

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
(list;
graph;
refs;
listen;
history;
text;
internal format)



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) = 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((N1)/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 HellstrÃ¶m, Sep 06 2015


CROSSREFS

Cf. A000040, A001223, A246776, A246777, A246779, A246780, A249669.
Sequence in context: A152047 A289341 A263096 * A101344 A152048 A046934
Adjacent sequences: A246775 A246776 A246777 * A246779 A246780 A246781


KEYWORD

nonn


AUTHOR

Farideh Firoozbakht, Sep 26 2014


STATUS

approved



