

A090512


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.


0



1, 2, 3, 12, 15, 30, 7, 128, 9, 50, 22, 36, 26, 28, 15, 1232, 85, 414, 19, 500, 84, 462, 69, 120, 200, 364, 81, 532, 29, 810, 31, 10784, 198, 272, 70, 1476, 148, 266, 39, 2160, 123, 1890, 43, 308, 855, 368, 94, 336, 49, 1550, 153, 52, 1484, 648, 220, 3416, 1026, 3886
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OFFSET

1,2


COMMENTS

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


LINKS



EXAMPLE

a(5) = 15 because if it were 5 or 10, a(6) would not exist.


CROSSREFS



KEYWORD

base,nonn


AUTHOR



EXTENSIONS



STATUS

approved



