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a(1) = 1; for n > 1 take the digits of a(n-1) one after the other and remove one occurrence of this digit from the digits of m = k*n (k > 0) if possible; a(n) is the smallest multiple m of n which has the least number of remaining digits (counted with multiplicity).
0

%I #5 Dec 05 2013 19:55:56

%S 1,2,3,4,5,6,7,8,9,90,99,96,26,28,285,528,85,18,19,20,21,22,23,24,25,

%T 52,27,28,29,90,93,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,

%U 49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71

%N a(1) = 1; for n > 1 take the digits of a(n-1) one after the other and remove one occurrence of this digit from the digits of m = k*n (k > 0) if possible; a(n) is the smallest multiple m of n which has the least number of remaining digits (counted with multiplicity).

%e For n = 10 we have a(n-1) = a(9) = 9; removing 9 from 9*n = 9*10 = 90 results in one remaining digit (i.e. 0) and for every smaller multiple of 10 (i.e. 10, 20, ..., 80) there are two remaining digits, so a(10) = 90. For n = 20 we have a(n-1) = a(19) = 19; for 1*n = 1*20 = 20 no digit is removed and two digits remain and there is no multiple of 20 which has less than two remaining digits (for 100 twice the digit 0), so a(20) = 20.

%K base,nonn

%O 1,2

%A _Amarnath Murthy_, Nov 21 2002

%E Edited, corrected and extended by _Klaus Brockhaus_, Dec 03 2002