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A320890 a(1) = 11. For all subsequent terms a(n), take a(n-1) and substitute for the k-th digit the binary number of times that digit has appeared in a(n-1), reading left to right from the 1st to k-th digit.

%I A320890 #27 Dec 01 2018 08:23:33
%S A320890 11,110,1101,110111,110111100101,11011110010111010111111001000,
%T A320890 1101111001011101011111100100010011010101101111011001101111011111000010001111100010010100110101011
%N A320890 a(1) = 11. For all subsequent terms a(n), take a(n-1) and substitute for the k-th digit the binary number of times that digit has appeared in a(n-1), reading left to right from the 1st to k-th digit.
%C A320890 Each term is an initial segment of all of its successors.
%C A320890 There are always more 1's than 0's in a term.
%C A320890 The proportion of 0's or 1's in the n-th term approaches 1/2 as n approaches infinity.
%C A320890 Starting with any binary integer apart from 0 or 1 and applying the same process to yield a sequence s(n), we have that, for a sufficiently large x, a(n) is always an initial segment of s(n+x). The constancy and uniqueness of the limiting behavior of initial segments in base 2 is unique among all bases, unless the tally system is considered as a degenerate case.
%H A320890 Chai Wah Wu, <a href="/A320890/b320890.txt">Table of n, a(n) for n = 1..8</a>
%e A320890 a(1) = 11
%e A320890 The first 1 is replaced with 1, and the second 1 is replaced with 10 (two), so a(2) = 110 (1|10)
%e A320890 The first 1 is replaced with 1, the second 1 with 10, and the first 0 with 1, so a(3) = 1101 (1|10|1)
%e A320890 The first 1 is replaced with 1, the second 1 with 10, the first 0 with 1, and the third 1 with 11 (three), so a(4) = 110111 (1|10|1|11)
%e A320890 The first 1 is replaced with 1, the second 1 with 10, the first 0 with 1, the third 1 with 11, the fourth 1 with 100, and the fifth 1 with 101, so a(5) = 110111100101 (1|10|1|11|100|101)
%e A320890 The first 1 is replaced with 1, the second 1 with 10, the first 0 with 1, the third 1 with 11, the fourth 1 with 100, the fifth 1 with 101, the sixth 1 with 110, the second 0 with 10, the third 0 with 11, the seventh 1 with 111, the fourth 0 with 100, and the eighth 1 with 1000, so a(6) = 11011110010111010111111001000 (1|10|1|11|100|101|110|10|11|111|100|1000)
%t A320890 FromDigits /@ Nest[Append[#, Flatten[IntegerDigits[#, 2] & /@ Table[Count[#, Last@ #] &@ #[[1 ;; k]], {k, Length@ #}]] &[#[[-1]] ] ] &, {{1, 1}}, 6] (* _Michael De Vlieger_, Oct 23 2018 *)
%o A320890 (PARI) eva(n) = subst(Pol(n), x, 10)
%o A320890 replace(v) = my(w=[], zeros=0, ones=0); for(k=1, #v, if(v[k]==0, zeros++; w=concat(w, binary(zeros))); if(v[k]==1, ones++; w=concat(w, binary(ones)))); w
%o A320890 terms(n) = my(v=[1, 1], i=0); while(i < n, print1(eva(v), ", "); i++; v=replace(v))
%o A320890 /* Print initial 7 terms as follows: */
%o A320890 terms(7) \\ _Felix Fröhlich_, Oct 23 2018
%o A320890 (Python)
%o A320890 A320890_list = [11]
%o A320890 while len(A320890_list)<10:
%o A320890     a0,a1,s = 0,0,''
%o A320890     for d in str(A320890_list[-1]):
%o A320890         if d == '0':
%o A320890             a0 += 1
%o A320890             s += bin(a0)[2:]
%o A320890         else:
%o A320890             a1 += 1
%o A320890             s += bin(a1)[2:]
%o A320890     A320890_list.append(int(s)) # _Chai Wah Wu_, Nov 30 2018
%Y A320890 Cf. A005150, A007651, A001387, A005151.
%K A320890 nonn,base
%O A320890 1,1
%A A320890 _Thomas Anton_, Oct 23 2018