Propagation of a $\{0,2\}$-block along the second column

For a sequence $a : \mathbb{N} \to \mathbb{N}$ write $(\Delta a)(n) = |a(n+1) - a(n)|$ and let $\Delta^j a$ be the $j$-fold iterate. This theorem is a propagation lemma for the second column of the difference triangle of $a$:

The mechanism is that $\{0,2\}$ is closed under absolute differences, $|0-0| = |2-2| = 0$ and $|0-2| = |2-0| = 2$, so a block of $0$s and $2$s occupying the indices $1, \dots, m+1$ of a row produces a block of $0$s and $2$s occupying the indices $1, \dots, m$ of the next row: the block shortens by one index per step but never leaves $\{0,2\}$. After $j$ steps the index $1$ is still covered, which is the assertion above.

Two features distinguish this from Odlyzko's propagation lemma for the leading column (Gilbreath.propagation), which concludes $(\Delta^j a)(0) = 1$ from $a(0) = 1$ together with the same kind of block. First, no hypothesis whatsoever is placed on $a(0)$: the entries of the triangle at indices $\ge 1$ are computed from entries at indices $\ge 1$ only, so the tail of the triangle evolves autonomously and the leading column is irrelevant to it. Second, the conclusion concerns the column of index $1$ rather than the column of index $0$.

The statement is about an arbitrary sequence of natural numbers, not specifically about the primes; applied to a row $d^k$ of the Gilbreath triangle, and combined with $\Delta^{j} d^{k} = d^{k+j}$, it says that a block of $0$s and $2$s at the indices $1, \dots, m+1$ of row $k$ forces $d^{k+m}(1) \in \{0, 2\}$.