/n) is a strictly decreasing function of n" is true if and only if a(n) - A001223(n) is nonnegative for all n. The conjecture is true for all primes p where p < 4.0*10^18. (See A. Kourbatov link.)
0, 1, 5, 7, 10 & 20 are not in the sequence. It seems that these six integers are all the nonnegative integers which are not in the sequence.
From Alexei Kourbatov, Nov 27 2015: (Start)
Theorem: if prime(n+1) - prime(n) < prime(n)^(3/4), then every integer > 20 is in this sequence.
Proof: Let f(n) = prime(n)^(1+1/n) - prime(n). Then a(n) = floor(f(n)).
Define F(x) = log^2(x) - log(x) - 1. Using the upper and lower bounds for f(n) established in Theorem 5 of J. Integer Sequences Article 15.11.2; arXiv:1506.03042 we have F(prime(n))-3.83/(log prime(n)) < f(n) < F(prime(n)) for n>10^6; so f(n) is unbounded and asymptotically equal to F(prime(n)).
Therefore, for every n>10^6, jumps in f(n) are less than F'(x)*x^(3/4)+3.83/(log x) at x=prime(n), which is less than 1 as x >= prime(10^6)=15485863. Thus jumps in a(n) cannot be more than 1 when n>10^6. Separately, we verify by direct computation that a(n) takes every value from 21 to 256 when 30 < n <= 10^6. This completes the proof.
(End)
REFERENCES
Paulo Ribenboim, The little book of bigger primes, second edition, Springer, 2004, p. 185.
LINKS
Robert G. Wilson v, Table of n, a(n) for n = 1..10000
A. Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015.
A. Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, arXiv:1506.03042 [math.NT], 2015.
A. Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, Journal of Integer Sequences, 18 (2015), Article 15.11.2.
Carlos Rivera, Conjecture 30
Wikipedia, Firoozbakht's conjecture.
Wikipedia, Prime gap.
FORMULA
a(n) = (log(prime(n)))^2 - log(prime(n)) + O(1), see arXiv:1506.03042. - Alexei Kourbatov, Sep 06 2015
MAPLE
N:= 10^4: # to get entries corresponding to all primes <= N
Primes:= select(isprime, [2, seq(2*i+1, i=1..floor((N-1)/2))]):
seq(floor(Primes[n]^(1+1/n) - Primes[n]), n=1..nops(Primes)); # Robert Israel, Mar 23 2015
MATHEMATICA
f[n_] := Block[{p = Prime@ n}, Floor[p^(1 + 1/n)] - p]; Array[f, 75]
PROG
(Magma) [Floor(NthPrime(n)^(1+1/n)) - NthPrime(n): n in [1..70]]; // Vincenzo Librandi, Mar 24 2015
(PARI) first(m)=vector(m, i, floor(prime(i)^(1+1/i)) - prime(i)) \\ Anders Hellström, Sep 06 2015
KEYWORD
nonn
AUTHOR
Farideh Firoozbakht, Sep 26 2014
STATUS
approved