Where $x = A001203$, $Pi = A002486$, $A002485$
$Pi(n) = Pi(n-2) + x\times [Pi(n-1)]$ for all $Pi > n+1 $
Hypothesis: This relation evaluates true for all $A002486$ and $A002485$.
Lemma: All "randomness" in $Pi$ decimal expansion is generated by the term "$x$" in the above relation.
See $x =$ https://oeis.org/A001203 See $Pi(n)$ = https://oeis.org/A002485
Per a commenter, an attempt at clarification.
The numerator and denominator convergents for the fractional expansion of pi, as with 22/7, 333/106, 355/113, 103993/33102, where [22, 333, 355, 103993, ...] is part of A002486 and [7, 106, 113, 33102, ...] are part of A002485. Starting with Pi(0) = 333 or 106, Pi(1) =Pi(n-1)+[x*(Pi(n-2)]= 333+(1*22) = 355, where x(n) = [1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...], or A001203 less the first two terms. So, 292*355+333=103993, thus defining the randomness in the decimal expansion as being limited to x (the truncated sequence taken from A001203).
This is just the standard recursive rule for continued fractions. If $a = a_0 + 1/(a_1 + 1/(a_2+ \ldots))$ is a continued fraction with coefficients $(a_n)$ and convergents $(p_n/q_n) = a_0 + 1/(a_1+ 1/(a_2+\ldots + 1/a_n))$, then $$ p_n = a_n p_{n-1} + p_{n-2} \text{ and } q_n = a_n q_{n-1} + q_{n-2} $$ for all $n \ge 2$. See e.g. Wikipedia.