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A133697
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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).
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0
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OFFSET
| 1,1
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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 (rgwv(AT)rgwv.com), Mar 01 2008
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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)
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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 (rgwv(AT)rgwv.com), Mar 01 2008
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CROSSREFS
| Sequence in context: A087567 A136629 A197525 * A122010 A180911 A084774
Adjacent sequences: A133694 A133695 A133696 * A133698 A133699 A133700
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KEYWORD
| nonn
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AUTHOR
| Philippe LALLOUET (philip.lallouet(AT)orange.fr), Jan 04 2008
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EXTENSIONS
| a(6) - a(8) from Robert G. Wilson v (rgwv(AT)rgwv.com), Mar 01 2008
a(9) > 120000000, Robert G. Wilson v (rgwv(AT)rgwv.com), Mar 01 2008
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