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A090512
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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.
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0
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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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 (wasserma(AT)spawar.navy.mil), Dec 20 2005
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EXAMPLE
| a(5) = 15 because if it were 5 or 10, a(6) would not exist.
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CROSSREFS
| Sequence in context: A173079 A173903 A154785 * A076175 A067780 A124486
Adjacent sequences: A090509 A090510 A090511 * A090513 A090514 A090515
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KEYWORD
| base,nonn
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AUTHOR
| Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Dec 06 2003
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EXTENSIONS
| More terms from David Wasserman (wasserma(AT)spawar.navy.mil), Dec 20 2005
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