A110303 - OEIS

%I #48 May 14 2021 03:51:51

%S 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,

%T 28,29,30,31,32,33,34,35,36,37,38,39,41,42,43,44,45,46,47,48,49,50,51,

%U 52,53,54,55,56,57,58,59,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75

%N Alternators.

%C 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.

%C 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

%H Michael De Vlieger, <a href="/A110303/b110303.txt">Table of n, a(n) for n = 1..10001</a> (adapted to offset by Michel Marcus)

%H 45th International Mathematical Olympiad (45th IMO), <a href="http://www.jstor.org/stable/30044168">Problem #6 and Solution</a>, Mathematics Magazine, 2005, Vol. 78, No. 3, pp. 247, 250-251.

%H The IMO Compendium, <a href="https://imomath.com/othercomp/I/Imo2004.pdf">Problem 6</a>, 45th IMO 2004.

%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.

%F Positive n, not congruent to 0 mod 20.

%F a(n + 19) = a(n) + 20. - _David A. Corneth_, Apr 13 2021

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

%t Select[Range[75], Mod[#, 20] != 0 &] (* _Michael De Vlieger_, Apr 13 2021 *)

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

%K base,easy,nonn

%O 1,2

%A _Walter Nissen_, Jul 18 2005

%E Offset 1 from _Michel Marcus_, May 12 2021