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A078295 Smallest multiple of n whose digits can be re-arranged to be a substring of the cyclic concatenation 1234567890123435678901234... 0
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 132, 12, 65, 56, 45, 32, 102, 54, 76, 120, 21, 132, 23, 120, 675, 78, 54, 56, 87, 120, 465, 32, 132, 34, 210, 324, 2109, 76, 78, 120, 123, 210, 43, 132, 45, 3542, 423, 432, 98, 123450, 102, 312, 901, 54, 1320, 56, 342, 3654, 354, 120 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
Numbers n such that n or some digit permutation of n is a substring of the cyclic concatenation of digits 1,2,3,4,5,6,7,8,9,0,1,2...
If a(n) contains more 1's than it contains 0's and n>10, then a(10n)=10a(n) - Sam Alexander, Oct 19 2003
LINKS
FORMULA
If m>1, a(10^m) is obtained by starting with a string of m-1 1's, then concatenating a string of m-1 2's, then a string of m-1 3's, ..., a string of m-1 9's, then finally a string of m 0's. So a(1000)=112233445566778899000. Also, to get a(5*10^m) from a(10^m), m>1, just take the rightmost 5 and move it in front of the 9's, so a(5000)=112233445667788995000 - Sam Alexander, Oct 19 2003
EXAMPLE
a(36) = 324, digits can be rearranged as 2,3,4. a(35) = 210, (0,1,2.)
CROSSREFS
Sequence in context: A044912 A061378 A004872 * A326302 A004883 A004894
KEYWORD
base,nonn
AUTHOR
Amarnath Murthy, Nov 26 2002
EXTENSIONS
More terms from Sam Alexander, Oct 19 2003
STATUS
approved

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Last modified April 25 12:33 EDT 2024. Contains 371969 sequences. (Running on oeis4.)