%I #29 Oct 21 2022 21:59:23
%S 0,1,1,2,2,2,3,3,3,3,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,6,7,7,7,7,7,7,
%T 7,7,8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,7,7,8,8,
%U 7,7,8,8,7,13,7,14,7,13,13,15,15,13,15,15,6,17,16,19,8,10,10,22,10,23
%N a(0) = 0; for n > 0, let S = concatenation of a(0)..a(n-1); a(n) is the number of times the digit at a(n-1) digits back from the end of S appears in S.
%H Michael De Vlieger, <a href="/A357930/b357930.txt">Table of n, a(n) for n = 0..9999</a>
%H Michael De Vlieger, <a href="/A357930/a357930.txt">Table of n, a(n) for n = 0..10^5</a>
%H Michael De Vlieger, <a href="/A357930/a357930_4.png">Scalar scatterplot of a(n)</a>, n = 0..2^16, with points assigned a color function as to digit d that is a(n-1) digits back. Black indicates d=0, red d=1, ..., magenta d=9.
%H Michael De Vlieger, <a href="/A357930/a357930_5.png">Log-log scatterplot of a(n)</a>, n = 0..2^20, with points assigned a color function as to digit d that is a(n-1) digits back. Black indicates d=0, red d=1, ..., magenta d=9.
%H Scott R. Shannon, <a href="/A357930/a357930_2.png">Image of the first 5000000 terms</a>.
%e a(7) = 3 as a(6) = 3 and the string concatenation of a(0)..a(6) = "0112223", and the digit 3 digits back from the end of the string concatenation is 2, and 2 has appeared three times in the string.
%o (MATLAB)
%o function a = A357930( max_n )
%o a = 0; s = '0'; c = zeros(1,10); c(1) = 1;
%o for n = 2:max_n
%o k = c(s(end-a(n-1))-47); sk = num2str(k);
%o c(sk-47) = c(sk-47)+1; s = [s sk]; a(n) = k;
%o end
%o end % _Thomas Scheuerle_, Oct 21 2022
%Y Cf. A117707, A356348, A000217, A351753.
%K nonn,base,look
%O 0,4
%A _Scott R. Shannon_, Oct 21 2022