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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
1, 7, 69, 420, 1796, 12073, 101397, 1139211, 5440508, 320620306, 2058187481, 36451609409, 54594153615, 4100904808215 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
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
LINKS
EXAMPLE
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).
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 *)
CROSSREFS
Sequence in context: A306386 A136629 A197525 * A224758 A219330 A122010
KEYWORD
nonn,more
AUTHOR
Philippe LALLOUET (philip.lallouet(AT)orange.fr), Jan 04 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
STATUS
approved

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Last modified March 29 05:48 EDT 2024. Contains 371265 sequences. (Running on oeis4.)