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A276309
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Integer part of the ratio of alternate consecutive prime gaps.
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7
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2, 2, 1, 1, 1, 1, 3, 0, 1, 2, 0, 1, 3, 1, 0, 1, 2, 0, 1, 2, 1, 2, 0, 0, 1, 1, 1, 7, 1, 0, 0, 1, 1, 0, 3, 0, 1, 1, 0, 1, 1, 0, 1, 3, 6, 0, 0, 1, 3, 0, 1, 3, 0, 1, 0, 1, 2, 0, 2, 7, 0, 0, 1, 7, 1, 0, 0, 0, 3, 2, 1, 0, 0, 1, 2, 0, 1, 2, 0, 1, 1, 0, 2, 1, 2, 0, 0, 1, 6, 2, 0, 1, 1, 0, 3, 0
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OFFSET
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1,1
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COMMENTS
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Conjectures: The most frequent ratio among alternate prime gaps is 1, while the most frequent ratios among consecutive prime gaps seems to be 2 and 1/2, both with nearly the same frequency (see links). It also appears that next four most frequent ratios are 2, 1/2, 3 and 1/3, all four with nearly the same frequency (see links).
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LINKS
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Table of n, a(n) for n=1..96.
Andres Cicuttin, Counts on first 200000 ratios of alternate prime gaps
Andres Cicuttin, Several histograms of the logarithm of the ratio of consecutive prime gaps (prime(n+2)-prime(n+1))/(prime(n+1)-prime(n)) obtained for the first 2*10^5 primes with different bin sizes
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FORMULA
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a(n) = floor((prime(n + 3) - prime(n + 2))/(prime(n + 1) - prime(n))) .
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EXAMPLE
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For n=2, the second prime is 3, and the next three primes are 5, 7, and 11. So the ratio of prime gaps is (11-7)/(5-3) = 4/2 = 2, and the integer part of this is a(2) = 2. - Michael B. Porter, Aug 11 2016
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MATHEMATICA
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Table[Floor[(Prime[j + 3] - Prime[j + 2])/(Prime[j + 1] - Prime[j])], {j, 1, 200}]
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PROG
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(MAGMA) [Floor((NthPrime(n+3)-NthPrime(n+2))/(NthPrime(n+ 1)- NthPrime(n))): n in [1..100]]; // Vincenzo Librandi, Aug 30 2016
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CROSSREFS
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Cf. A001223, A274263, A272863, A274225.
Sequence in context: A156072 A215788 A060990 * A165031 A286634 A099245
Adjacent sequences: A276306 A276307 A276308 * A276310 A276311 A276312
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KEYWORD
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nonn
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AUTHOR
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Andres Cicuttin, Aug 06 2016
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STATUS
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approved
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