A075365 Smallest k such that (n+1)(n+2)...(n+k) is divisible by the product of all the primes up to n.

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

Offset

1,2

Links

T. D. Noe, Table of n, a(n) for n=1..1000

Formula

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.

Example

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.

Mathematica

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]]]]

Crossrefs

Cf. A075366, A075367, A075368.

Sequence in context: A072969 A139712 A175856 * A268466 A075274 A331309

Adjacent sequences: A075362 A075363 A075364 * A075366 A075367 A075368

Keyword

nonn

Author

Amarnath Murthy, Sep 20 2002

Extensions

Edited by Dean Hickerson, Oct 28 2002

Status

approved