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%I #11 Aug 17 2015 11:24:45
%S 2,11,29,2999
%N a(1) = 2; a(n+1) = smallest number > a(n) whose sum of digits is a(n).
%C a(5) = 3*10^333-1 = 2999...999 with 333 9's and contains 334 digits.
%C Surprisingly enough the first four terms are all primes and match those of A062802, but a(5) is divisible by 65033 and is different from A062802(5).
%C Sequences with other seeds: 3,12,129,399999999999999,...; 4,13,139,4999999999999999,...; 5,14,149,59999999999999999.
%F For n>=3, a(n) = (a(n-1) mod 9 + 1)*10^floor(a(n-1)/9) - 1. - _Max Alekseyev_, Aug 13 2015
%F For n>=3, a(n) = 3*10^b(n) - 1, where b(3)=1 and for n>=4, b(n)=(10^b(n-1)-1)/3. In other words, decimal representation of b(n) is formed by digit 3 repeated b(n-1) times. - _Max Alekseyev_, Aug 13 2015
%Y Differs from A062802 starting at a(5).
%K nonn,base
%O 1,1
%A _Zak Seidov_, Feb 17 2005
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