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Let $d_k$ be the $k^{th}$ difference sequence of the primes; that is, $$d_k = \sum_{i=0}^{k} (-1)^i {k \choose i} P_{k+1-i},$$ where $P_i$ denotes the $i$-th prime number. Let $(s_n)$ be the increasing sequence of all $k$ satisfying $d_k > 0.$ Is $s_{n+1} - s_n \in \{1,2,3\}$ for every $n$?

Here's some initial evidence: $$d_1 = 3 - 2 = 1$$ $$d_2 = (5-3) - (3-2) = 1$$ $$d_3 = 7 - 3\cdot5 + 3\cdot3 - 2 = -1$$ $$d_4 = 11 - 4\cdot7 + 6\cdot5 - 4\cdot3 + 2 = 3,$$ and so on, so that $(s_n) = (1,2,4,6,8,10,12,14,15,17,19,...).$ For $n$ up to $2000$, my computer finds $s_{n+1} - s_n = 1$ for $15$ values of $n$$; $ $s_{n+1} - s_n = 2$ for $976$ values; and $s_{n+1} - s_n = 3$ for $10$ values.

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  • $\begingroup$ I may be wrong, but I think a positive answer to your question could be a first step towards a solution of the so-called Proth-Gilbreath conjecture. $\endgroup$ May 25, 2015 at 19:17
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    $\begingroup$ There is either a sign disagreement or some notational ambiguity. Where do the minus signs come from? what is P_{k+1-r}? What is r? $\endgroup$ May 25, 2015 at 20:30
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    $\begingroup$ There is some discussion of $d_k$ at oeis.org/A007442 $\endgroup$ May 25, 2015 at 23:21

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