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Hay posibilidad de que en un futuro puedan venir nuevos equipos partners para esta entrega, como paso con la SS Lazio. A estos se los conoce como equipos partner. De Wikipedia, la enciclopedia libre.

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Giant trees mod 1-3 2-4 betting system

I'm a bit surprised: today I find it not so readable as I recalled from my last visit about - maybe it is due to new versions and more abstraction, but he is linked in many pages. The above is now the correct link. Actually, the introductory paper consists of two files. I have linked to the first. The B. Pls excuse I didn't place the links directly to the right place in the wiki -article: I don't want to possibly damage its html.

Gottfried: I was going to do this, but then I had second thoughts. You should do it. You found the references. Don't worry about messing up the wiki. If you mess it up, somebody will fix it. You should just jump in and do it without fear!

I don't think it's possible for it to fall into a loop that excludes 1. I think you have successfully proven that the loop is the only one with exactly one tripling step. Unless you also include the negative integer domain. Here you will find the sequence -1 -2 -1 which also contains exactly one tripling loop. All issues of structure based on patterns of halving and tripling span the entire integer domain, not just the positive integers. In fact, the conjecture is simply false in the negative integers and is usually not discussed.

Nevertheless, to properly understand the patterns, you cannot ignore the negative integers because the math takes you there whether or not you want to go there. Every possible pattern of halving and tripling occurs infinitely many times on the Collatz tree and from those patterns one can derive a formula that determines whether of not said pattern forms a loop. But that loop can be either positive or negative. Mensanator: In this case I think we should cut Bart some slack, since he did specify "positive" when he stated what he was trying to prove.

Now, continuing just for fun down the path you mention: it turns out that some patterns of halving and tripling can only occur in loops if you generalize to rational numbers. Here, you have to regard a fraction as being even if its numerator is even, and odd if the numerator is odd.

Ah, you are actually getting the same generalization by replacing 1 with C above. Sure, Bart can have all the slack he needs. I was just trying to point out that things are sometimes clearer when you look at the whole picture.

The sequence is trivially an integer and thus a loop if X-Y is 1 or If you restricted your research to positive integers, you may not ever discover that interesting bit. Any additional loops would be non-trivial. A non-trivial loop exists at Sure it's negative, but the very fact it exists means we can't conclude that non-trivial loops are impossible.

I was wondering if somebody could add some research paths that people have taken to try to prove it. And speaking of translating, even I know better than to use "Y" as a variable if I'm translating Spanish to English you might want to go patch up your translated equations. As to content, is that supposed to be some kind of proof? Aren't you using circular logic? Don't the convergences you speak of assume the conjecture is true? Try the convergences in the negative numbers, specifically, Don't work, do they?

You are assuming, without proof, that there are no non-trivial loops in the positive integers. Don't you need to prove that before you can trot out the convergences? Or did I not understand what you're saying quite possible since I don't speak Spanish and am relying on the dubious translation. Just looked at your link and I have one question: have you tried making the graph G using negative integers?

You say that. The only possibilities for this would be that a cycle exists somewhere in the tree and forms a "whirlpool" which integers cannot leave. Yes, it's in the negative domain instead of the positive, but the underlying cause of such "whirlpools" is independent of the domain. So there is nothing preventing such a "whirlpool" from occuring in the positive domain. The fact that integers are determined by their leaving connection, which is oriented toward 1, seems to provide an argument against the existence of such cycles.

Whoa, dude, you got a serious problem. What's with all the repeated values? Why do I see 27 on the 11th level? Because in this line,. For 84, it's not, so you should NOT append 27 to your tree at this point. And why the infinite recursion? There's no need for that, but I won't quibble about that absurdity. That's as far as I got in your paper, but I think it's enough to disqualify its inclusion in the Wikipedia. What you should do is fix up that paper so that it actually accomplishes something, submit it to a journal for publication and don't be surprised when it's rejected , and then you can reference it here.

There are going to be repeated values. This simple algorithm is not the focus of the paper; it's just an example from which the paper branches. Don't worry and just keep reading. Why do you think the 27 comes from 82? There's no 82 in the previous tree. If you line up the numbers, you'll see that 27 is a left child of But your program doesn't do that, does it?

How did you come to the conclusion that Program 2. Do you acually think numbers appear on the Collatz tree more than once? Is there any point in reading the rest of the paper having demonstrated that a you don't understand how to program, b you don't understand the Collatz problem and c you make no effort at quality control?

What do you think an actual peer reviewer would do with your paper? If you do it properly and save your trees into text files to prevent running out of memory, you'll find that the files are 3 GB by the time you reach level That's where I stopped because the level files exceeded the capacity of a CD even when zipped. And why not register with Wikipedia and get a user account. Then you can sign your posts. Right now I'm just talking to an IP address. Ok, I read the rest of the paper.

They are, however, incomplete. You state that they "creates and follows a general binary tree". Your programs do NOT create a binary tree. They don't even create any Collatz numbers because you neglected to assign any values to "n" in. All your programs do is create the parameters a,b,i,r. This can be easily rectified by wrapping a,b,i,r in a for loop once a,b,i,r are calculated:.

Should I start the n-loop at 1? Ignoring that, let's look at the numbers actually produced. Plotting the numbers on the Collatz tree:. But not the entire branch, just the first element. To see the second element, we have to re-calculate a,b,i,r:.

The numbers created by any given set of a,b,i,r are cousins, not children. Program 2 when done correctly generates every child of every element of a given level. By induction, all nodes are created. And even if I discount the 0's and negative numbers, there's another problem: duplicates. If you are properly generating a Collatz tree, there won't be any duplicate values.

Did you even know this was happening? Do you want to know why? If we start at 21 and do 8 even steps and 2 odd steps instead of 6 evens and 1 odd, we get:. Oh, and by the way, there's nothing inherrently wrong about the ideas presented in Program 5, I use a similar technique in my own programs. But I don't claim it's something it's not. Wouldn't the most obvious optimization be to stop if the current value of the function is less than the original input, since all numbers below the input must have terminated, or the program would have screamed, "i've found a counterexample!

If so, perhaps it should be added? I used the standard found here: [6] What standard are you referencing? Rolling snake-eyes on a pair of dice does not rebut the claim that the average outcome of a roll of dice is 7. But such patterns are not representative of the average. For a binary number of n bits chosen at random, the distribution of consecutive 1's will be a Negative Binomial, specifically, a Geometric Distribution.

In such a distribution, there will be twice as many 1-bit patterns as 2-bit, twice as many 2-bit as 3-bit, twice as many 3-bit as 4-bit, etc. The mean of a Geometric Distribution is the inverse of the probability. Thus, for such a randomly chosen number the mean bit pattern length will be 2 regardless of how large n is. From this relationship we can predict that there will be.

So the probalistic evidence is still correct despite the sequence beginning with consecutive increasing odd numbers. Suppose n is odd. In other words, the probabilistic evidence suggests that the next odd number should be slightly larger. I checked one of the references and found a page [7] on this heuristic argument.

I don't exactly agree with Mensanator or this page which are essentially the same argument , but the disagreement is over a technicality. Expectation must be calculated as a sum of the possible outcomes weighted by their probabilities. The other arguement uses a product of the outcomes, weighted by the probalities using power to define the weight. Formally speaking, this is not expectation, per se. But of course it has meaning, and is good heuristic evidence.

Of course, it can be converted to an expectation by applying a logarithm. Note that the logarithm of expectation is not the same as expectation as the logarithm. In other words, we're both right. I think the article could be corrected to avoid this confusion from arising again.

Just to be sure, I ran some tests on the first 20, odd numbers. The average difference between the next odd number and the current for these was 0. In particular, note that, the next odd was slightly larger, on average. However, exponentiating the average logarithm of the ratio of the next odd to the current odd gave 0.

This difference arises because of different meanings of average. Maybe this is worth adding to Paradox , provided that there is some published literature about such kinds of counterintuitive phenomena, which is surely more general than the Collatz problem. Then again, maybe it's not worth it. Here's a smaller example of what I did.

Take the first four odd numbers: 1,3,5,7. For Collatz sequences beginning with these numbers, the next odd number is 1,5,1,11, respectively. The differences are , , , , that is, 0,2,-4,4. At least we are slowly starting to communicate. But I'm still confused. You said the AM of the first odd numbers is 0. I get that now also, but I fail to see what the significance is. The AM is completely unstable for the odds, so of what use is this when comparing to the AM of a Collatz sequence?

Which is also somewhat stable although it bears no resemblence to that of the odd numbers. And yes, the AM is negative. Really, really, really negative. But maybe the tree crawler algorithm wasn't the best choice for a Collatz sequence generator. A tree crawler creates a path that minimizes the factors of 2. Perhaps a similar magnitude number created from random decimal digits would be more representative. Makes you even wonder why we're making such a fuss over the heuristics, eh?

Mensanator, thank you for running those tests. I also ran a few more tests too. Taking random odds n instead of taking the first bunch of odds , I found that the AM of s n -n fluctuates quite wildly. Probably this is because of the fairly large standard deviation. In similar tests for the GM, I found less fluctuation, similar to what you found.

I don't have a good explanation for any of this. Anyway, back to the article. I don't think that the AM stuff needs to mentioned in the main article, and here's why. I went to Lagarias's annotated bibliography, and couldn't find mention of it. I then asked him. He responded quickly, saying he was unaware of people studying tha AM. Therefore, whether or not the AM is relevant, significant or interesting, it would be original research, which disqualifies form inclusion in a wikipedia article.

When I first read the article, I interpreted average to mean AM. There was no hint the average meant geometric mean of the ratios. Only on reading your contributions to the discussion page, did I learn that what was meant was the GM. A very simple and general solution to Collatz-type problems is presented here: [8].

It's worth looking into I've seen it. It's rubbish. I concur. Sr13 , 28 September UTC. I am removing the above link from the External links section because the page it links to is incorrect. It claims to present a breathtakingly simple proof of the conjecture which is, as it turns out, not a proof at all. I just thought I'd leave a note here in case there's any question. Couldn't you have posted this on a web page and just made a link to it? Sadly, it's easier to read in the editor, a sure sign that someone doesn't understand Wikipedia's formatting.

And what do all the little square boxes mean? Want me to host it on my web site? If so, e-amil me a. Laurence A. Park, Kotagiri Ramamohanarao. Mismatch Sampling. Self-indexing Natural Language. Nieves R. New Perspectives on the Prefix Array. Indexed Hierarchical Approximate String Matching. Russo, Gonzalo Navarro, Arlindo L. Clique Analysis of Query Log Graphs. Alexandre P.

Out of the Box Phrase Indexing. Oren Kapah, Gad M. Landau, Avivit Levy, Nitsan Oz. Pattern Matching with Pair Correlation Distance. Christina Boucher, Daniel G. Brown, Stephane Durocher.


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