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%I #22 Dec 05 2023 09:03:28
%S 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,
%T 19,18,29,28,31,30,29,28,27,26,37,36,35,34,41,40,43,42,41,40,47,46,45,
%U 44,43,42,53,52,51,50,49,48,59,58,61,60,59,58,57,56,67,66,65,64,71,70
%N Smallest k such that (n+1)(n+2)...(n+k) is divisible by the product of all the primes up to n.
%H T. D. Noe, <a href="/A075365/b075365.txt">Table of n, a(n) for n=1..1000</a>
%F 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.
%e 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.
%t 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]]]]
%Y Cf. A075366, A075367, A075368.
%K nonn
%O 1,2
%A _Amarnath Murthy_, Sep 20 2002
%E Edited by _Dean Hickerson_, Oct 28 2002
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