4 ms·
My take, but correct me if I'm wrong. Take the relation to the GCD triangle http://oeis.org/A054521 http://oeis.org/A054521 At GCD(n,k)!=1 all numbers are div
by szemet 8y ago
My take, but correct me if I'm wrong.
Take the relation to the GCD triangle http://oeis.org/A054521 http://oeis.org/A054521
At GCD(n,k)!=1 all numbers are divisible by GCD(n,k) therefore contains no prime
At GCD(n,k)==1 we have https://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions https://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arith... - so those series contain infinitely many primes - and seems like they actually contain at least one prime in all the pixels of the first N rows (but this should be explained/proved, if it is always true for any chosen N, or just happen to be true for the N-s tried by the OP)
- tomsmeding 8y agoYou're onto it: what you say would still need to be proven is actually not even true, see anderskaseorg's comments, e.g. https://news.ycombinator.com/item?id=17104624 https://news.ycombinator.com/item?id=17104624. So their picture is nice, and the gcd!=1 cells are all red, but the gcd=1 cells need not be black. For odd N this fails loads of times (always?), and for even N you quickly find failing cells when you start looking for it (see linked comment).
- deleted 8y ago[deleted]