

A075365


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


4



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
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OFFSET

1,2


LINKS



FORMULA

If p <= n < q, where p and q are consecutive primes, then a(n) = 2pn, unless n=10. Sketch of proof: a(n) >= 2pn, 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 > 2pn 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



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



