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A235400 a(1)=1. For n>1, assume a(n-1) has decimal expansion qrstuvwxyz (with at most ten digits, where some of q, r, s, ... may be zero). Then a(n) = sum 99...9 (with z 9's) + 88...8 (with y 8's) + 77...7 (with x 7's) + ... For example, if a(n-1) were 243, we would sum 77 + 8888 + 999 and get 9964.

%I A235400 #32 Jun 19 2017 07:54:44
%S A235400 1,9,999999999,4999999995,4000099995,2888988885,800099995,2900099996,
%T A235400 3000999996,3334333329,1000006195,889655560,368199996,2923444555,
%U A235400 111394795,1452329335,666792195,1449655625,334781195,948992565,1145513395,889070335,372323885,168990195
%N A235400 a(1)=1. For n>1, assume a(n-1) has decimal expansion qrstuvwxyz (with at most ten digits, where some of q, r, s, ... may be zero). Then a(n) = sum 99...9 (with z 9's) + 88...8 (with y 8's) + 77...7 (with x 7's) + ... For example, if a(n-1) were 243, we would sum 77 + 8888 + 999 and get 9964.
%C A235400 The sequence is periodic: a(470) = a(312) = 8005, etc.
%C A235400 See A235460 for the analogous sequence where we number the digits starting with 1 at the leftmost digit.
%D A235400 Rodolfo Kurchan, Problems with repdigits, to appear in Proceedings of Gathering for Gardner 2014 conference
%H A235400 Alois P. Heinz, <a href="/A235400/b235400.txt">Table of n, a(n) for n = 1..1000</a>
%e A235400 1 leads to one 9, then 9 leads to 999999999, then we have to sum 111111111 + 222222222 + ... + 999999999 = 4999999995, then we have to sum 1111 + 222222222 + ... + 888888888 + 99999 = 4000099995, and so on.
%t A235400 nxt[n_]:=Total[FromDigits/@(Table[#[[1]],{#[[2]]}]&/@Thread[{ Range[ 9,10-IntegerLength[n],-1],Reverse[IntegerDigits[n]]}])]; NestList[ nxt,1,30] (* _Harvey P. Dale_, Apr 24 2015 *)
%Y A235400 Cf. A235460.
%K A235400 nonn,base,nice
%O A235400 1,2
%A A235400 _Rodolfo Kurchan_, Jan 09 2014