A110303 Alternators.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75

Offset: 1

Author

Walter Nissen, Jul 18 2005

Keywords

• base
• easy
• nonn

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.

This sequence is the answer to the 6th problem proposed the 2nd day by Iran during the 45th International Mathematical Olympiad, in Athens (Greece), 2004 (see links). - Bernard Schott, Apr 12 2021

Examples

			11 is an alternator and in the sequence because it has a multiple which is alternating. The least of these multiples is 121.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10001 (adapted to offset by Michel Marcus)
• 45th International Mathematical Olympiad (45th IMO), Problem #6 and Solution, Mathematics Magazine, 2005, Vol. 78, No. 3, pp. 247, 250-251.
• The IMO Compendium, Problem 6, 45th IMO 2004.
• Index to sequences related to Olympiads.

Crossrefs

Cf. A030141, A030142, A110304, A110305, A008602 (complement), A343335, A343336.

Programs

Mathematica

Select[Range[75], Mod[#, 20] != 0 &] (* Michael De Vlieger, Apr 13 2021 *)

Formula

Positive n, not congruent to 0 mod 20.

a(n + 19) = a(n) + 20. - David A. Corneth, Apr 13 2021

Extensions

Offset 1 from Michel Marcus, May 12 2021