This obviously isn't a very efficient formula but I was wondering if it is possible to find a larger prime with algebra given only smaller primes and yep it is! :D
For any number x that has prime(3)=5 as a factor, values on row_3 give the minimum distance to the next prime > x. ie. numbers with 5 as a factor: 5,10,15,20,25,30,35,40,45,50,55,60,.. using row3: 1,0,1,2,3,0; all numbers >5 up to 5+1 aren't prime all numbers >10 up to 10+0 aren't prime all numbers >15 up to 15+1 aren't prime all numbers >20 up to 20+2 aren't prime all numbers >25 up to 25+3 aren't prime all numbers >30 up to 30+0 aren't prime This pattern can be repeated indefinitely using the same repeating values on row3 ie: all numbers >35 up to 35+1 aren't prime all numbers >40 up to 40+0 aren't prime all numbers >45 up to 45+1 aren't prime all numbers >50 up to 50+2 aren't prime all numbers >55 up to 55+3 aren't prime all numbers >60 up to 60+0 aren't prime.
I really don't want to discourage you -- and I recognize that everyone has to start somewhere -- but you really need to improve your number theory game. Or just your playing-around-with-equations game.
I believe you if you say you're not trolling, but part of the reason that I thought you were trolling was that you had missed a very elementary simplification, something like combining 3*B + 2*B to get 5*B. It looked so obvious to me (once I had re-written your formula) that I assumed you were being intentionally obscure -- hence, trolling.
It's true that, the way you had written it, it was much harder to see than that, but part of what you need to improve is re-writing things to help you see such things. (see also: Whitespace Is Our Friend)
---- I came up with what seemed to me a considerable simplificiation ( 6*EulerPhi(primorial(n-1)) - primorial(n-1) )*(x-1) = 4*primorial(n-1) - 2*primorial(n-2) And you asked me how to solve it for x. I think you asked for x = something. That is what I mean by "solve for x".
I can do that, and I will. But this is a good example to show what I consider an essential technique for playing around with equations. I don't know if this has an official name, but I call it "chunking". One takes a big, messy part of an even bigger, messier formula and replaces it with something simple looking. You haven't really changed the formula, because the simple-looking thing still represents the big, messy thing.
Here, we can re-write ( 6*EulerPhi(primorial(n-1)) - primorial(n-1) )*(x-1) = 4*primorial(n-1) - 2*primorial(n-2) as B*(x-1) = C if we set B = ( 6*EulerPhi(primorial(n-1)) - primorial(n-1) ) and C = 4*primorial(n-1) - 2*primorial(n-2) Then we can solve for x: x = 1 + C/B and we can, if we wish, put the big messy things back: x = 1 + ( 4*primorial(n-1) - 2*primorial(n-2) ) / ( 6*EulerPhi(primorial(n-1)) - primorial(n-1) )
There might be further simplifications that can be made at this point. Or not, I haven't looked.
Chunking gets much more valuable when the same big, messy thing shows up more than once. This is something to keep in mind when you're think about how to slice apart your bigger, messier formula into chunks.
For example, in your primorial(n) - (2*((EulerPhi(primorial(n)))+(EulerPhi(primorial(n))-(1/2*(primorial(n-1))))+(EulerPhi(primorial(n))-(primorial(n-1))))) = 2*primorial(n-2) we could chunk p(n) = primorial(n) p(n-1) = primorial(n-1) p(n-2) = primorial(n-2) E(n) = EulerPhi(primorial(n)) and, doing nothing else, your formula gets much easier to understand. p(n) - (2*((E(n))+(E(n)-(1/2*(p(n-1))))+(E(n)-(p(n-1))))) = 2*p(n-2)
Of course, Whitespace is still Our Friend, though finding out how to use whitespace to help _you_ see how something big and messy is put together is something _you'll_ need to work out. p(n) - ( 2*( ( E(n) ) + ( E(n) - (1/2*(p(n-1))) ) + ( E(n) - (p(n-1)) ) ) ) = 2*p(n-2)
---- So: 1. Whitespace Is Our Friend. 2. Chunking Is Our Friend.
---- Elsethread, Kevin Aylward made a joke (an old one, but a good one) All odd numbers are prime:
The point of the joke is that we don't do "experimental error" in math. If I say "all odd numbers are prime", I am taken to mean *ALL* odd numbers are prime, without exception. So, the thing said is definitely NOT proven.
The techniques that we use in order to be able to _say_ something _like_ "all odd numbers are prime" (but something else, something correct) -- to say it and be correct without exception -- are what most people refer to when they say "mathematics".
Just checking some of *ALL* is not one of those techniques. I think you can see why.
x slowly moves back towards 0, or else some limit close to -0 I guess, but it doesn't seem like it would go positive again. Any ideas on what that means? I had kind of expected x to just get larger.
I don't want to discourage anyone either, with that being said here is a new example to calculate the nth prime (Mathematica code), however unfortunately it requires the nth prime, but I'd like to hope the formula can be fiddled with to take in nth-1 and give the nth prime!
If you plug in the nth prime ie n=125, it outputs 691.97, and the nth prime is 691. So this requires to put in the same prime that it gives, not much use, but maybe it can be simplified :D
Here is the Mathematica output of running that code:
then, since x only occurs as "x times some known coefficient", I'd collect all the x-multiple-terms on one side of '=' and all the constant-terms on the other.
That seems to be what you'd intended to do below, more or less, but with some x-dependent terms left on both sides.
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