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A164798 a(n) = the smallest integer >= n such that a(n)!/(n-1)! is divisible by every prime from 2 to the largest prime divisor of a(n)!/(n-1)!. (a(1)=1.)

%I A164798 #7 Mar 11 2014 01:32:46
%S A164798 1,2,4,4,6,6,10,8,10,14,15,12,22,15,16,16,26,18,34,21,38,38,38,24,46,
%T A164798 46,46,46,46,30,58,32,62,62,36,36,62,74,74,74,74,82,82,86,86,86,86,48,
%U A164798 94,94,94,94,94,54,106,106,106,106,106,60,118,122,66,64,122,122,122,134
%N A164798 a(n) = the smallest integer >= n such that a(n)!/(n-1)! is divisible by every prime from 2 to the largest prime divisor of a(n)!/(n-1)!. (a(1)=1.)
%C A164798 a(n) = A164799(n) + n -1.
%e A164798 Consider the products of consecutive integers, m!/9!, m >= 10. First, 10 is divisible by 2 and 5, but there is a prime gap since 3 is missing from the factorization. 10*11 is divisible by 2, 5, and 11, but 3 and 7 are missing. 10*11*12 is divisible by 2, 3, 5, and 11, but 7 is missing. 10*11*12*13 is divisible by all primes up to 13, except 7. But 10*11*12*13*14 is indeed divisible by every prime from 2 to 13. So a(10) = 14.
%Y A164798 Cf. A164799
%K A164798 nonn
%O A164798 1,2
%A A164798 _Leroy Quet_, Aug 26 2009
%E A164798 Terms beyond a(13) from _R. J. Mathar_, Feb 27 2010