A133697 a(n) = smallest number k such that P(k)/P(k+1) > P(k+1)/P(k+2) > ... > P(k+n+1)/P(k+n+2), where P(k) = k-th prime = A000040(k).

1, 7, 69, 420, 1796, 12073, 101397, 1139211, 5440508, 320620306, 2058187481, 36451609409, 54594153615, 4100904808215

Offset: 0

Author

Philippe LALLOUET (philip.lallouet(AT)orange.fr), Jan 04 2008

Keywords

• nonn
• more

Comments

In other words, the rank of the smallest prime number such that the ratio between each prime and the following one is decreasing for at least n+2 consecutive ratios.

The sequence of primes P[a(n)] begins 2,17,347,2903,15373,128981,... - Robert G. Wilson v, Mar 01 2008

a(9) > 120000000. - Robert G. Wilson v, Mar 01 2008

If 113 is, as conjectured, the last term of A124129, then P(a(n)) = A158939(n+2). Proof: Let x and y be the prime gaps following the prime p = P(j) > 113, so that P(j+1) = P(j) + x and P(j+2) = P(j) + x + y. The inequality P(j)/P(j+1) > P(j+1)/P(j+2) can be written as p/(p+x) > (p+x)/(p+x+y), which simplifies to y > x+x^2/p. By assumption, x^2 < p, so this holds if and only if y > x. So the condition P(j)/P(j+1) > P(j+1)/P(j+2) is equivalent to increasing prime gaps, P(j+2) - P(j+1) > P(j+1) - P(j). (In fact, since all prime gaps except the first are even, it is enough to assume the weaker conjecture that 7 is the only prime P(j) such that (P(j+1)-P(j))^2 >= 2*P(j).) - Pontus von Brömssen, Nov 19 2021

Examples

			P(1)=2, P(2)=3, P(3)=5; 2/3 > 3/5, hence a(0)=1.
17/19 > 19/23 > 23/29 is the first double inequality satisfied by consecutive primes, hence a(1)=7 as 17=P(7).
347/349 > 349/353 > 353/359 > 359/367 is the first triple inequality satisfied by consecutive primes, hence a(2)=69 as 347=P(69).
		

Crossrefs

Cf. A000040, A124129, A158939.

Programs

Mathematica

(* for the 6th term *) n = 12000; While[ Prime[n]/Prime[n + 1] < Prime[n + 1]/Prime[n + 2] || Prime[n + 1]/Prime[n + 2] < Prime[n + 2]/Prime[n + 3] || Prime[n + 2]/Prime[n + 3] < Prime[n + 3]/Prime[n + 4] || Prime[n + 3]/Prime[n + 4] < Prime[n + 4]/Prime[n + 5] || Prime[n + 4]/Prime[n + 5] < Prime[n + 5]/Prime[n + 6] || Prime[n + 5]/Prime[n + 6] < Prime[n + 6]/Prime[n + 7] || Prime[n + 6]/Prime[n + 7] < Prime[n + 7]/Prime[n + 8], n++ ]; Print[n] (* Robert G. Wilson v, Mar 01 2008 *)

Extensions

a(6)-a(8) from Robert G. Wilson v, Mar 01 2008

a(9)-a(13) (based on data for A158939) from Pontus von Brömssen, Nov 19 2021

Edited to make name and offset consistent by Pontus von Brömssen, Nov 19 2021