Actually every six consecutive terms(after the second term)couldn’t be in this type:$ a,b,a,b,a,b$ Note that $ x_i$ is unic number that $ 0\leq{x_i}\leq9$ and $ x_i = 2x_{i + 6} - x_{i + 7} \mod 10$ because: $ x_i + x_{i + 1} + x_{i + 2} + x_{i + 3} + x_{i + 4} + x_{i + 5} = x_{i + 6} \mod 10 \\ x_{i + 1} + x_{i + 2} + x_{i + 3} + x_{i + 4} + x_{i + 5} + x_{i + 6} = x_{i + 7} \mod 10 \\ \Longrightarrow{ x_i = 2x_{i + 6} - x_{i + 7} \mod 10}$ Therefore if $ x_i,x_{i + 1},x_{i + 2},x_{i + 3},x_{i + 4},x_{i + 5},a,b,a,b,a,b$ appearance in the sequence,then: $ x_i = x_{i + 2} = x_{i + 4} = 2a - b \mod 10 \\ x_{i + 1} = x_{i + 3} = x_{i + 5} = 2b - a \mod 10$ therefore the six concecutive terms which appearance before $ a,b,a,b,a,b$ is also in this type:$ c,d,c,d,c,d$ and the six concecutive terms which appearance before $ c,d,c,d,c,d$ is also in the same type and so on … contradiction!(note that the start of sequence is: $ 1,0,1,0,1,0,3,5,0,9,8,5,...$

Actually, I think the above proof is flawed, because some of the indices are off by one. Working backwards from 0,1,0,1,0,1, I get the sequence would have to be $ \ldots, 2, 9, 2, 9, 9, 0, 1, 0, 1, 0, 1$, which doesn't match the pattern claimed. Here's my solution: Define the function $ Q(i) = 2 x_i + 4 x_{i+1} + 6 x_{i+2} + 8 x_{i+3} + 2 x_{i+5} \mod 10$. From the beginning of the sequence, compute $ Q(1) = 8$. Next, I claim that $ Q(i+1) = Q(i)$. To see this, we have: $ Q(i+1) = 2 x_{i+1} + 4 x_{i+2} + 6 x_{i+3} + 8 x_{i+4} + 2 x_{i+6} \mod 10$ $ = 2 x_{i+1} + 4 x_{i+2} + 6 x_{i+3} + 8 x_{i+4} + 2 (x_i + x_{i+1} + x_{i+2} + x_{i+3} + x_{i+4} + x_{i+5})$ $ \mod 10$ $ = 2 x_i + 4 x_{i+1} + 6 x_{i+2} + 8 x_{i+3} + 10 x_{i+4} + 2 x_{i+5} \mod 10$ $ = 2 x_i + 4 x_{i+1} + 6 x_{i+2} + 8 x_{i+3} + 2 x_{i+5} \mod 10$ $ = Q(i)$ We must therefore have $ Q(i) = 8$ for all $ i$. Now suppose that the sequence contained the terms "0, 1, 0, 1, 0, 1"; that is, there is some $ k$ so that $ (x_k, x_{k+1}, x_{k+2}, x_{k+3}, x_{k+4}, x_{k+5}) = (0,1,0,1,0,1)$. Then we compute $ Q(k) = 4$. But this is impossible; we conclude that "0, 1, 0, 1, 0, 1" never appears.

juggle5 wrote: Actually, I think the above proof is flawed, because some of the indices are off by one. Working backwards from 0,1,0,1,0,1, I get the sequence would have to be $ \ldots, 2, 9, 2, 9, 9, 0, 1, 0, 1, 0, 1$, which doesn't match the pattern claimed. yes...you are right...sorry