20190118, 09:20  #1 
May 2016
7·23 Posts 
is this a demonstrable formula?: (Extension of the independent variable)
Good morning,
is this a demonstrable formula?: (Extension of the independent variable) this formula has found prime numbers of different value A=((n*n)*99)/2 B=sqrt(A) C=sqrt(A)1 In The Code .Py "B=Func" "C=Func1" "N= n+1" For n = 1 to 10000 N=1229 prime numbers Func=997 prime numbers Func1=982 prime numbers TOTAL PRIME NUMBERS OF DIFFERENT VALUE = 3208N=1979 For n = 1 to 100000 N=9592 prime numbers Func=8052 prime numbers Func1=8131prime numbers TOTAL PRIME NUMBERS OF DIFFERENT VALUE = 25775N=16183 etc... Code:
aa=1 ii=1 i2=0 GG=0 FF=0 VV =0 e = 0 AA=0 c = 1 a =3 bb=0 dd=0 I=1 II=1 III=1 i=1 i1=0 b=0 p=1 equTwo=0 equTwoo=0 equTwooo=0 mod=0 equ=0 moda=0 modaa=0 print('Helloword \n\n'); bb=input('Inserire numero: ') while p<=bb: equTwo= ((p*p)*99)//2 equTwoo=((equTwo**(1/2.0))//1) equTwooo=equTwoo1 print(' N= >' , p) print(' FUNC= >' , equTwoo) print(' Func1=>', equTwooo) while I<=p: I=I*1 I=I+1 if p==I : print(' P is prime ', p) AA=AA+1 break if p%I==0 : break while II<=equTwoo: II=II*1 II=II+1 if equTwoo==II : print(' Func is prime ', equTwoo) modaa=modaa+1 break if equTwoo%II==0 : break while III<=equTwooo: III=III*1 III=III+1 if equTwooo==III : print(' FUNC1 is prime ', equTwooo) moda=moda+1 break if equTwooo%III==0 : break p=p+1 aa=aa+1 I=1 II=1 III=1 print('\n\n\nTOTALE NUMERI P TROVATI : ' , AA) print('\n\n\nTOTALE NUMERI Func TROVATI : ',modaa) print('\n\n\nTOTALE NUMERI FUNC1 TROVATI : ',moda) Last fiddled with by Godzilla on 20190118 at 09:43 
20190118, 15:46  #2  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
59×157 Posts 
Quote:
And so is this  A=((n*n)*98)/2 B=sqrt(A) Can you take a wild guess what B equals to? We are trying to understand how much do you know (if anything at all)? 

20190118, 16:20  #3  
May 2016
7·23 Posts 
Quote:
EDIT n=88 >>> ((88*88)*99)//2 383328 >>> ((383328**(1/2.0))//1) 619.0 >>> 619 is prime and..I do not understand ... Last fiddled with by Godzilla on 20190118 at 16:43 

20190118, 18:24  #4  
Aug 2006
3·1,987 Posts 
Quote:


20190118, 18:46  #5  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
5·1,201 Posts 
Quote:
Such beautiful variable naming Such consistent spacing and alignment Such excellent and useful comments Clearly so much effort expended to make it concise and easy to understand The work of a code god /sarc Couldn't you at least try to make it readable. Or even tell us what interpreter/language it is? 

20190118, 19:21  #6 
May 2016
10100001_{2} Posts 
@GRGreathouse could it be a new alghorithm?
@retina language is Python language, copy the code to a text file and save it with filename.py, open python and run it. 
20190118, 19:22  #7 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
242F_{16} Posts 
Ok, so you do understand some things. That is: squaring something and then taking a sqrt is a waste of time. But you prefer to waste time (yours and others').
What is so difficult, then, in taking sqrt(99/2) ? It is a constant, if you take it with enough digits of precision you will be set for doing simply: B = floor(n * 7.03562363973514433184) C = B  1 Your function is a line through numbers. Granted, its slope is ugly, but so what. What makes you think that it will be better than most other functions to "generate primes"? For comparsion: B = 2310*n+1 C = 2310*n1 will generate more primes than "A=sqrt(((n*n)*99)/2)". How do you arrive at your "wild and never before seen" "functions"? Do you just randomly press some buttons? 
20190118, 19:34  #8  
May 2016
7×23 Posts 
Quote:


20190118, 19:47  #9 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6005_{10} Posts 

20190118, 22:04  #10  
Sep 2016
331 Posts 
Quote:


20190206, 15:10  #11 
Dec 2018
Miami
29 Posts 
what a strange math forum this is. People are always bickering. and when you ask something serious, no response, but BS always get answers.

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