|
|
A110304
|
|
Least alternating multiple of alternators.
|
|
5
|
|
|
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 121, 12, 52, 14, 30, 16, 34, 18, 38, 0, 21, 418, 23, 72, 25, 52, 27, 56, 29, 30, 341, 32, 165, 34, 70, 36, 74, 38, 78, 0, 41, 210, 43, 616, 45, 92, 47, 96, 49, 50, 561, 52, 212, 54, 165, 56, 456, 58, 236, 0, 61, 434, 63, 256, 65, 858, 67, 272, 69
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
An alternating integer is a positive integer for which, in base-10, the parity of its digits alternates. E.g., 121 is alternating because its consecutive digits are odd-even-odd, 1 being odd and 2 even. Of course, 1234567890 is also alternating. An alternator is a positive integer which has a multiple which is alternating.
For n congruent to 0 mod 20, a(n) is shown as zero to indicate that n is not an alternator.
|
|
LINKS
|
45th International Mathematical Olympiad (45th IMO), Problem #6 and Solution, Mathematics Magazine, 2005, Vol. 78, No. 3, pp. 247, 250-251.
|
|
EXAMPLE
|
a(11) = 121 because 121 is the least multiple of 11 which is alternating.
|
|
CROSSREFS
|
|
|
KEYWORD
|
base,easy,nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|