A Pattern Built by Flipping Its Copy

Start with a tile marked zero. Make a second tile, change its zero to one, and place it after the first. You now have 01. Copy those two tiles, flip both copied values to obtain 10, and append that pair. The row is now 0110. Repeating the same construction gives the rows above.

Tiles wrap onto another line when the screen is narrow, but the string still reads left to right, then from the end of one line to the start of the next. The line breaks are a drawing choice, not another mathematical operation. Reset returns to the single starting zero.

Every zero in the old half has a one at the corresponding position in the new half, and every one has a zero. Consequently each completed doubling contains exactly as many zeroes as ones. This statement concerns complete power-of-two rows. A shorter prefix such as 011 need not be balanced.

Turning off the flip produces 0, then 00, then 0000, so the control is a different sequence. Its simplicity helps isolate which operation creates the structure. The Thue-Morse claims on this page do not apply to that altered rule.

Number the positions from zero. Write a position in binary, count its ones and keep the parity of that count: an even count gives zero and an odd count gives one. Position six is 110 in binary, with two ones, so the bit there is zero. Position seven is 111, with three ones, so its bit is one.

The binary-parity table compares this direct calculation with the copied construction. These are two descriptions of one sequence, rather than two unrelated patterns that happen to look alike. The proof below explains why their agreement continues beyond the displayed rows.

Let $w_0=0$ and $w_{m+1}=w_m\overline{w_m}$, where the bar exchanges zero and one in every position and juxtaposition means concatenation. Each row starts with the previous row, so every position eventually has a fixed value. Those values define the infinite sequence $t$.

If $s_2(n)$ counts the ones in the binary expansion of $n$, then

The recurrence is equivalent to the substitution $0\mapsto 01$ and $1\mapsto 10$. The finite construction stops at 256 bits here for readability, not because the mathematical rule ends.

The first row agrees with the parity rule at position zero. Suppose a row of length $2^m$ agrees at all its positions. Every new position is $2^m+j$, with $0\le j<2^m$. Its binary expansion is that of $j$, padded to $m$ places, with an additional leading one. Its digit-sum parity is therefore the opposite of the parity at $j$. Appending the complement produces exactly those opposite bits. Induction proves the parity description at all positions.

Balance has a shorter argument: if the old row has $z$ zeroes and $o$ ones, its complement has $o$ zeroes and $z$ ones. Their concatenation has $z+o$ of each. Neither proof depends on how the tiles are coloured or how many fit on a screen.

The sequence certainly repeats short pieces: 01 and 10 occur many times. A periodic sequence requires one fixed positive shift to work at every position, or at every sufficiently late position for eventual periodicity. Testing a chosen shift can find a mismatch, but passing a finite test does not establish that the shift works forever.

For example, 0110 has only one comparison at shift three, and it agrees. In 01101001, positions two and five disagree, ruling out period three from the beginning. The period view states both the tested range and any witness rather than calling a short agreement a repeating infinite sequence.

The infinite Thue-Morse sequence is known to be overlap-free, a stronger theorem stated in the linked Allouche-Shallit survey. It follows that it has no three consecutive identical non-empty blocks, whereas any eventually periodic tail contains such a triple. That proves non-periodicity using a published theorem, not using this page's finite period checks. The deeper result is cited rather than presented as something established by the animation.