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Least m >= a(n-1)+n such that m!/(m-n)! is a multiple of a(n-1)!/(a(n-1)-(n-1))!.
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%I #9 Mar 13 2015 19:10:00

%S 1,3,6,10,16,52,1176,687378,236241851626,

%T 2197451321740962081109754668237130,

%U 2414396155710550624720051944524837499100253655538086242554948251375

%N Least m >= a(n-1)+n such that m!/(m-n)! is a multiple of a(n-1)!/(a(n-1)-(n-1))!.

%C a(1)=1. Main diagonal of triangle A094270.

%C If p is a prime factor of any of a(n-1)-(n-1) to a(n-1), then a(n) mod p < n.

%H Martin Fuller, <a href="/A094272/b094272.txt">Table of n, a(n) for n = 1..12</a>

%H Martin Fuller, <a href="/A094272/a094272.gp.txt">PARI/GP program for A094272</a>

%e Product of the terms of the 4th row = 7*8*9*10 = 5040. Product of the terms of the 5th row = 12*13*14*15*16 = 524160 = 104*5040.

%Y Cf. A094270, A094271, A094273, A094274.

%K nonn

%O 1,2

%A _Amarnath Murthy_, Apr 27 2004

%E Edited by _Martin Fuller_, Jun 13 2007