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Let b(1)=1, b(2)=2, b(n) = sum of digits of b(1)+b(2)+b(3)+...+b(n-1), sequence gives values of n such that b(n)=3.
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%I #25 Jun 29 2014 22:27:25

%S 3,5,7,9,17,19,27,29,87,95,97,159,591,599,601,663,1143,4609,4617,4619,

%T 4681,5161,8993,13165,38277,38279,38341,38821,42653,46825,75043,79223,

%U 327015,327023,327025,327087,327567,331399,335571,363789,367969,642981,647153,2847029,2847031,2847093,2847573

%N Let b(1)=1, b(2)=2, b(n) = sum of digits of b(1)+b(2)+b(3)+...+b(n-1), sequence gives values of n such that b(n)=3.

%C The {b(n)} sequence is A084228. - _N. J. A. Sloane_, Jun 26 2014

%C Note that b(k)==0 (mod 3) for n>2.

%H Robert G. Wilson v, <a href="/A084229/b084229.txt">Table of n, a(n) for n = 1..107</a>

%F Conjecture : a(n)/n^3 is bounded.

%t k = 3; lst = {}; a = 3; While[k < 100000001, b = a + Total@ IntegerDigits@ a; If[b == a + 3, AppendTo[lst, k]; Print@ k]; a = b; k++]; lst (* _Robert G. Wilson v_, Jun 27 2014 *)

%o (PARI) // sumdig(n)=sum(k=0,ceil(log(n)/log(10)),floor(n/10^k)%10) // an=vector(10000); a(n)=if(n<0,0,an[n]) // an[1]=1; an[2]=2; for(n=3,5300,an[n]=sumdig(sum(k=1,n-1,a(k)))) // for(n=1,5300,if(a(n)==3,print1(n,",")))

%Y Cf. A065075, A084228.

%K base,nonn

%O 1,1

%A _Benoit Cloitre_, Jun 21 2003

%E a(23) onward from _Robert G. Wilson v_, Jun 27 2014