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 A090512 Least multiple of n such that the n-th concatenation is a multiple of n. The (previous) (n-1)-th term is so chosen that the n-th term exists. 0

%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 n-th concatenation is a multiple of n. The (previous) (n-1)-th term is so chosen that the n-th 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

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Last modified July 14 02:38 EDT 2024. Contains 374291 sequences. (Running on oeis4.)