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%I #18 Dec 02 2020 17:51:30
%S 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 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 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 "Fibonacci in digits": start with a(0)=0, a(1)=1; repeatedly adjoin the digits of the sum of the next two terms.
%C 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 All digits appear infinitely often, although the sequence is not periodic.
%H Hakan Icoz, <a href="/A093086/b093086.txt">Table of n, a(n) for n = 0..20000</a>
%e After S_6 = {0,1,1,2,3,5,8} we have 5+8 = 13, so we get
%e S_7 = {0,1,1,2,3,5,8,1,3}. Then 8+1 = 9, so we get
%e S_8 = {0,1,1,2,3,5,8,1,3,9}. Then 1+3 = 4, so we get
%e S_9 = {0,1,1,2,3,5,8,1,3,9,4}, and so on.
%p 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
%Y Cf. A093087-A093098, A105967, A102085, A214365.
%K nonn,base,easy
%O 0,4
%A _Bodo Zinser_, Mar 20 2004
%E Edited by _N. J. A. Sloane_, Mar 20 2010
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