4 ms·
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 Poly
by oefrha 7y ago
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_gaps_between_primes 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.
- war1025 7y agoNeat. 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.
- oefrha 7y agoActually I replied without thinking too much, sorry... Your question is (probably a fair bit) stronger than the one I related to. Edited.
- chairmanwow1 7y agoBig 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.
- war1025 7y agoTo 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 https://en.wikipedia.org/wiki/Gray_code
- cygaril 7y agoI think you were right the first time.
- oefrha 7y agoNah, 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.