In this question on MSE, Enzo Creti asks for a prime number formed by concatenating the Mersenne numbers $2^n-1$ and $2^{n-1}-1$, for example, 40952047. For all residues modulo 7, he found primes except for the residue 6. This is somewhat surprising because the residue 1 occurs only with half frequency.

Is there any hidden structure forcing a non-trivial factor in the case of residue 6, or was it just "bad luck" that no prime was found despite an enormous search range?

I invite everyone to join in the search for a prime. I posted the necessary details on github.

The following vector contains all numbers n<=366800 leading to a prime

[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770]

Exponent $541456$ leads to another probable prime with residue 5 mod 7 and 325990 digits, but it need not be the next in increasing order. More details can be found on the github-site. Heuristically, for every k>=3 , the range [10^k..10^(k+1)] should contain 5.4 numbers leading to a prime, so the range [10^4..10^5] with 8 primes is "above average", whereas the range [10^3..10^4] is within the expectation. The sequences of exponents and associated primes are here: A301806 and here: A298613. I noticed these things: let's call ec(n)=$2^n-1$||$2^{n-1}-1$ where || denotes the concatenation in base 10.

ec(43*5) is prime. 5 (odd) is congruent to -8 mod 13. ec(43*1620) is prime. 1620 (even) is congruent to 8 mod 13. ec(43*2140) is prime. 2140 (even) is congruent to 8 mod 13. ec(43*12592) is prime. 12592 (even) is congruent to 8 mod 13. Moreover 43*5=41*5+10, 43*1620=41*1699+1, 43*2140=41*2244+16, 43*12592=41*13206+10. This implies that $\frac{43*5}{41}$, $\frac{43*1620}{41}$, $\frac{43*2140}{41}$, $\frac{43*12592}{41}$ will have a repeating term $\overline{02439}$ or in other words that they are of the form $41s+r$, where r is an integer in the set $(1,10,16,18,37)$