<s>[INST] Starting with $1,9,9,3$ the sequence continues as the last digit of the sum of preceding 4 terms. Will subsequence $7, 3, 6, 7$ ever appear in it? Formally, $a_n = (a_{n-1} + a_{n-2} + a_{n-3} + a_{n-4})$ mod $10$.
Is there a mathematical way of knowing whether $7, 3, 6, 7$ will ever occur in the sequence?
One idea I was delving upon was to prove the sequence is analogous to a decimal expansion of an irrational number, that way any arbitrary four numbers would appear.
 [/INST] If $X_n = \pmatrix{a_n\cr a_{n+1}\cr a_{n+2}\cr a_{n+3}\cr}$, we have $X_{n+1} = M X_n \mod 10$ where $M = \pmatrix{0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr 1 & 1 & 1 & 1\cr}$.  That matrix has characteristic polynomial $P(\lambda) = \lambda^4 - \lambda^3 - \lambda^2 - \lambda - 1$, which is irreducible both mod $2$ and mod $5$.  The period of
$M$ mod $p$ must be a divisor of $p^4-1$.  We find the period mod $2$ is $5$ and the period mod $5$ is $312$.  Therefore the period mod $10$ is $5 \cdot 312 = 1560$.
With $X_0 = \pmatrix{1\cr 9\cr 9\cr 3\cr}$ I find $X_{1552} = \pmatrix{7\cr 3\cr 6\cr 7\cr}$.
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