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

Author

Amarnath Murthy, Sep 20 2002

Keywords

• nonn

Examples

			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.
		

Links

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

Crossrefs

Cf. A075366, A075367, A075368.

Programs

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]]]]

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.

Extensions

Edited by Dean Hickerson, Oct 28 2002