Does $\exists\ n$ such that the first $2n$ digits of Thue Morse, $X_{2n},$ is the concatenated sequence $X_n X_n?$ If not then why not?

Question

Background:

The Thue–Morse sequence is the binary sequence (an infinite sequence of $0$s and $1$s) obtained by starting with $0$ and successively appending the Boolean complement of the sequence obtained thus far. The first few steps of this procedure yield the strings $0$ then $01, 0110, 01101001, 0110100110010110,$ and so on, which are prefixes of the Thue–Morse sequence. The full sequence begins:

$01101001100101101001011001101001....$

For each $n\in\mathbb{N},$ let $X_n$ be the (string of) first $n$ digits of the Thue-Morse sequence.

Does $\ \exists\ n\ $ such that the (string of) first $2n$ digits of T-M is the concatenated sequence/string $X_n X_n\ ?$

If yes, then obviously by definition of T-M, $n\neq 2^k,\ k\in\mathbb{N}$. I also know that the T-M sequence contains no cubes, i.e. concatenated contiguous subsequences of the form $XXX.$ Other than this observation, I don't see how to make progress on the problem. I have checked this up until $n=2^{13}\approx 8000$ and I have found no such $n,$ which makes me suspect the answer is "no." But the fact that the T-M sequence does contain "squares" in general (not necessarily from the first digit like in the question above), does make me wonder why the answer to the above question could be no, if that is in fact the answer.

Below is some code. I found the code that generates the T-M sequence with a quick google search, and the rest of the code - which is trying to find an example to the problem above - I wrote myself.

# Python3 Program to find nth term of
# Thue-Morse sequence.

# Return the complement of the
# binary string.
def complement(s):
    comps = "";

    # finding the complement
    # of the string.
    for i in range(len(s)): 

        # if character is 0, append 1
        if (s[i] == '0'):
            comps += '1';

        # if character is 1, append 0.
        else:
            comps += '0';

    return comps;

# Return the nth term of 
# Thue-Morse sequence.
def nthTerm(n):

    # Initializing the string to 0
    s = "0";

    # Running the loop for n - 1 time.
    for i in range(1, n): 

        # appending the complement of 
        # the string to the string.
        s += complement(s);
 
    return s;

# Driver Code


n = 15;
print(nthTerm(n))
tM_List = nthTerm(n)
print(tM_List)

for i in range(2**(n-2)):
    #print(i, list(tM_List[:i+1])),
    #print(i, list(tM_List[i+1:2*i+2])),
    if list(tM_List[:i+1]) == list(tM_List[i+1:2*i+2]):
        print('success')

Answer 1

It cannot happen, for reasons I discussed in my comment. Let $T_n$ be the $n$th bit of the Thue-Morse sequence $T$.

First, we have the self-similarity property: that replacing $0$ with $01$ and $1$ with $10$ in $T$ will produce $T$ once more. Looking at it in reverse, it also implies that $T$ is a sequence of $01$s and $10$s, the order of which is once again essentially $T$. Also note that this implies $T_{2k-1}\neq T_{2k}$ for any $k$. $(\ast)$

Now, if $n$ is even, we can rewrite as $X_{4n} = X_{2n}X_{2n}$, by reversing this similarity property, we then see that $X_{2n} = X_n X_n.$ By successively eliminating factors of $2$ from $n$, we can assume without loss of generality that $n$ is odd.

As you have verified, and as can be clearly seen, $n \ge 3$.

Since $X_{2n} = X_nX_n$ and $T$ begins $01$, we have $T_{n+1}=T_1=0$ and $T_{n+2}=T_2=1$. Now $(\ast)$ with $2k=n+3$ implies $T_{n+3}\neq T_{n+2}$, so $T_{n+3}=0$. Since $n\ge 3$, we have $T_3=0$.

In other words, $T$ begins $010$. But, of course, this is simply not how the Thue-Morse sequence begins. It begins $011$, which means we have our contradiction. No such $n$ exists.