Gilbreath's conjecture says that for every positive integer $n$, if we write out the first $n$ primes
$2,3,5,7,11,13,\ldots,p_n$
take the differences between consecutive terms
$1,2,2,4,2,\ldots,p_n-p_{n-1}$
take the absolute values of the differences between these terms
$1,0,2,2,\ldots,\lvert(p_n-p_{n-1})-(p_{n-1}-p_{n-2})\rvert$
and then repeat the last step until only one term remains, then after the initial $2$, every term in the first column is $1$.
I considered a similar question, where instead of writing the absolute difference between each term and the following term, we consider the difference between that term and the next term greater than or equal to it.
For $n=22$,
$2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79\\1,2,2,4,2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6\\1,0,2,0,2,0,2,2,0,4,0,0,2,2,0,0,4,0,2,4\\1,2,0,2,0,2,0,2,4,0,0,2,0,2,0,4,0,2,2\\1,0,2,0,2,0,2,2,0,0,2,0,2,2,4\\1,2,0,2,0,2,0,0,0,2,0,2,0,2\\\vdots$
The question is whether or not the first column (after $2$) consists only of $1$s.
Edit: I wrote a short program following this procedure and used it to test the conjecture for the first $10000$ primes, from which we get $2128$ rows (after the initial row), all beginning with $1$. However, it's clear that this will be the case by the $17$th row, which consists only of the initial $1$ followed by $0$s and $2$s.
Edit 2: Upon reflection, it could take considerable work just to prove the existence of this sequence, as while it's well known that gaps between prime are unbounded, this certainly doesn't imply that for every positive integer $k$ there exists a prime gap $a$ such that the next prime gap $b$ with $b\geq a$ is such that $b-a\geq k$, and this would only guarantee the existence of the $4$th row.