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%I #19 Feb 10 2016 07:25:35
%S 321,444,675,680,370,268,949,206,851,208,132,444,567,68,37,826,994,
%T 620,185,820,213,444,756,806,703,682,499,62,518,82,321,444,675,680,
%U 370,268,949,206,851,208,132,444,567,68,37,826,994,620,185,820,213,444,756,806
%N Three-digit numbers, starting with 321, such that when written in a table with three columns, there are arithmetic progressions modulo 10 down the diagonals, with steps of 1, 2 and 3 respectively.
%C Repeats after 30 terms. - _Chai Wah Wu_, Feb 08 2016
%D GCHQ Director's Christmas Puzzles for 2015.
%H GCHQ, <a href="http://s3-eu-west-1.amazonaws.com/puzzleinabucket/GCHQ_Puzzle_2015_-_Solutions.pdf">Solutions to Director's Christmas Puzzles for 2015</a>
%e The table begins
%e 3 2 1
%e 4 4 4
%e 6 7 5
%e 6 8 0
%e 3 7 0
%e 2 6 8
%e 9 4 9
%e 2 0 6
%e 8 5 1
%e 2 0 8
%e ...
%e and one can see the three arithmetic progressions down the diagonals: 3, 4, 5, 6, ...; 2, 4, 6, 8, ...; 1, 4, 7, 0, ... .
%o (Python)
%o from __future__ import division
%o A268517_list = [321]
%o for i in range(10**4):
%o a = A268517_list[-1]
%o A268517_list.append(((a+1+(2-i)%3) % 10)*100 + ((a//100+1+(-i)%3) % 10)*10 + ((a//10+1+(1-i)%3) % 10)) # _Chai Wah Wu_, Feb 08 2016
%K nonn,base
%O 0,1
%A _N. J. A. Sloane_, Feb 08 2016
%E More terms from _Chai Wah Wu_, Feb 08 2016
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