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Search: id:A075365
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| A075365 |
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Smallest k such that (n+1)(n+2)...(n+k) is divisible by the product of all the primes up to n. |
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4
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| 0, 2, 3, 2, 5, 4, 7, 6, 5, 5, 11, 10, 13, 12, 11, 10, 17, 16, 19, 18, 17, 16, 23, 22, 21, 20, 19, 18, 29, 28, 31, 30, 29, 28, 27, 26, 37, 36, 35, 34, 41, 40, 43, 42, 41, 40, 47, 46, 45, 44, 43, 42, 53, 52, 51, 50, 49, 48, 59, 58, 61, 60, 59, 58, 57, 56, 67, 66, 65, 64, 71, 70
(list; graph; listen)
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OFFSET
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1,2
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LINKS
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T. D. Noe, Table of n, a(n) for n=1..1000
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FORMULA
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If p <= n < q, where p and q are consecutive primes, then a(n) = 2p-n, unless n=10. Sketch of proof: a(n) >= 2p-n, to make (n+1)...(n+a(n)) divisible by p. If r is a prime less than p and r does not divide (n+1)...(2p), then r > 2p-n and 2r <= n, so 4p < 3n < 3q. But q/p is known to be < 4/3 for all primes p >= 11.
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EXAMPLE
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a(6) = 4 as (6+1)*(6+2)*(6+3)*(6+4) is divisible by 2*3*5 but (6+1)*(6+2)*(6+3) is not.
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MATHEMATICA
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a[n_] := Module[{div, k, pr}, div=Times@@Prime/@Range[PrimePi[n]]; For[k=0; pr=1, True, k++; pr*=n+k, If[Mod[pr, div]==0, Return[k]]]]
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CROSSREFS
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Cf. A075366, A075367, A075368.
Sequence in context: A055573 A072969 A139712 this_sequence A075274 A135737 A125179
Adjacent sequences: A075362 A075363 A075364 this_sequence A075366 A075367 A075368
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KEYWORD
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nonn
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AUTHOR
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Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Sep 20 2002
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EXTENSIONS
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Edited by Dean Hickerson (dean(AT)math.ucdavis.edu), Oct 28 2002
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