Have you taken the DeuceHound (mod 2) for a test ride?
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Have you taken the DeuceHound (mod 2) for a test ride?
I believe I can do the same thing for 3 that I did for 2. Getting all primes together under one roof for the DeuceHound (mod ∞) will be a monster job. I do not know if it is possible. I suspect if it were, it would already have been done.
Keep your eye peeled for unsolved number theory problems for my collection where only a finite number of solutions are known. For Fermat's Last theorem there were two known solutions, 1 and 2. For Brocard's problem there are three known solutions, (4!, √25) (5!, √121) (7!, √5041). There must be quite a few unsolved problems with this setup where a few solutions are known and it is not known if there are any more.
Then there is the Beal conjecture where infinite solutions are known, but all of one variety--where A, B and C share a common factor. He offers $1,000,000 for a proff that only this type of solution is possible. He is a billionaire banker to the big cat oil men, and worth, they say) about $8 billion, who jusy happens to be a quality mathematician.
I don't think he would pay anything for my DeuceHound. I would likely be laughed out of the king's cour for bringing something common. But for a solution to Brocard's problem, especially one he could help tidy up with his technical knowledge, he might shell out a million, as well, or some amount of conch anyway. Personally, I do not have any way at all to approach his conjecture. He loves number theory. I do have a way to approach Brocard's Problem, and the approach is improving. It seems like all the work done on Brocard is of the bounding type, or monsterous computer calculations that are able to promise no more soultions up to a certain limit. I will be trying to eliminate particular forms.
If one could eliminate particular forms that would be something. I haven't tried implementing the DeuceHound 2 yet. I will try to build a jupyter notebook and send you the link to it. It looks like I would need to know the largest power of 2 that divides n! and then I can work with the remainder.
The DeuceHound (mod 2) is extraordinarily simple to implement. In plain English, you take the F2 of the perfect power of 2, which we know is always 2k-1, then add to it the F2 of Q, with Q being the amount you have factorialized beyond the perfect power. For instance
F2{(26+31)!}=
F2{26!}+F2{31!}=
F2{64!}+F2{31!}=
63+26=89.
The DeuceHound (mod 2) works for any two numbers one puts in for k and Q.
The formula again is (and I would like to dispense with writing so many factorial signs, so I will introduce the new notation of F2! to mean all the factors of 2 in a factorialized number):
F2!{2k+Q)}=F2!{2k}+F2!{Q}.
If anyone has not seen this perfectly, they only need to gaze at the Ruler Sequence for a while. I do not have to look at it to write it down. Neither will you, if you gaze at it carefully after what I have to say about it. From every virgin power, it simply starts all over again. There are all kinds of clever ways the DeuceHound could be used, but just stick with the basics for now, until you see it really does start over from any perfect power of 2. I will make it really easy, and show the Ruler Sequence up to 95 (26+31), which means it will end on an invisible number 95. That's okay. 94 is just fine for our purposes.
1 2 1 3..1 2 1 4..1 2 1 3..1 2 1 5..1 2 1 3..1 2 1 4..1 2 1 3..1 2 1 6..1 2 1 3..1 2 1 4.. 1 2 1 3..1 2 1
The Ruler Sequence up to 94. From 64 to 95, the sequence is exactly the same as it is from 1 to 31. This is its nature. The Ruler Sequence is an awesome Mandelbrotian Object.
Here's a pdf export of a jupyter notebook that confirms your result. I would like to make it more general. That is given n = 2343542, compute the number of factors of 2 in n! This just checks your example: https://drive.google.com/file/d/0B96...ew?usp=sharing
All you have to do is plug your own numbers into the DeuceHound (mod 2). You have to figure a way to code that formula into Python, so you can plug values of your choosing into the DeuceHound for k and Q. I can also provide a formal proof, which is only to bring the formulas and reasoning from former posts together. However, the need for a lot of additions became unnecessary when I noticed that The Ruler Sequence repeated itself from the beginning every time a new power of 2 showed up. But as a throwback, remember when I said the difference between F2{(2k)!} and F2{(2k-2)!} would be k itself. That seems obvious enough, doesn't it? Through such reverse extrapolation we proved the entire premise before. Once my mind has moved on, it is difficult to recall, exactly. But I do know this, writing a formula was going to prove horrendously difficult the way every other measure is 1 2 1 3 and the others keep changing, until I noticed that other symmetry in the Ruler Function--that it repeats itself exactly from the beginning every time it reaches a virgin power, i.e. a power for the first time.
I will show you how it might be done for a fairly difficult number, how one one might produce a Result through Descent.
Suppose we are asked to find F2! for (210+150)!
The first term is always easy. We know that contains 210-1 factors of 2.
For the value 150, we will need to break it up again. We could break it up several times if we wanted to, but that is our choice. We could just take the value of the 150 through the familiar Floor Function, and add it to what we already have, or we could continue with the DeuceHound method, which of course we will do.
F2!{150}=F2!{128}+F2!{22}.
Again. it is easy for us to add up the 11 Ruler Sequence values for 22, but we prefer to decend further:
F2!{150}=F2!{128}+F2!{16}+F2!{4}+F2!{2}.
Now we have descended all the way to the bottom. What it suggests is that we could build the whole thing forward instead of backwards, in as tiny of increments as we felt like handling. We sum
1023+127+15+3+1=1169
Notice that these summands are all powers of 2 minus the quantity one.
Okay, that is it.
Now let us look at some more wonderments within the Ruler Sequence. Let us look at just the pure powers of 2 in the sequence. What do you suppose we would get if we added up all those powers to, say, 26, since we have been working with that as an example.
2+4+8+16+32=64.
64-1=F2!{26}.
The above is why the DeuceHound works.We could use this method, too, but we would still need the DeuceHound (mod 2) to caculate for values of Q.
I think one way to do this is to take the number n represented in binary and then apply the rule to each bit that is set. I may get this ready tomorrow.
It could be done this way.
F2!{2k+Q}=
[k-11∑ [2n]]-1+ F2!{Q}.
Yes, it is intimately related to binary. However, I do not know if that is the best way to go about it. Maybe though. I do not know Python or OOP. I would simply do it like my last post, if I were programming a computer to work it out, which may be precisely what you meant. I would use the Descent Technique showed earlier to calculate F2! for Q.Quote:
Originally Posted by YesNo
I think it would be the same as what you are suggesting to do. The binary representation of the number picks out the powers of 2 in that number.
Python is an OOP language. There are classes and inheritance. You don't need to know much about it to make it work. I started using it when my wife was working on machine learning concepts. We built clusters and then created models. I know (next to) jack about how these things are done, but with the Python packages, such as, pandas, sympy, scikit-learn, and so on, one can get results very quickly with very little actually programming. The programming work has all been done in the packages and it has all been optimized by others. You don't have to program these things, only use what others have packaged to explore your ideas. You are giving me ideas on how to use other packages I have not explored before. It is more a matter of looking up what is available and using that to implement an idea. Also you get mathjax which allows you to format text.
I don't know when I will get a general implementation done. I need to look for the right packages that deal with bit maps or representations of numbers to different bases so as to get factors other than 2. I don't want to have to implement that myself because I will probably implement it wrong and it will not be as efficient.
Your idea of filling bytes would work. You would actually be adding each member of the Ruler Sequence individually that way. One shorter way is to figure the nearest power of 2, then get Q with the method you suggest. That would cut down on a lot of the work. Perhaps you are just looking to empirically verify through a number of different examples. I am so confident I am resting now, watching movies. In a day or two I will be coming back to see if I can do anything about 3's. I am trying to get a good sense of all the figurate connections too.Quote:
Originally Posted by YesNo
One way to speed up the process is to "vectorize" the number. That is represent each bit as an element in a vector and then perform component-wise replacement of the 1 with the desired number. I am still working on that, however, Python's numpy or pandas packages should provide a way to do that. For the moment I have updated the notebook with a function that will do arbitrary numbers using a for loop and numpy's binary_repr to split out the individual bits. It not not an efficient way to do that.
Next steps: (1) vectorize the solution and (2) include other bases besides binary.
https://drive.google.com/file/d/0B96...ew?usp=sharing