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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)/P(k+1+n), where P(k) = k-th prime = A000040(k). 0
 7, 69, 420, 1796, 12073, 101397, 1139211, 5440508 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 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 1+n consecutive primes. The sequence of primes P[a(n)] begins 17,347,2903,15373,128981,... - Robert G. Wilson v, Mar 01 2008 a(9) > 120000000. - Robert G. Wilson v, Mar 01 2008 LINKS EXAMPLE P(1)=2, P(2)=3, P(3)=5, P(4)=7; 2/3 > 3/5 but 3/5 < 5/7, hence 1 is not in the sequence 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 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 Adjacent sequences:  A133694 A133695 A133696 * A133698 A133699 A133700 KEYWORD nonn AUTHOR Philippe LALLOUET (philip.lallouet(AT)orange.fr), Jan 04 2008 EXTENSIONS a(6)-a(8) from Robert G. Wilson v, Mar 01 2008 STATUS approved

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Last modified February 24 10:37 EST 2020. Contains 332209 sequences. (Running on oeis4.)