Saudi Arabia booklet 2024

The answer is NO. Consider all words of length 13 and continue them periodically in both directions to create the sequences. For details, take $x_1{x}_2\dots x_{13}$ as some original word of length 13 then add $x_1,x_2,x_3,\dots$ to the right and add $x_{13},x_{12},x_{11},\dots$ to the left will result: $$\dots x_{11}{x}_{12}{x}_{13}(x_1{x}_2\dots x_{13})x_1{x}_2{x}_3\dots$$ There are two identical original words as $AAA\dots A$ and $BBB\dots B$ will generate the unique sequences. For the other words, each of them will generate the same sequences with another 12 words as follows: $$(x_2{x}_3{x}_4\dots x_{13}{x}_1),(x_3{x}_4{x}_5\dots x_{13}{x}_1{x}_2),\dots ,(x_{13}{x}_1{x}_2\dots x_{12})$$ So in total, the number of distinct sequences is $$\frac{2^{13}-2}{13}+2>600.$$ By the definition of the sequences, one can check that any word of length at least 13 may appear only in 1 sequence. Each word of length $n<13$ may appear in at most $2^{13-n}$ sequences, since we fix $n$ letters and choose another $13-n$ other letters with $2^{13-n}$ ways to form a word of length 13 (some of them may coincide). And note that $$(2^8+2^7+\cdots +2^0)+17=2^9+16<600$$ so we will have available sequences.