Is There a Proven Method for Finding Prime Numbers?

[Sariaht]

It was really close, perhaps the ways you can wright n on is >= the n-1:th prime. But how could i ever prove it?

 

[chronon]

7=3+2+2 It's rather a shame really

 

[Sariaht]

Strange...

 

[matt grime]

You're claiming the n'th partition number into non-empty sets is the n-1st prime? That'd be nice, but it isn't true. p_n is, however, the coefficient of x^n in

$$\prod_{k \geq 1} \frac{1}{1-x^k}$$

 

[arildno]

No need to get embarassed, Sariaht!
I thought it was a really cool idea which just happened to be wrong.

 

[Sariaht]

Perhaps if you...

Lets say the n'th partition number into non-empty sets is >= the n-1st prime!

That the ways you can wright n on is >= the n-1:th prime.

 

[matt grime]

There are many bounds for primes you could look up.

 

[Sariaht]

I made a simple equation for the ways you can wright n on:

nodpf(1 to a)*a - nodpf(1 to a)((1 to a) - 1) and add to this a

Primes and 1 excluded in (1 to a), if the current term is a prime or 1 then skip

nodpf = number of different prime factors, for instance 12 has the factors 2,3 and 3. The different prime factors i define as 2 and 3.

There is another way also, I will post it as soon as i figure it out. The one above is simpler to express.

For six the equation becomes:

(1)*6 - (1)(4 - 1)
+
(2)*6 - (2)(6 - 1)
+
6
=
11

For seven the equation becomes:

(1)*7 - (1)(4 - 1)
+
(2)*7 - (2)(6 - 1)
+
7
= 15

For eight the equation becomes:

(1)*8 - (1)(4 - 1)
+
(2)*8 - (2)(6 - 1)
+
(1)*8 - (1)(8 - 1)
+
8
=
20

And that is correct. the equation can be simplified into:
nodpf(1 to a)(a - ((1 to a) - 1)) and add to this a

For nine it looks like this:

nodpf(4)(9 - ((4) - 1))
+
nodpf(6)(9 - ((6) - 1))
+
nodpf(8)(9 - ((8) - 1))
+
nodpf(9)(9 - ((9) - 1))
+
9
=
1(9 - ((4) - 1))
+
2(9 - ((6) - 1))
+
1(9 - ((8) - 1))
+
1(9 - ((9) - 1))
+
9
=
9 - 3
+
18 - 10
+
9 - 7
+
9 - 8
+
9
=
54-28
=
26

 

[matt grime]

Bits of that don't make sense: what does primes excluded in (1 to a) mean?

 

[Sariaht]

I hope I answered the question in the last post after you asked.

 

[matt grime]

No, it still makes no sense in many places as a piece of English.