A263023 Largest integer k such that prime(n+1) < prime(n)^(1+1/k).

1, 2, 4, 4, 14, 9, 25, 15, 13, 50, 19, 35, 77, 42, 32, 37, 122, 43, 72, 153, 54, 88, 63, 52, 113, 235, 121, 252, 130, 40, 156, 108, 339, 71, 375, 128, 134, 210, 144, 151, 466, 96, 504, 256, 523, 90, 96, 304, 618, 313, 214, 657, 134, 233, 240, 247, 755, 255

Offset: 1

Author

Alexei Kourbatov, Oct 08 2015

Keywords

• nonn

Comments

Firoozbakht's conjecture: prime(n+1) < prime(n)^(1+1/n).

Firoozbakht's conjecture restated for this sequence: a(n) >= n.

I further conjecture that n = 1,2,4 are the only values of n with a(n) = n.

Record values of a(n) occur when prime(n) and prime(n+1) are twin primes.

Upper bound for all n: a(n) < (1/2)*(prime(n)+2)*log(prime(n)).

Examples

			prime(1)=2; a(1)=1 because k=1 is the largest k for which 3 < 2^(1+1/k).
prime(2)=3; a(2)=2 because k=2 is the largest k for which 5 < 3^(1+1/k).
prime(10)=29; a(10)=50 because k=50 is the largest k for which 31 < 29^(1+1/k).
		

References

• Paulo Ribenboim, The little book of bigger primes, 2nd edition, Springer, 2004, p. 185.

Links

• 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.
• Carlos Rivera, Conjecture 30

Crossrefs

Cf. A000040, A002386, A182514, A182519, A245396, A246776, A246778, A249669.

Programs

Magma

[Floor(Log(NthPrime(n))/(Log(NthPrime(n+1))-Log(NthPrime(n)))): n in [1..60]]; // Vincenzo Librandi, Oct 08 2015

Mathematica

Table[Floor[Log@ Prime@ n /(Log@ Prime[n + 1] - Log@ Prime@ n)], {n, 58}] (* Michael De Vlieger, Oct 08 2015 *)

PARI

a(n) = floor(log(prime(n))/(log(prime(n+1)) - log(prime(n)))) \\ Michel Marcus, Oct 10 2015

Formula

a(n) = floor(log(prime(n))/(log(prime(n+1)) - log(prime(n)))).

Extensions

More terms from Vincenzo Librandi, Oct 08 2015