is this a demonstrable formula?: (Extension of the independent variable)

Godzilla · 2019-01-18, 09:20

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=Func-1"
"N= n+1"

For n = 1 to 10000
N=1229 prime numbers
Func=997 prime numbers
Func-1=982 prime numbers
TOTAL PRIME NUMBERS OF DIFFERENT VALUE = 3208-N=1979

For n = 1 to 100000
N=9592 prime numbers
Func=8052 prime numbers
Func-1=8131prime numbers
TOTAL PRIME NUMBERS OF DIFFERENT VALUE = 25775-N=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=equTwoo-1
    
    
    
    
    print(' N= ---->' , p)
    print(' FUNC= ---->' , equTwoo)
    print(' Func-1=---->', 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(' FUNC-1 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 FUNC-1 TROVATI : ',moda)

Batalov · 2019-01-18, 15:46

Quote:

Originally Posted by Godzilla

Good morning,

is this a demonstrable formula?
A=((n*n)*99)/2
B=sqrt(A)
C=sqrt(A)-1

Yes, this is demonstrable formula.

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)?

Godzilla · 2019-01-18, 16:20

Quote:

Originally Posted by Batalov

Yes, this is demonstrable formula.

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)?

With number 98 B= n*7 ?

--EDIT--

n=88

>>> ((88*88)*99)//2
383328
>>> ((383328**(1/2.0))//1)
619.0
>>>
619 is prime

and..I do not understand ...

CRGreathouse · 2019-01-18, 18:24

Quote:

Originally Posted by Batalov

Yes, this is demonstrable formula.

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)?

Quote:

Originally Posted by Godzilla

With number 98 B= n*7 ?

--EDIT--

n=88

>>> ((88*88)*99)//2
383328
>>> ((383328**(1/2.0))//1)
619.0
>>>
619 is prime

and..I do not understand ...

99 isn't 98.

retina · 2019-01-18, 18:46

Quote:

Originally Posted by Godzilla

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=equTwoo-1
    
    
    
    
    print(' N= ---->' , p)
    print(' FUNC= ---->' , equTwoo)
    print(' Func-1=---->', 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(' FUNC-1 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 FUNC-1 TROVATI : ',moda)

This code is totally awesome

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?

Godzilla · 2019-01-18, 19:21

@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.

Batalov · 2019-01-18, 19:22

Quote:

Originally Posted by Godzilla

With number 98 B= n*7 ?

and..I do not understand ...

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*n-1
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?

Godzilla · 2019-01-18, 19:34

Quote:

Originally Posted by Batalov

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*n-1
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?

I try with small tests with the electronic calculator and pen and paper, and later with the computer, if I do not understand I ask.

retina · 2019-01-18, 19:47

Quote:

Originally Posted by Godzilla

@retina ... copy the code to a text file and save it with filename.py, open python and run it.

With names like equTwooo that ain't gonna happen.

Mysticial · 2019-01-18, 22:04

Quote:

Originally Posted by retina

This code is totally awesome

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?

Obligatory:

jrsousa2 · 2019-02-06, 15:10

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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2019-01-18, 19:21	#6
Godzilla May 2016 7×23 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.

2019-02-06, 15:10	#11
jrsousa2 Dec 2018 Miami 35₈ 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.