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Valentine's Gold - Goldbach's_conjecture

Feb 14, 2024 - 19:47:16 (153d 16h)

Feb 14, 2024 - 19:47:16 (153d 16h)

Dear Challengers,

Recently i thought about the Goldbach's conjecture.

I wanted to play with numbers a little and tehron and i eventually made a challenge out of it.

We hope you will like this little programming excercise and wish you...

Happy Challenging!

- gizmore and tehron

PS: My proof to the conjecture:

1) Adding two odd numbers yields an even number. Adding 0.5 to the probabilty that an even number is covered.

2) Something with log(n)

3) ????

4) Profit ;)

Recently i thought about the Goldbach's conjecture.

I wanted to play with numbers a little and tehron and i eventually made a challenge out of it.

We hope you will like this little programming excercise and wish you...

Happy Challenging!

- gizmore and tehron

PS: My proof to the conjecture:

1) Adding two odd numbers yields an even number. Adding 0.5 to the probabilty that an even number is covered.

2) Something with log(n)

3) ????

4) Profit ;)

The geeks shall inherit the properties and methods of object earth.

Global Rank: 251

Totalscore: 87349

Posts: 1648

Thanks: 1344

UpVotes: 896

Registered: 16y 153d

Last Seen: 11h 8m

The User is Offline

RE: Valentine's Gold - Goldbach's_conjecture

Feb 18, 2024 - 16:42:12 (149d 19h)

Feb 18, 2024 - 16:42:12 (149d 19h)

I thought more about a proof and i just came up with an idea to disproof it, which i need to calculate.

Let us assume the number of even numbers called "eve(N)" is N/2

Let us assume the number of primes called "p(N)" is N/log(N)

Then, the number of possible combinations called "comb(N)" is p(N)^2

maybe there is a spot somewhere in N where the follwing happens: eve(N) > comb(N)

This would mean that there are simply more even numbers than possible prime combinations for a large N.

This would disproof the conjecture.

But i guess eve(N) is always smaller than comb(N)...

- gizmore

Let us assume the number of even numbers called "eve(N)" is N/2

Let us assume the number of primes called "p(N)" is N/log(N)

Then, the number of possible combinations called "comb(N)" is p(N)^2

maybe there is a spot somewhere in N where the follwing happens: eve(N) > comb(N)

This would mean that there are simply more even numbers than possible prime combinations for a large N.

This would disproof the conjecture.

But i guess eve(N) is always smaller than comb(N)...

- gizmore

The geeks shall inherit the properties and methods of object earth.

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