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.

11, 110, 1101, 110111, 110111100101, 11011110010111010111111001000, 1101111001011101011111100100010011010101101111011001101111011111000010001111100010010100110101011

Offset: 1

Author

Thomas Anton, Oct 23 2018

Keywords

• nonn
• base

Comments

Each term is an initial segment of all of its successors.

There are always more 1's than 0's in a term.

The proportion of 0's or 1's in the n-th term approaches 1/2 as n approaches infinity.

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.

Examples

			a(1) = 11
The first 1 is replaced with 1, and the second 1 is replaced with 10 (two), so a(2) = 110 (1|10)
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)
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)
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)
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)
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..8

Crossrefs

Cf. A005150, A007651, A001387, A005151.

Programs

Mathematica

FromDigits /@ Nest[Append[#, Flatten[IntegerDigits[#, 2] & /@ Table[Count[#, Last@ #] &@ #[[1 ;; k]], {k, Length@ #}]] &[#[[-1]] ] ] &, {{1, 1}}, 6] (* Michael De Vlieger, Oct 23 2018 *)

PARI

eva(n) = subst(Pol(n), x, 10)
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
terms(n) = my(v=[1, 1], i=0); while(i < n, print1(eva(v), ", "); i++; v=replace(v))
/* Print initial 7 terms as follows: */
terms(7) \\ Felix Fröhlich, Oct 23 2018

Python

A320890_list = [11]
while len(A320890_list)<10:
    a0,a1,s = 0,0,''
    for d in str(A320890_list[-1]):
        if d == '0':
            a0 += 1
            s += bin(a0)[2:]
        else:
            a1 += 1
            s += bin(a1)[2:]
    A320890_list.append(int(s)) # Chai Wah Wu, Nov 30 2018