Hacker News Reader

Comment by oefrha | Hacker News Reader

This is just a weaker version of the twin prime conjecture. You’re asking for primes that differ by 2^k instead of 2. Last I heard the best result was from Polymath 8a.

http://michaelnielsen.org/polymath1/index.php?title=Bounded_...

Edit: Actually I was wrong, differing by a binary digit is much stronger than a difference of 2^k. But that’s a starting point.

View on HN

Neat. Thanks for tying it in to something real. Makes sense now that you phrase it that way. Not something I've ever looked into much.

Actually I replied without thinking too much, sorry... Your question is (probably a fair bit) stronger than the one I related to. Edited.

Big side note, but what you just exhibited is one of my favorite things about this community: admitting being wrong.

It wasn’t really until the workplace where I encountered level-headed people that would just shrug their shoulders and say “You are right. I was wrong” and then move on.

It’s so simple, but is amazingly powerful.

To take it a little further, what I was even more specifically interested in was a sequence of primes following the principle of a gray code [1] where each subsequent prime would differ from the previous prime by precisely one bit.

So basically, what is the longest sequence of primes you could find following that pattern?

[1] https://en.wikipedia.org/wiki/Gray_code

I think you were right the first time.

Nah, 3 (0b11) and 5 (0b101) differ by 2^1 but differ by two binary digits. Or 7 (0b111) and 11 (0b1011). But as I said it’s a starting point.

Reply on news.ycombinator.com