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A093086 "Fibonacci in digits": start with a(0)=0, a(1)=1; repeatedly adjoin the digits of the sum of the next two terms.

%I A093086 #22 Aug 18 2025 11:36:58
%S A093086 0,1,1,2,3,5,8,1,3,9,4,1,2,1,3,5,3,3,4,8,8,6,7,1,2,1,6,1,4,1,3,8,3,3,
%T A093086 7,7,5,5,4,1,1,1,1,6,1,0,1,4,1,2,1,0,9,5,2,2,2,7,7,1,1,5,5,3,3,1,9,1,
%U A093086 4,7,4,4,9,1,4,8,2,6,1,0,8,6,4,1,0,1,0,5,1,1,1,1,8,1,3,1,0,5,1,2,1,0,8,7,1,8
%N A093086 "Fibonacci in digits": start with a(0)=0, a(1)=1; repeatedly adjoin the digits of the sum of the next two terms.
%C A093086 Formally, define strings of digits S_i as follows. S_0={0}, S_1={0,1}. For n >= 1, let S_n={t_0, t_1, ..., t_z}. Then S_{n+1} is obtained by adjoining the digits of t_{n-1}+t_n to S_n. The sequence gives the limiting string S_oo.
%C A093086 All digits appear infinitely often, although the sequence is not periodic.
%H A093086 Hakan Icoz, <a href="/A093086/b093086.txt">Table of n, a(n) for n = 0..20000</a>
%e A093086 After S_6 = {0,1,1,2,3,5,8} we have 5+8 = 13, so we get
%e A093086 S_7 = {0,1,1,2,3,5,8,1,3}. Then 8+1 = 9, so we get
%e A093086 S_8 = {0,1,1,2,3,5,8,1,3,9}. Then 1+3 = 4, so we get
%e A093086 S_9 = {0,1,1,2,3,5,8,1,3,9,4}, and so on.
%p A093086 with(linalg): A:=matrix(1,2,[0,1]): for n from 1 to 100 do if A[1,n]+A[1,n+1]<10 then A:=concat(A,matrix(1,1,A[1,n]+A[1,n+1])) else A:=concat(A,matrix(1,2,[1,A[1,n]+A[1,n+1]-10])) fi od: matrix(A); # _Emeric Deutsch_, May 31 2005
%t A093086 Fold[Join[#, IntegerDigits[Total[#[[#2;; #2+1]]]]] &, {0, 1}, Range[100]] (* _Paolo Xausa_, Aug 18 2025 *)
%Y A093086 Cf. A093087-A093098, A105967, A102085, A214365.
%K A093086 nonn,base,easy
%O A093086 0,4
%A A093086 _Bodo Zinser_, Mar 20 2004
%E A093086 Edited by _N. J. A. Sloane_, Mar 20 2010