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I know what you mean. I hope that all of the grim projections are wrong, but at the moment it's hard to feel hopeful. I believe that, in he end, things will be alright. Just look to the people around you; friends, family, strangers, and treat them as kindly as you can. It will spread from there, so long as you keep to it. It won't always be easy, but it's worth it. In the meantime, I know that a lot of people are scared and stressed out. I hope that by just having a nice, relaxing time playing games and hanging out, maybe it will take the pressure off for a little while.

Well, thank you, I am glad I did not come off as arrogant, and I am glad it was enjoyable to watch as well :) As far as naming goes, I had the same idea, and I was trying to avoid "numbers of the form 6n ± 1", "the number to which a given hexa is adjacent" and "the number to which 6 is multiplied to get an anchor". I guess it made more sense to me as I was working on this. I will keep this in mind, and I encourage anyone who feels similarly to say as much, but I think I would prefer to keep it as-is for now unless others say they had the same issue.

- "all hexas bigger than 3 must be hexas" is a tautology Yes, I have since corrected that mistake. It should (and now does) read "all _primes_ greater than 3 are hexas". - If so he is pathetically wrong: 385 (6*64+1) is not. If by hexadjacent you mean 6n+1 OR 6n-1 you are right is fairly simple to prove in a way far easier than you try (and your proof of this not valid for that part). You'll have to explain this more. I don't really follow what you mean. - Your language is flawed (see above) and you claim lemmas that you don't prove despite of being true. I prove my statements in the paper. I said at the beginning that this video is an overview, and that more details are provided in the linked paper. - Sorry the rest does not worth reading the rest. Life is too short. Sorry for being blunt but I am not being 10% as blunt as a mathematician will be. This isn't being blunt, this is kinda just being rude. Pointing out a mistake I made (and have already corrected) is one thing. Saying I am "pathetically wrong", assuming you are referring to me in that sentence, goes beyond bluntness.

The proof that every prime number greater than 3 is either on more or less than a multiple of 6 is correct, actually. Also, mathematicians are really nice people, and not blunt at all. The fact that you write a "blunt" comment to somebody who is neither arrogant nor mean, is incredibly low. Do better.

you are not arrogant. Mathematics is about precision. There is simply no other way to communicate in math. I liked the Ramanujan reference :3 <3

@@Kraflyn Well, thank you :) I think that many would understand that I am trying to speak confidently rather than arrogantly, but I nevertheless prefer to clarify these sorts of things, especially since I have had cases in the past where my tone is misconstrued. From what I can tell, that has not happened here, and it's nice to see.

​@@twilightstar7781 It's about Twin Primes, you were an angel actually in your very mathematical and specific critique :D :3 What do they expect? It's Twin Primes ffs.... One of the ddepest questions in math.

Throwing out unprovoked insults is, frankly, an immature way to respond to such a project. If you disagree, or if I did not clarify something sufficiently, that's all you need to say. If you aren't willing to put in some effort to meet me halfway, I am afraid there is only so much I can do for you. If you aren't willing to put in that effort and would rather just move on, then feel free to move on. I'd rather have you simply pass me by than go out of your way to insult me without actually engaging with the work.

Here's hoping! I believe I emailed them too, so maybe something will come of it.

For the first problem, that is something to address. I will have to think more on that. For the second,The IN() function counts the instances of either one particular hexa, or one pair of corresponding instances in the entire hexorial cycle. These two versions are only used to give the bounds on the approximation. Using it to count instances of hexas gives the error as being no more than n, and using it to count instances over the whole cycle ensures that that remains the case after we remove the redundant instances. You are right that essentially the only point where H_n(x) is _exactly_ the correct value for valid referents is at the end of the hexorial cycle. But that isn't what we're using; in the paper, I show that at ANY x, the correct number of valid referents can be no more than n away from H_n(x). I will be honest, I don't know what you were doing with the Python code. But perhaps an example would help. Let's go with the case where n = 3. When I used n at this point in the paper, it is the number of prime hexas under consideration, so in this case we are considering the first 3 prime hexas, which will be 5, 7, and 11. The critical area is ((11^2) - 1) / 6 = 20. Our approximation is the lesser over the greater hexorials, so in this case it is (3 * 5 * 9 ) / (5 * 7 * 11)(x) = (0.35064935064) * x. We plug in the critical area and get (0.35064935064) * (20) ~7.013. So now we know that the ACTUAL number of valid referents less than 20 is no more than 7.013 + 3 and no less than 7.013 - 3, or between 10.013 and 4.013. On the chart in the paper (the second one, after it has been converted to referents on the vertical axis instead of anchors), you can see there are 7 rows where the first 3 cells are uncolored, i.e. 7 valid referents, those being r = 3, 5, 7, 10, 12, 17, and 18. So indeed, it is well within our predicted range. It doesn't matter that this fails to give us the exact value, all that matters is that we have a lower bound, which we can then show to increase. As noted earlier, I need to put in a bit more work to show that this increasing approximation does not approach some value, but that's for another time.

​@@twilightstar7781 "IN() function counts the instances of ... one particular hexa." I don't get that, can you use IN's parameters in this sentence? What are "instances" here? Wrt. the second half: Ok, I thought you are claiming H_n*x - n <= #{valid referents up to x} for all x, but you only claim that for x=((h_n)^2 +- 1)/6, i.e. the upper bound of your critical area? I modified my code, it came up with these counter-examples: Failed for n = 96 and x = 45240 . coprime_count = 2739 and H(n) * x = 2844.939988966829 Test failed for n = 96 Failed for n = 98 and x = 48780 . coprime_count = 2917 and H(n) * x = 3044.5270994316384 Test failed for n = 98 Failed for n = 99 and x = 49868 . coprime_count = 2962 and H(n) * x = 3101.0529002722756 Test failed for n = 99 ... ------- def generate_primes_up_to(n): sieve = [True] * (n + 1) sieve[0] = sieve[1] = False for start in range(2, int(n**0.5) + 1): if sieve[start]: for multiple in range(start * start, n + 1, start): sieve[multiple] = False return [num for num in range(n + 1) if sieve[num]] primes = generate_primes_up_to(1000)[2:] # print(primes) def H(n): p = 1.0 for i in range(0, n): p *= (primes[i] - 2) / primes[i] return p def coprime(y, n): for i in range(0, n): referent = (primes[i] + 1) // 6 if y % primes[i] == primes[i] - referent or y % primes[i] == referent: return False return True def test(n): H_n = H(n) coprime_count = 0 crit_area_end = (primes[n - 1] ** 2 - 1) // 6 for x in range(1, crit_area_end + 1): if coprime(x, n): coprime_count += 1 # print("x =", x, "is coprime with {5, ...,", primes[n - 1], "}") if ( coprime_count < H_n * crit_area_end - n or coprime_count > H_n * crit_area_end + n ): print( "Failed for n = ", n, " and x = ", crit_area_end, ". coprime_count = ", coprime_count, " and H(n) * x = ", H_n * x, ) return False return True for i in range(len(primes)): if not test(i + 1): print("Test failed for n = ", i + 1) # break

@@philippwei3352 - An instance of a hexa is a referent which is invalid WRT that hexa. So for example, the cell (4, 5), that is, row 4 and column 5, is colored in, so the referent 4 has an instance of the hexa 5. 4 is thus an instance of 5, but it is NOT an instance of 7, because it is valid WRT 7. For an example with parameters: IN(x, 1, 5) only increases when x, interpreted as a referent, is invalid with respect to 5, i.e. it is flat everywhere except at those values where x is congruent to +-1 mod 5. What this means is that 6x + 1 or 6x - 1 is divisible by 5, and therefore cannot be prime. - RE: "I thought you are claiming H_n*x - n <= #{valid referents up to x} for all x, but you only claim that for x=((h_n)^2 +- 1)/6, i.e. the upper bound of your critical area?" Yes, this is a more accurate description. That is the only region where the approximation is accurate to find twin primes. It will count referents which are valid WRT your chosen set of prime hexas, but outside of the critical area, it isn't guaranteed that a referent is valid with respect to some prime hexa you weren't considering - You'll have to forgive me, I am a bit rusty on my python abilities. I don't know if it is generating primes correctly; in particular, I don't know if it is including 2 and 3, which it should not, because those are not hexas.

@@twilightstar7781 Thanks your response. Could you please write a clear definition that starts like this? "IN(m, d, x) counts ...". And then actually use m, d and x in that sentence. I'm still unsure how x plays into this, and I think you just switched the order of your parameters in your last answer? This makes it even more time consuming to check all plausible interpretations. " I don't know if it is including 2 and 3." It does not, the line "primes = generate_primes_up_to(1000)[2:]" has this [2:] in the end, which means "Take the list starting at the element 2, (skipping element 0 and 1), up until the end of the list". So I'm skipping the first two primes since they are not hexas. I figured out a way for you to double-check this without believing my code: Within the critical area, invalid rows are just twin primes, right? Except for the trivial invalids, like 2 being invalid wrt 11, because 6*2-1 is divisible by 11. To my understanding, your IN function also counts those "small rows" as invalid, but if we include them, the number of valid rows is exactly the number of twin primes, and your lower bound is still too large, starting at n=98. Failed for n = 98 and x = 48780 . coprime_count = 2941 and H(n) * x = 3044.5270994316384 Test failed for n = 98 Failed for n = 99 and x = 49868 . coprime_count = 2986 and H(n) * x = 3101.0529002722756 Test failed for n = 99 Failed for n = 100 and x = 51708 . coprime_count = 3085 and H(n) * x = 3203.9280313614927 Test failed for n = 100 If you take n=98 primes, the 98th hexa is the 100th prime, which is 541. 541*541-1 is 292680, so x=48780. H(n)*x=3044.5..., that should be checkable with Excel. And for the number of valid rows, just ask ChatGPT how many twin primes there are below 292680. For me it produced this very simple python code (see below), along with the answer 2942. This matches my 2941, since I'm not counting (3,5). In any case, this is less than your bound of 3044.5-98=2946.5. -------- import sympy as sp # Find all prime numbers less than 292680 limit_new = 292680 primes_new = list(sp.primerange(1, limit_new)) # Find all twin primes (p, p+2) twin_primes_new = [(p, p + 2) for p in primes_new if sp.isprime(p + 2)] # Get the count of twin primes twin_prime_count_new = len(twin_primes_new) twin_prime_count_new -------- I now also have a guess where your proof went wrong: When you count invalid rows, you are counting invalid instances and subtract some stuff "for each redundancy". You never quite define what a redundancy is, but you talk about "one pair of corresponding instances" in your first reply to me. I think you are miscounting rows with three or more invalid instances. A row of three invalid instances has 3 pairs of instances, not just the 2 that you need to subtract. You would need to add triples of instances, and then subtract quadruples... This is known as the "Inclusion-exclusion principle", (see Wikipedia) and I don't see you doing anything like that anywhere.

@@philippwei3352 Sorry for the confusion, I admit I didn't check the order of arguments, and that was careless of me. IN(m, d, x) counts the number of positive integers less than or equal to x which are congruent to d or -d mod m. For example, IN(5, 1, 6) = 3 because there are 3 positive integers less than or equal to 6 which are congruent to either 1 or -1 mod 5, those being 1, 4, and 6. - "Within the critical area, invalid rows are just twin primes, right?" It is the VALID rows that correspond to twin primes. I suspect that this was a typo based on the rest of that paragraph, but I thought I should clarify just to be sure. I must admit it feels suspicious that this allegedly breaks down at the 100th prime, with no discernable cause. That feels like an awfully convenient place for that to happen, honestly. I still suspect it's an error in the code rather than the theory, but I was never very good at diagnosing code even when I was studying it, so I can't be sure. Could you give the results for smaller values of n so that we can check that it is doing these calculations correctly? - "You never quite define what a redundancy is" I guess I didn't explicitly write it out, but I did say this: "This sum counts all instances of the hexas being considered, but there are some referents which are invalid with respect to more than one hexa. In order to count only the invalid referents [as opposed to all invalid instances], we will need to subtract them [the extra instances] out relative to the whole hexorial cycle [...] where k is the number of redundant instances, and lj is the jth such redundancy." I figured this made it clear enough what I meant be "redundancy", by I guess not. - "I think you are miscounting rows with three or more invalid instances." The process you describe in this section is exactly the process I was describing in the paper. The example I used only had two hexas under consideration for simplicity, but in a case such as you describe, you would indeed subtract out multiple copies of the function. I included this under the broad header of "redundancy". For example, if we were to look at 5, 7, and 11, x = 64 should be invalid with respect to all three of them, because 6(64) + 1 = (5 * 7 * 11). This essentially means that in the basic approximation with redundancies, there are three copies of IN(385, 64, x) that are added together, one for each of the three hexas. In order to remove them correctly, you would have to subtract two of them so that you're only left with one.

I don't really follow your reasoning that my argument boils down to "the heuristic argument". It is heuristic, but it is not the argument you seem to be describing. My method does not vaguely gesture at the infinitude at play, but instead gives a specific lower bound on how many twin primes exist in a critical area, and shows that that bound increases as you look at larger critical areas. Nor is it a probabilistic argument; it just turns out that using the probability as an approximation works out pretty well. It's not "I expect that about this proportion of referents in the critical area are valid because that's the probability", it's "I can use the probability as an approximation because I know that the true value is pretty close to that probability". If this is a genuine flaw, I would like to understand it better, and I think I would benefit from a more detailed description of how you understand my argument and where it goes wrong.

@@twilightstar7781 Well you're stating patterns that rely on direct observation. Can you demonstrate that, without bound, that the patterns you see will hold? From what I've noticed, it looks as if all of your methods---your lower bounds, your invalid anchors, etc---only represent candidates for twin primes, not a guarantee that there will exist one, especially as the prime numbers continue to become sparser as magnitude increases. For example, I've found a very clear and distinct pattern with implications for a problem I'm working on and have verified its exact calculability up to n = 4*10^17, though am still working on a rigorous proof as to why this connection exists. If you can find a rigorous connection between your observations of these patterns which are so clear and undeniable truth, then this could be a worth publishing. Otherwise, keep pushing forward with your endeavors! By the way, just out of curiosity, how far have you verified your results to?

@@Monkala2 I feel I should reiterate that this video is meant to be an overview; it is deliberately light on the details, as it is meant to give more of a general sense of how it works. The paper gives more detail as to how I justify my claims. - "Can you demonstrate that, without bound, that the patterns you see will hold?" Yes; I discuss in the paper how these patterns naturally arise from properties of modular arithmetic. - "it looks as if all of your methods---your lower bounds, your invalid anchors, etc---only represent candidates for twin primes, not a guarantee that there will exist one" This is not the case. If a referent is valid within the appropriate critical area, it IS the host of a pair of twin primes. You have to be careful about the critical area, though. So if you're only looking at 5 and 7, then 3 is valid with respect to both of them. The critical area is at 8, and 3 < 8, so this referent is within the critical area. This means that 6*3 hosts a pair of twin primes (17 and 19); it doesn't say "there might be a pair here", and that happens to be true. These facts alone are sufficient to conclude that there is a pair of twin primes there. However, if you picked some x > 8, this is not enough information; you'd need to expand the critical area and the primes under consideration.

I have, in fact. I emailed him and a few other channels for which I could find contact info for. I hoped to just spread it around as much as I could

​@@twilightstar7781 anyone who sees this should consider commenting on matt's vids plus some other mathematicians if possible to try get more eyes on it!

I am not one who can prove or dismiss what you have here, but when you think you have something you have to share it. There's no shame if this turns out to be wrong. Sharing this was absolutely the right thing to do, so no need to feel nervous about it. I hope smarter people than I can verify this for you. Very exciting and best of luck!

@@taylorschaerer9854 Thank you! While I sorta know that deep down, it is always nice to hear from someone else :)

Yes, I do, when I get the time. Welcome!

Yeah I always forget that, I don't think that's a common feature anymore. Use it or lose it, as they say, and I fear I have lost it.

It's-a me! I'm back :)

@@twilightstar7781 yippee

It's always here, so no worries :)

@@twilightstar7781 u havent uploaded in a while! Are you ok???

@@Thewhiteplagueteam I'm ok! I am just taking a bit of a break so I can work on some other stuff. I have a big project in the works. I hope to start a new series soon as well.

@@twilightstar7781 okie

@@twilightstar7781 my yt has grown so much in 10 days hope ur yt grows too

Bless you

Thank you! I'm glad I saw it through.

It's called rain world

@@Thewhiteplagueteam Maybe someday! I think if there's enough support, I might start a suggestion box for future videos

On disco

Rd

I'm one of ur first subscribers :3

Glad you enjoy it! My discord is Cosmic Clockwork, but I don't use it too often. Maybe I'll start a channel if there's enough interest though :)

​​@@twilightstar7781 i couldn't find it mine is .cheez_it.

@@twilightstar7781 I couldn't find it

@@Thewhiteplagueteam I will look in to starting a channel and put the link in the description. Just keep an eye out for that, I suppose!

@@twilightstar7781 alright:D

Avast, ye sea dogs, this village be mine!

Thank you! No plans to stop any time soon :)

I do, yes :)

Did I do a goof somewhere?

So I hear. Though I do have a bit of a backlog right now, so I probably won't get the opportunity to play it any time soon.

@@twilightstar7781 alright if you ever get the chance I feel like it would be a good chance, but they're releasing a new one on the 17 right now

​@@twilightstar7781oof

I've gotten into a bit of a rhythm now, so keeping up the pace isn't too bad. So long as someone enjoys it, then it's barely a bother to me :)

Yup

I think he would be a lot less inclined to help us if that was what he had to deal with! Glad you like the intro :)

Hello :)

I dragged him for like 5 minutes through the snow only for my ultra-butter fingers to drop him down a hill. I felt so bad :(

Welcome back :)

Thank you! I don't really have social media, but I might start a presence if for no other reason than promotion and interaction with my audience.

Thank you! I'm glad you're enjoying it :)

Thank you!

Thank you! I must admit I don't recognize the username, but I suspect I do indeed know who you are :)

Ok

Thanks for the support :)

The gamer's eternal struggle

Yeah, it's taken me some getting used to, but I think I have since gotten the hang of it.

Happy to hear it! I hadn't thought about the fact that she might be trying to not distract herself with the temptation of returning to the world of light. I think the only problem is that her trips to the world of light could be limited, and theoretically disguised as diplomacy missions if need be. But even then, it's not a bad point to bring up, especially since she clearly shows that this is a priority for her throughout the story, so the idea that she thought it might distract her from that isn't illogical. As for why she went through the mirror with her own power, I suppose that was what I was trying to get across, even if I didn't explain it thoroughly; since she damaged the mirror, it was only strong enough to keep the portal open, but not send her through, though you also make a good point that she may be doing it just because it's safer.

@@twilightstar7781 I think the crux of the matter regarding diplomacy between worlds was that deep down she knew it wasn't that simple. Those native to the world of light can't exist normally inside the world of twilight, and vice-versa. Even if the two people were able to somehow diplomatically put their differences aside, co-existance between them still wouldn't be possible.

@@Crow_Rising Well sure, I just mean that she could use that as a pretext or a "I didn't come just because I missed you, baka" kind of reason if she really wanted to. I guess I just mean that it's possible for her to be committed to being a good ruler while also being able to visit her friends. But even then, it wouldn't be beyond her character to go to such extreme lengths for something she feels is important, so really it's a minor point.