# Prime factors of a sequence of integers which differ by consecutive prime differences

[Edited following Gerhard's answer. I had forgotten to state a crucial assumption; my apologies for the confusion. I have reworded slightly to try to make things clearer.]

Let $p$ be the $k$-th odd prime. Let $n$ be an even integer. Let $p_{1}$, $\ldots$, $p_{k+1}$ be consecutive odd primes. Suppose that $p$ divides $n - p_{i}$ for all $1 \leq i \leq k+1$. For instance, take $p=3$, take $n=100$, take $p_{1}=31$, and take $p_{2}=37$. Is there then at least one $i$ such that $n - p_{i}$ has a prime factor strictly greater than $p$?

Anything anybody can say (proof or reference) is welcome.

It is easy to prove that this holds when $k=1$. Indeed, using the fact that $n$ is even, so that both $n - p_{1}$ and $n - p_{2}$ are odd, the claim is immediate unless both $n-p_{1}$ and $n-p_{2}$ are powers of 3. In that case, the difference between $p_{1}$ and $p_{2}$ is $2p_{1}$, and this is impossible by Bertrand's postulate.

There are also plenty of examples when $k=2$. The one with lowest $n$ is $n=1662$, and the primes $1627$, $1637$, $1657$. Experimental evidence suggests that the question can be answered in the affirmative in this case too.

I would guess that there are examples for all $k$, but that very large numbers will be required; I do not actually have an example for any $k > 2$. Anything that can be said about the existence of examples is welcome too.

• Please proofread: some of your $p$s are missing subscripts, and your example doesn't satisfy the property described. – Greg Martin Jun 12 '18 at 23:12
• I think you are still leaving stuff out. Your proof for k=1 does not hold, and while the specific examples I provide (now at this writing) do not hold either, the general idea behind my construction still holds. Gerhard "Will Wait For Some Stability" Paseman, 2018.06.13. – Gerhard Paseman Jun 13 '18 at 21:15
• By the way, n=550 is part of a counterexample for k=1. Gerhard "Because Primes Can Be Large" Paseman, 2018.06.13. – Gerhard Paseman Jun 13 '18 at 23:10

Consider $k=1$ (so $p=3$), $n=32$, $p_1=23$ and $p_2=29$. This is a counterexample to your statement.
I believe your example to consider is still out of sync with your stated problem. If you changed it to $n+ p_1,n+p_{k+1}$, it would line up.
Your result is not true for $k=1$. As an example , consider 24 and 36. All that is needed is to have a prime gap of 12 for your statement to fail, and indeed 199 , 211 is one such example.
For more counterexamples, consider smooth numbers whose relative differences are divisible by the $k+1$st primorial, then go hunting for consecutive primes whose differences match that. Much like searching for primes in arithmetic progressions, you will (eventually) be rewarded (with short examples).