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Tinacity

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  1. I assure you that I wasn't being sarcastic when I replied to you.....and I do thank you for your input. I just feel that it seemed unlikely to me, though a good potential explanation.....Unlikely; considering the age of the post and that looking down the list of posts sorted by view number.....that kind of aberration seemed less likely due to the other posts looking unaffected by those kind of issues. I can't think of any "good" explanation as yet..
  2. And you feel this would account for such a large number of "views" within two months of a post? @michel123456
  3. Any idea @wtf , why this post has received the third highest number of views on this part of the forum? this topic has
  4. Do you envisage or presume a case where Euler's totient would be imprecise @wtf and inclusion/exclusion would be more precise?
  5. I hadn't noticed yet that they were giving different answers in the examples used. Certainly there are 8 relatively prime pairs for n=60 as you used incl/excl and using the totient function @wtf
  6. @wtf Thank you for your inputs thus far. I am not sure why it is important for "the totient function will always give the same answer as inclusion/exclusion" unless one of the two (or both) gives an inaccurate answer in some cases. Are you arguing that? Reviewing my work on the assumption I wanted to make about (3-2)(5-2)(7-2).......(p-2)................I am pretty satisified that this is a perfectly accurate way to calculate the number of relative prime pairs as argued by me so far.......I had simply forgotten that I had proven this to myself.
  7. Does the silence mean I goofed up on my whole narrative too many times? Or discomfort with the implication that I am withholding my main project? That isn't a matter of disrespect......Just protecting a LOT of work. Or do community members feel that my fifteen minutes of attention are up? Or is it that I simply make no sense, still? Or that my presentation style is just too casual, prosaic and/or inaccurate. Or is it boredom.....lol !!?
  8. Relative to 210 (the number of pairs being considered) I would say that this is calculated using Euler's Totient. I am using the less formal expression of the Totient formula. I am NOT using (1-1/2)(1-1/3) etc........but the more accessible.... (3-1)(5-1)(7-1) version. So....for 210 pairs (x,y) of n= 454, where x and y are natural numbers.......We use Euler's Totient to calculate the number of relative prime pairs (relative to the number of pairs) thus (3-2)(5-2)(7-2) = 15........."-2" being used, instead of "-1" in Euler's Totient, because each of the factors [3,5 and 7] knock out two pairs instead of only one. They would "knock out" only 1 pair at a time, when n shares any of those factors.
  9. I think that we just use Eulers Totient to calculate this kind of thing. Phi(30) = (3-1)(5-1) = 8 If we are looking at the thirty pairs for n = 60 then it is phi(60) divided by 2....... or 8 pairs. But what if we consider the first 210 sum pairs (x,y) of a number n? Where n is greater than 420. And n does not share an odd factor with 210 e.g n = 454 How many relatively prime pairs should we find? You see....this is much closer to my problem space.......I have been trying to focus on the "most difficult" numbers for so long, that I had become unfamiliar with the problem space we were first discussing.....where n was a product of primes...... I am so unused to describing any aspect of my problem space that I made big mistakes in presenting it the way I did.....I can see that now and can only apologise for that....and vow to try harder to avoid that issue.
  10. You have been kind and patient @wtf Let me think about what you have sent. Again....thank you for your time and consideration.
  11. My apologies....I was thinking about a different aspect of my problem space when I talked about order there....Ignore the order reference, if you would. 6 pairs in n=60 is correct. If we wish to call (1,59)...as one of those pairs....then, for 60 it is 7.....but this detail is irrelevant to any generalisation I seek for this matter. I got 13/47 but not 11/49........hmmm..... Oh yes.....I can see the potential reason for that issue arising.....It's because N itself, must not share any of the odd primes as a factor either. I am getting into parts of the problem space I didn't want to reveal atm....... I chose artificial examples....but chose them poorly. The numbers (N) I have been working with....automatically share none of the prime factors less than the sqrt(N/2) I think that the unexpected extra pairs you found, such as 11/49 are outcomes from N = 60 being divisible by all of the prime factors less than sqrt(N/2) where N=60 I will have to go and think about this forum approach......Thank you for your awesome input there.... @wtf I know that what I just said may seem confusing and even inappropriate but this is a very personal project I have worked on for some considerable time...Please bare with me. But you folks have certainly helped me see that I will have to think much harder about how I define my questions and answers..... I might assume that I could ad the caveat "at least" to this general formula; "p1 to pn.....the number of prime/relatively prime pairs is found by (p1-2)*(p2-2)......(pn-2)" To cover for N which may share prime factors less than sqrt (n-2)
  12. If we look at the number 2*3*5 =30 We can form 15 natural number pairs (x,y) which can sum to 30 1+29, 2+28 , 3+27, 4+26, 5+25, 6+24, 7+23, 8+22, 9+21,10+20, 11+19, 12+18, 13+17, 14+16, and 15+15 If I double 30...to make 60 Then I double the number of pairs (x,y) (members of the natural numbers) which form this number. I have 30 unique natural number pairs which sum to 60 (unique, as in order is also important...) Question: How many of these pairs (x,y) are there, where neither x or y is divisible by the the primes of its product 4*3*5 ? I want to assume that this can be calculated thus......(4-2)*(3-2)*(5-2) = 6 This could be from the following configuration of sums to 60 30/30 29/31 28/32 27/33 26/34 25/35 24/36 23/37 22/38 21/39 20/40 19/41 18/42 17/43 16/44 15/45 14/46 13/47 12/48 11/49 10/50 9/51 8/52 7/53 6/54 5/55 4/56 3/57 2/58 1/59 Here are the six: 29/31, 23/37, 19/41, 17/43, 13/47, and 7/53 Sorry if that factor of 2 error keeps creeping in and out of my narrative.....I am simply unused to actually describing my problem space. Before moving to describe the focus of the issue I want to examine....I need this method of calculating....confirmed or not. And it is pairs that I am referring to .......not primes or relative primes less than N. But pairs which sum to N.
  13. I have worked on smaller examples..... It's any general result I am looking to establish. I can see that for 2*3*5....If I multiply that by 2......The number of relative prime pairs, as it were......are indeed calculated from (3-2)*(5-2) = 3. Similarly for 4*3*5*7.....Finding the number of prime or relatively prime pairs from (3-2)*(5-2)*(7-2) = 15 and 135 pairs for 4*3*5*7*11 . Found by (3-2)*(5-2)*(7-2)*(11-2) Is this the general case then.....? p1 to pn.....the number of prime/relatively prime pairs is found by (p1-2)*(p2-2)......(pn-2) Or something around that....!! I am generally aware that I am not being super exact in everything I am saying here......I hope that isn't too annoying for everyone.
  14. More ambiguous than ever, since I just recently amended the question by a factor or 2, as it were. But yes.....Your simpler example does encompass my intentions. And (1,209) and (209,1) are two answers. Yes.....the pairs with 209 = 11*19 are relative prime pairs (if we count 1 as a prime that is; and it wouldn't detract from the quality of answer I am hoping for anyway) I can see the confusion in the phrase "... where neither is divisible by any of the primes which make its product?" Possibly because I didn't refer to each pair of numbers (x,y) which sum to N . Neither of those should be divisible by any of the factors of N. @wtf Here is my attempted amendment to my original question. I have realised that I formulated my question incorrectly, let's say, by a factor of two, as it were. And can only apologise humbly for wasting good folks time on here, attempting to advise me on the wrong question. I am not used to attempting to describe my space of interest and I was so close to it, that I miss_described a basic aspect of my problem space. Instead of asking folk to "consider 2*3*5*7*11*13*17*19*23 = 223092870 ( the product of the first 9 primes).. Call this number: N" I should have doubled that.....Because it is the first 223092870 PAIRS of natural numbers summing to N I am interested in examining..... Thus N in this case should be 4*3*5*7*11*13*17*19*23 = 446185740 so that I can form 223092870 pairs which sum to 446185740 from that. And hoping therefore.......that I can then assume that the number of pairs (x,y) which sum to N, but where neither x or y has any of the first 9 primes as an exact factor, is given by the product: (3-2)*(5-2)*(7-2)*(11-2)*(13-2)*(17-2)*(19-2)*(23-2) ? And can also assume that the number of pairs (x,y) which sum to N, but where x and y are even but neither x or y has any of the first 8 odd primes as an exact factor, is also given by the product (3-2)*(5-2)*(7-2)*(11-2)*(13-2)*(17-2)*(19-2)*(23-2) ? I think it better if I at least establish this first....before attempting any expansion of my questions from there
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