%I #4 Dec 05 2013 19:56:38
%S 1,2,3,12,15,30,7,128,9,50,22,36,26,28,15,1232,85,414,19,500,84,462,
%T 69,120,200,364,81,532,29,810,31,10784,198,272,70,1476,148,266,39,
%U 2160,123,1890,43,308,855,368,94,336,49,1550,153,52,1484,648,220,3416,1026,3886
%N Least multiple of n such that the nth concatenation is a multiple of n. The (previous) (n1)th term is so chosen that the nth term exists.
%C If n+1 has any prime divisors other than 2 or 5, then a(n) is to be chosen so that the concatenation of first n terms is a multiple of those primes (including multiplicity).  _David Wasserman_, Dec 20 2005
%e a(5) = 15 because if it were 5 or 10, a(6) would not exist.
%K base,nonn
%O 1,2
%A _Amarnath Murthy_, Dec 06 2003
%E More terms from _David Wasserman_, Dec 20 2005
