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%I #8 Oct 08 2012 15:52:18
%S 1,2,4,8,16,32,63,126,252,504,1008,2016,4032,8001,16002,32004,63000,
%T 126000,252000,504000,1008000,2016000,4032000,8001000,16002000,
%U 32004000,63000000,126000000,252000000,504000000,1008000000,2016000000
%N a(n)=2a(n-1) but when sum of digits of 2a(n-1) is greater than 9 take a(n) = largest number < 2a(n-1) which has sum of digits = 9.
%e a(12)=4032; 2*4032 = 8064; digit sum is 18; decrease until we get a number with digit sum 9, which is 8001; so a(13) = 8001. - From _N. J. A. Sloane_, Oct 08 2012
%Y Cf. A062259.
%K nonn,base
%O 0,2
%A _Paul Curtz_, Jun 15 2008
%E Keyword base added by _Reinhard Zumkeller_, Feb 25 2010
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