%I #4 Mar 31 2012 20:17:55

%S 1,9,9,3,9,9,3,9,9,1,99,9,9,9,9,9,9,9,6,9,54,6,9,9,6,9,9,4,9,9,33,36,

%T 99,3,9,9,3,9,9,3,27,27,24,99,9,5,9,9,9,9,63,6,27,45,2,9,9,9,9,9,33,

%U 18,18,5,18,99,3,9,9,3,72,45,7,18,18,15,99,9,9,9,63,5,27,18,6,27,36,15,9,9,11

%N a(n) = smallest k >= 1 such that k and k*n have the same digit sum.

%H D. W. Wilson, <a href="/A180012/b180012.txt">Table of n, a(n) for n=1..10000</a>

%e For all 1 < k < 99, k and 11*k have distinct digit sum. 99 and 99*11 = 1089 have the same digit sum 18. Therefore a(11) = 99.

%K base,easy,nonn

%O 1,2

%A _David W. Wilson_, Aug 06 2010