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A058989
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Largest number of consecutive integers such that each is divisible by a prime <= the n-th prime.
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2
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1, 3, 5, 9, 13, 21, 25, 33, 39, 45, 57, 65, 73, 89, 99, 105, 117, 131, 151, 173, 189, 199, 215, 233, 257, 263, 281, 299, 311, 329, 353, 377, 387, 413, 431, 449, 475, 491, 509, 537, 549, 573, 599, 615, 641, 657, 685, 717, 741
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OFFSET
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1,2
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COMMENTS
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Marty Weissman conjectured that a(n)=2q-1, where q is the largest prime smaller than the n-th prime. The conjecture holds for the first few terms, but then a(n) is larger than 2q-1. Phil Carmody proved a(n)>=2q-1. Terms were calculated by Weissman, Carmody and McCranie.
a(n)=A048670(n) - 1, A049300(n) is the smallest value of the mentioned consecutive integers. - Reinhard Zumkeller, Jun 14 2003
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REFERENCES
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Dickson, L. E., History of the Theory of Numbers, Vol. 1, p. 439, Chelsea, 1952.
H. Iwaniec, On the error term in the linear sieve, Acta. Arith. 19 (1971), pp. 1-30.
J. D. Laison and M. Schick, "Seeing dots: visibility of lattice points", Mathematics Magazine, Vol. 80 (2007), pp. 274-282. See page 281 reference 13.
János Pintz, Very large gaps between consecutive primes, Journal of Number Theory 63 (1997), pp. 286-301.
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LINKS
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Table of n, a(n) for n=1..49.
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FORMULA
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Iwaniec proved that a(n) << n^2 log^2 n. - Charles R Greathouse IV, Sep 08 2012
a(n) >= (2e^gamma + o(1)) n log^2 n log log log n / (log log n)^2, see A048670. - Charles R Greathouse IV, Sep 08 2012
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EXAMPLE
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The 4th prime is 7. Nine is the maximum number of consecutive integers such that each is divisible by 2, 3, 5 or 7. (Example: 2 through 10) So a(4)=9.
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CROSSREFS
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This sequence is the same as A048670 - 1. See that entry for additional information.
Sequence in context: A007042 A178415 A076274 * A049691 A206297 A136252
Adjacent sequences: A058986 A058987 A058988 * A058990 A058991 A058992
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KEYWORD
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nice,nonn
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AUTHOR
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Jud McCranie, Jan 16 2001
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EXTENSIONS
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Laison and Schick reference from Parthasarathy Nambi, Oct 19 2007
More terms from Max Alekseyev, Feb 07 2008
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STATUS
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approved
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