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A247107 a(0) = 0, a(n) = previous term + repunit of length of previous term for n > 0.

%I A247107 #81 Sep 20 2024 06:03:42
%S A247107 0,1,2,3,4,5,6,7,8,9,10,21,32,43,54,65,76,87,98,109,220,331,442,553,
%T A247107 664,775,886,997,1108,2219,3330,4441,5552,6663,7774,8885,9996,11107,
%U A247107 22218,33329,44440,55551,66662,77773,88884,99995,111106,222217,333328,444439
%N A247107 a(0) = 0, a(n) = previous term + repunit of length of previous term for n > 0.
%H A247107 Alois P. Heinz, <a href="/A247107/b247107.txt">Table of n, a(n) for n = 0..1000</a>
%H A247107 <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,0,0,0,0,0,0,10,-20,10).
%F A247107 a(0) = 0, a(n) = a(n-1) + A002275(A055642(a(n-1))) for n>0.
%F A247107 From _Jon E. Schoenfield_, Nov 30 2014: (Start)
%F A247107 For n > 1, a(n) = a(n-1) + (10^(floor(log_10(a(n-1))) + 1) - 1) / 9.
%F A247107 For n > 0, a(n) = ((n-1) mod 9 + 1) * (10^D - 1) / 9 + 1 - D, where D = floor((n-1)/9) + 1. (There are exactly D digits in a(n).) (End)
%F A247107 G.f.: -(10*x^10-10*x^9+1)*x/((10*x^9-1)*(x-1)^2). - _Alois P. Heinz_, Nov 30 2014
%e A247107 98 = 9*10 + 8 -> 10*10 + 9 = 109.
%e A247107 109 = 1*100 + 0*10 + 9*1 -> 2*100 + 1*10 + 10*1 = 220.
%e A247107 a(42) = 44440 + (10^(floor(log_10(44440))+1)-1) / 9 = 44440 + (10^(4+1)-1) / 9 = 44440 + 99999/9 = 44440 + 11111 = 55551.
%t A247107 a[0]=0; a[n_]:=FromDigits[IntegerDigits[a[n-1]]+1]; Array[a,50,0] (* _Stefano Spezia_, Sep 19 2024 *)
%Y A247107 Similar to A158699, but with simpler rules.
%Y A247107 Cf. A002275, A055642.
%K A247107 nonn,base,easy
%O A247107 0,3
%A A247107 _Dhilan Lahoti_, Nov 30 2014