# A new prime serie?

This is strange... I can sort of proove this.

( n(1/2 + 1/3 + ... + 1/pa) - (1/3 + 2/5 + ... + (a-1)/pa) minus all whole queries ) <= ½

--> n = p

If it's true and I was the first to find the serie; can I name it after me?

In that case i would like to name my equation the EO-equation.

But pa is the last existing prime between the two numbers 1 and n.

Thereby it's a bit difficult to find really big primes, but anyway.

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## Answers and Replies

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Originally posted by Sariaht
This is strange... I can sort of proove this.

mod1( n(1/2 + 1/3 + ... + 1/pa) - (1/2 + 2/3 + 3/5 + ... + (a-1)/pa) ) <= ½
I must say that I don't really understand what you are trying to say.

What do you mean with mod1( ... )?

By the way, are you aware that both the series 1/2 + 1/3 + 1/5 + ... + 1/pa and the series 1/2 + 2/3 + 3/5 + ... + (a-1)/pa diverge in the limit to infinity? Actually, the second series diverges considerably faster, so the total series ( n(1/2 + 1/3 + ... + 1/pa) - (1/2 + 2/3 + 3/5 + ... + (a-1)/pa) ) goes to $$-\infty$$ in the limit $$a\rightarrow\infty$$...

I ment modulus.

mod1(23.5) = .5

If the first prime is two, how ever you think, the risk is fifty fifty for the next two primes to have a factor two.

If the risk is higher than... Well I can't explain it without a new small numbertheory

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Originally posted by suyver
I must say that I don't really understand what you are trying to say.

What do you mean with mod1( ... )?

modulus 12.34 = .34

(Am I not right in this?)

Anyway what i ment was everything after the dot in tu.vxyz....

So .342563448 in 23.342563448 is modulus 23.342563448.

Originally posted by suyver
I must say that I don't really understand what you are trying to say.

What do you mean with mod1( ... )?

By the way, are you aware that both the series 1/2 + 1/3 + 1/5 + ... + 1/pa and the series 1/2 + 2/3 + 3/5 + ... + (a-1)/pa diverge in the limit to infinity? Actually, the second series diverges considerably faster, so the total series ( n(1/2 + 1/3 + ... + 1/pa) - (1/2 + 2/3 + 3/5 + ... + (a-1)/pa) ) goes to $$-\infty$$ in the limit $$a\rightarrow\infty$$...

The n don't cover the second serie, it only covers the first.

Change sign on $$-\infty$$

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Originally posted by Sariaht
the n don't cover the second serie, it only covers the first.
But the first series diverges for n->oo! If you won't take the limit in the second series as well, then the result will just be +oo.

Originally posted by Sariaht
Change sign on $$-\infty$$
I do not understand this.

Anyway: can't you just give an example with the first 10 (or so) primes?

Originally posted by Sariaht
modulus 12.34 = .34

(Am I not right in this?)

Edit: I have been informed that this is an accepted notation (see below).

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Originally posted by suyver
Sorry, I was wrong.

Good try anyway.

Good night Erik-Olof Wallman

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Hurkyl
Staff Emeritus
Gold Member
Actually, I've seen "mod 1" used in this way somewhat frequently.

Originally posted by Hurkyl
Actually, I've seen "mod 1" used in this way somewhat frequently.
Thank you, I stand corrected. (I had never seen it.)

HallsofIvy