2016-01-10, 00:01 | #1 |
"Turki Ismael"
Jan 2016
Tunisia /France/ Belgum
2^{2} Posts |
Primal numbers formula
Hello
Sorry for bad english I have found the formula to the primal numbers, it's a conjecture but it work very well Im making a video to show that. There is 3 formula every one make a set of number . The first are the biger and contain primal numbers, numbers divisible by 5 (easy to soustract) and intruders numbers THe second formula make just the intruder numbers* The last make the numbers divisible by 5 In clear there is 3 formula but it there are som tool to soustract 2 or 3set of number like A-U=Z (A=U+Z) i need it to make a one explicite formula ********************************************************************************** First formula : (im not going to write the explicite one becaus to long) 6k+(3+((-1)^L)*2) k=1 L=1and2 K=2 L=1and2 etc Second is : 18(k+k')+36kk'+[12k(-1)^L']+[12k'(-1)^L]+[9]+[((-1)^L')*6)+((-1)^L)*6)]+[((-1)^(L+L'))*4] If i make the explicite forme is very very long k=1 k'=1,2,3,4......+infini k=2 k'=1,2,3,4......+infini etc The third its just a suit a formula to make 35.45.55.65.85.95 etc Just make the set of number with the first formula, soustract the intruder you have all primal number and 35.45.55.65.85.95 etc i making a formua for the third and it will be done Thats of sorry for my horrible english My name is TURKI Ismael |
2016-01-10, 05:32 | #2 |
Romulan Interpreter
Jun 2011
Thailand
3·47·67 Posts |
Your first formula generates all numbers of the form \(6k\pm 1\). When L is 1, you get a \(6k+1\) number, and when L is 2, you get a \(6k+5\) number. We already know that all primes are of this form, and the composites covered by this formula are not divisible by 3.
Then the second formula (which is written in a very stupid way, who the hell is L and L'? I assumed L is like in formula 1, and L' is its reverse, i.e. 3-L) generates some composites of the form \(6k+5\) (because you add a 9, but subtract \(4*(-1)^{odd}\) at the end), i.e 77, 119, etc. So, you just write down all the numbers which are 1 or -1 mod 6, then sieve out all which are multiple of 5 (the third formula) then you sieve out few other composites which are multiple of 7, but not all. How do you sieve out 49, for example? After you apply your "sieve", there are still a lot of composites in the string. Back to grade 5 arithmetic books... |
2016-01-10, 14:50 | #3 |
"Turki Ismael"
Jan 2016
Tunisia /France/ Belgum
2^{2} Posts |
I have difficulty expressing myself il french
Second my second formula : K=Σ(1,n) [1- || cos(n*(pi/2)) || ] L=||cos((n+1)*(pi/2))|| K'=Σ(1,n') [1- || cos(n'*(pi/2)) || ] L=||cos((n'+1)*(pi/2))|| The second formula are in the picture http://nsa38.casimages.com/img/2016/...5801462726.jpg If you developpe the second formula and calcule intruder numbers up to were you want, make same with the third you can elimiate all intruder and just have primal number. I have made this with very big number and it work TURKI Ismael |
2016-01-10, 15:51 | #4 |
Romulan Interpreter
Jun 2011
Thailand
3×47×67 Posts |
So, essentially you say that any composite number which is 1 or -1 (mod 6) is either a multiple of 5, or it is of the form:
18*(a+b)+ 36*a*b+ 12*a*(-1)^v+ 12*b*(-1)^u+ 6*(-1)^v+ 6*(-1)^u+ 4*(-1)^(u+v)+ 9 for some natural a, b, and some u,v in {0,1}, and that no prime number has this form. If this is true, wouldn't we have a very simple primality test, by working viceversa? Like for example taking a prime candidate, subtract 9, add or subtract 4 (to make it divisible to 6, from which we guess the parities of u and v), then dividing it by 6, or by 12 (if u and v have different parities, the "6*.." terms cancel each other) then solving the diofantic equation in a and b? Last fiddled with by LaurV on 2016-01-10 at 15:54 |
2016-01-10, 16:03 | #5 |
"Turki Ismael"
Jan 2016
Tunisia /France/ Belgum
4_{10} Posts |
Sorry
This formula if you see generated 49,77,91 etc its the first intruder THe third formula generate 25,35,55,65 etc First formula generate this number 4+2+4+2 etc 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 53 55 59 61 65 67 71 73 77 79 83 85 89 91 95 Now eliminate 49,77,91,119,121 etc And 25,35,55,65 What do you have ???? Primal number |
2016-01-10, 20:24 | #6 |
"Turki Ismael"
Jan 2016
Tunisia /France/ Belgum
4_{16} Posts |
Note :
K={1,1,2,2,3,3,4,4,5,5,......} =Σ(1,n)[1-|cos(n*(pi/2)|) cos is the absolut value ! L={1,0,1,0,1,0,1,0,1,0....... } =|cos(n*(pi/2)| there to : absolut value ! |
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