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A246778
a(n) = floor(prime(n)^(1+1/n)) - prime(n).
10
2, 2, 3, 4, 6, 6, 8, 8, 9, 11, 11, 12, 13, 13, 13, 14, 15, 15, 16, 16, 16, 17, 17, 18, 19, 19, 19, 19, 19, 19, 21, 21, 22, 21, 22, 22, 22, 23, 23, 23, 24, 23, 24, 24, 24, 24, 25, 26, 26, 26, 26, 26, 26, 27, 27, 27, 27, 27, 27, 27, 27, 28, 29, 29, 28, 28, 29, 30, 30, 30
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
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, Prime gap.
FORMULA
a(n) = A249669(n) - A000040(n). - M. F. Hasler, Nov 03 2014
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