12th Irish 1999 Problem A5 The sequence u 0 , u 1 , u 2 , ... is defined as follows. u 0 = 0, u 1 = 1, and u n+1 is the smallest integer > u n such that there is no arithmetic progression u i , u j , u n+1 with i < j < n+1. Find u 100 .
981 Solution The sequence is just the numbers which have no 2 when written in base 3. See USAMO 95/1 . If we write the numbers in base 3 and interpret them in base 2, then we just have the sequence of all the non- negative integers. We have 100 = 1100100 2 . So u 100 = 1100100 3 = 981. 12th Irish 1999 © John Scholes jscholes@kalva.demon.co.uk 22 Dec 2003 Last updated/corrected 22 Dec 03