## Search found 91 matches

- Mon Jun 18, 2012 3:55 pm
- Forum: Number Theory
- Topic: Both primeness
- Replies:
**14** - Views:
**8142**

### Re: Both primeness

Factoring this number Name: RSA-2048 Digits: 617 Digit Sum: 2738 2519590847565789349402718324004839857142928212620403202777713783604366202070759555626401852588078440691829064124951508218929855914917618450280848912007284499268739280728777673597141834727026189637501497182469116507761337985909570009733...

- Sun Jun 17, 2012 4:10 pm
- Forum: Game Theory
- Topic: Blackjack dice game variant
- Replies:
**4** - Views:
**12483**

### Re: Blackjack dice game variant

Little up for this game.

I still think that the game is deeply abstract : if we move smartly we win.

I still think that the game is deeply abstract : if we move smartly we win.

- Fri Jun 15, 2012 2:34 pm
- Forum: Number Theory
- Topic: Both primeness
- Replies:
**14** - Views:
**8142**

### Re: Both primeness

Is there something wrong ?

Thank you for any comment.

Thank you for any comment.

- Wed May 30, 2012 3:57 pm
- Forum: Number Theory
- Topic: Both primeness
- Replies:
**14** - Views:
**8142**

### Re: Both primeness

Thousand thanx Mister Hk, I do not know how to thank you. You helped me to solve a big problem : the factorization of large numbers. Here is my idea where all it starts : It is easy to compute any n odd semi-prime as : m^2 + r where m=int(sqrt(n)) and r=remainder Example : n=1633=40^2 + 33 Let us as...

- Wed May 30, 2012 1:41 pm
- Forum: Number Theory
- Topic: Both primeness
- Replies:
**14** - Views:
**8142**

### Re: Both primeness

No clue yet?

Maybe the solution is unique.

Maybe the solution is unique.

- Sat May 26, 2012 4:18 pm
- Forum: Number Theory
- Topic: Both primeness
- Replies:
**14** - Views:
**8142**

### Re: Both primeness

Let n=(a^2+1)*(a^4-a^2+1)*x^2 - (2*b*a^3)*xy + (b+1)*(b-1)*y^2

For a=2 and b=4

n=65*x^2 - 64*xy + 15*y^2=(5x-3y)*(13x-5y)

Can you find other values of (a,b) such as n could be factorized.

a,b >=1

a not equal to b

Thank you for any clue.

For a=2 and b=4

n=65*x^2 - 64*xy + 15*y^2=(5x-3y)*(13x-5y)

Can you find other values of (a,b) such as n could be factorized.

a,b >=1

a not equal to b

Thank you for any clue.

- Wed May 23, 2012 4:54 pm
- Forum: Number Theory
- Topic: Both primeness
- Replies:
**14** - Views:
**8142**

### Re: Both primeness

You are right.

Anyway finding coefficients that gives 100% semi prime is certainly impossible.

So I will try to find coefficients such as for any values of x and y we will obtain the maximal density of semi-prime at least for some large interval.

Still thinking for now.

Anyway finding coefficients that gives 100% semi prime is certainly impossible.

So I will try to find coefficients such as for any values of x and y we will obtain the maximal density of semi-prime at least for some large interval.

Still thinking for now.

- Wed May 23, 2012 3:05 pm
- Forum: Number Theory
- Topic: Both primeness
- Replies:
**14** - Views:
**8142**

### Re: Both primeness

If we can find k sets of 4 coefficients covering all the odd semi prime numbers then we can then produce k form of p+q. And if we can prove that all even numbers are covered by such form then Goldbach conjecture will be solved. In my case the coefficients were (5,-3,13,-5) covering 1/3 of all semi p...

- Tue May 22, 2012 3:48 pm
- Forum: Number Theory
- Topic: Both primeness
- Replies:
**14** - Views:
**8142**

### Re: Both primeness

Thank you very much to you both.

Now if we add p+q=5x-3y+13x-5y=2*(9x-4y) how many even numbers can we write as sum of 2 prime ? (link with Goldbach conjecture)

How many odd semi-prime numbers can we write as (5x-3y)*(13x-5y)?

Thank you for any help.

Now if we add p+q=5x-3y+13x-5y=2*(9x-4y) how many even numbers can we write as sum of 2 prime ? (link with Goldbach conjecture)

How many odd semi-prime numbers can we write as (5x-3y)*(13x-5y)?

Thank you for any help.

- Sat May 19, 2012 3:36 pm
- Forum: Number Theory
- Topic: Both primeness
- Replies:
**14** - Views:
**8142**

### Both primeness

Hi everybody, p=5x-3y q=13x-5y x,y negative or positive numbers p and q both odd prime Find all the couples p,q Are they infinitely many? Can we write all the prime numbers the way above? Example : x=7 y=4 p=5*7-3*4=35-12=23 q=13*7-5*4=91-20=71 p=23 q=71 The couple (p,q)=(23,71) is valid x=3 y=2 p=5...

- Fri May 04, 2012 2:26 pm
- Forum: Number Theory
- Topic: Prime of the form p=(n+2k+1)! - n! + (2k+1)
- Replies:
**3** - Views:
**3742**

### Re: Prime of the form p=(n+2k+1)! - n! + (2k+1)

No answer yet ?

- Fri May 04, 2012 2:24 pm
- Forum: Combinatorics
- Topic: Word combination
- Replies:
**3** - Views:
**4574**

### Re: Word combination

Thank you for your comment.

- Thu May 03, 2012 6:52 pm
- Forum: Combinatorics
- Topic: Word combination
- Replies:
**3** - Views:
**4574**

### Word combination

Hi everybody, We have 5 words A=blue (4 letters) B=often (5 letters) C=memory (6 letters) D=glad (4 letters) E=curve (5 letters) So 24 letters How many ways can we place them in 24 squares in a row such as : - each word could be read from left to right - all the squares filled - one letter per squar...

- Wed Apr 25, 2012 9:31 pm
- Forum: Number Theory
- Topic: Mertens function
- Replies:
**11** - Views:
**9194**

### Re: Mertens function

There is one thing to keep in mind :

the number of elements of the sequence A(n) is = n(n+1)/2

Example n=5

Card (A(n))=(5*6)/2=30/2=15

I did not know how to express it.

the number of elements of the sequence A(n) is = n(n+1)/2

Example n=5

Card (A(n))=(5*6)/2=30/2=15

I did not know how to express it.

- Wed Apr 25, 2012 9:28 pm
- Forum: Number Theory
- Topic: Mertens function
- Replies:
**11** - Views:
**9194**

### Re: Mertens function

Grouping terms in pairs of two easily gives $a(2n) = -n$, and then it is immediate that $a(2n+1)=n+1$ (you omitted a minus 1 in your $a(5)$, so this is in fact correct). Not sure what $a(n)$ represents, though. Thank you for your comment. A(n) is a sequence simulating the worst case of Mertens func...

- Wed Apr 25, 2012 6:17 pm
- Forum: Number Theory
- Topic: Mertens function
- Replies:
**11** - Views:
**9194**

### Re: Mertens function

Here is my Mertens-clone sequence. A(n)=+1-1-1+1+1+1-1-1-1-1........+1+1+1....+1((n-1) times)-1-1-1...-1(n times) We start by +1 then we change the sign (-1) 2 times then we rechange the sign (+1) 3 times and so on. Example A(5) = +1-1-1+1+1+1-1-1-1-1+1+1+1+1+1=+3 (corrected) In fact the idea is tha...

- Sun Apr 22, 2012 8:56 pm
- Forum: Number Theory
- Topic: Mertens function
- Replies:
**11** - Views:
**9194**

### Re: Mertens function

Thank you for your comment. I tried without success to reorder the remaining 2 sets such as I could remove other couples. I changed my mind so now I think I have found a way to calculate the absolute bounds of M(n) : inf and sup. I mean by absolute : M(n) can reach them but never never exceed them n...

- Sat Apr 21, 2012 6:33 pm
- Forum: Number Theory
- Topic: Mertens function
- Replies:
**11** - Views:
**9194**

### Re: Mertens function

Here is how I found the formula. Let us split n in infinite sets. Set 1 : 1,3,5,7,......,2k+1 (set of odd numbers) Set 2 : we multiply set 1 by 2 so 2,4,6,10,12,........2(2k+1) Set 3 : we multiply set 1 by 4 Set 4 : we multiply set 1 by 8 and so on All the natural numbers will be used only once. Her...

- Mon Apr 16, 2012 1:15 am
- Forum: Number Theory
- Topic: Prime of the form p=(n+2k+1)! - n! + (2k+1)
- Replies:
**3** - Views:
**3742**

### Re: Prime of the form p=(n+2k+1)! - n! + (2k+1)

What is really strange is that we can conjecture there are infintely many.

I need confirmation anyway.

Maybe some expert among you will help me understand why?

Thank you.

I need confirmation anyway.

Maybe some expert among you will help me understand why?

Thank you.

- Sun Apr 15, 2012 11:43 pm
- Forum: Number Theory
- Topic: Prime of the form p=(n+2k+1)! - n! + (2k+1)
- Replies:
**3** - Views:
**3742**

### Prime of the form p=(n+2k+1)! - n! + (2k+1)

Let p=(n+2k+1)! - n! + (2k+1)

How many solutions (n,k)?

Are they infinitely many?

How many solutions (n,k)?

Are they infinitely many?