### Friday, September 10, 2004

## Amateur math, neat relation

I'm going to give a derivation again, and talk about what I face with today's mathematicians so you can see how they operate and how they make certain that amateur mathematicians are cut out.

The simplest way to understand what I did is to consider the count of odd composites up to and including N that are divisible by 3, as

3C <= N

where C is the count of all composites, so C <= N/3, but every odd natural besides one can be written as 2k+1, where k is a natural, so you can replace C with 2k+1, to get

3(2k+1) <= N,

and solving you get

k <= (N-3)/6, and using floor(x) = [x], you have

k = [(N-3)/6] as the count of composites.

When I did my original research over two years ago, knowing that N was to be even, I only cared about odds so I used the equivalent of

3(2k+1) <= N-1

which gives k = [(N-4)/6], which does work as long as N is even.

Here the brackets mean to just take the integer part, like [1.3333...] = 1.

Now, notice that instead of 3 I can use any odd prime, so I have

p(2k+1)<=N, so k = [(N-p)/2p]

is the count of

BUT, you can also get that count by using [N/p] - [N/2p] - 1, so I have as a fundamental relationship that

[N/p] - [N/2p] - 1 = [(N-p)/2p]

and in general

[N/j] - [N/2j] - 1 = [(N-j)/2j]

where j is a natural greater than 1.

You can of course tweak it a bit so that you have

j(2k-1)<=N, so k = [(N+j)/2j]

and then the formula becomes

[N/j] - [N/2j] = [(N+j)/2j]

which was actually guessed by an anonymous poster a while back when I was talking about my "tidbits".

Some poster on sci.math made a post like that was a big deal, but in mathematics guessing is one thing, proving is another.

I've seen a lot of talk about these simple relations already, as if…as if mathematicians knew them before my work, and the discussions I started.

But that's how today's mathematicians operate.

They don't care about the math, only about what's said about the math.

They are all about prestige and holding on to their social position and the math be damned.

Go out, look in number theory texts or other math books and find that relation.

I hope you do find it, as if this is new then mathematicians are worse than pathetic, because if the people of sci.math who are still trying to push the idea that my work is useless or old never learned that relation but STILL are trying to downgrade it, then there's just no doubt that they hate mathematics.

Math is not a social game. People who truly love mathematics don't stop to first decide if they like someone before they accept a result.

That's the beauty of mathematics, but there are these people who call themselves mathematicians who other people call mathematicians who get paid, and they probably figure that if they don't play these social games then maybe they can't get their wife that car, or maybe they can't put their kid through school.

But they are willing to sacrifice you for their needs.

To them amateur mathematicians are worse than scum, and scarier than nuclear bombs.

But the math doesn't care about their mortgages. It doesn't care about their political needs. And it doesn't care about them.

Today's mathematicians have to hate mathematics because mathematics doesn't look out for them. It doesn't pay attention to their needs. It doesn't worry about their bills.

Mathematics just doesn't care, so they don't care about mathematics.

They may SAY they care, but that's just to get money and prestige.

Don't believe me? Then find

[N/j] - [N/2j] - 1 = [(N-j)/2j]

where j is a natural greater than 1, in a math textbook.

When you don't find it, understand the people you're dealing with, no matter what label they have on themselves.

They may call themselves mathematicians, but they hate mathematics.

The simplest way to understand what I did is to consider the count of odd composites up to and including N that are divisible by 3, as

3C <= N

where C is the count of all composites, so C <= N/3, but every odd natural besides one can be written as 2k+1, where k is a natural, so you can replace C with 2k+1, to get

3(2k+1) <= N,

and solving you get

k <= (N-3)/6, and using floor(x) = [x], you have

k = [(N-3)/6] as the count of composites.

When I did my original research over two years ago, knowing that N was to be even, I only cared about odds so I used the equivalent of

3(2k+1) <= N-1

which gives k = [(N-4)/6], which does work as long as N is even.

Here the brackets mean to just take the integer part, like [1.3333...] = 1.

Now, notice that instead of 3 I can use any odd prime, so I have

p(2k+1)<=N, so k = [(N-p)/2p]

is the count of

**odd**composites up to and including N that have p as a factor, where p is an odd prime.BUT, you can also get that count by using [N/p] - [N/2p] - 1, so I have as a fundamental relationship that

[N/p] - [N/2p] - 1 = [(N-p)/2p]

and in general

[N/j] - [N/2j] - 1 = [(N-j)/2j]

where j is a natural greater than 1.

You can of course tweak it a bit so that you have

j(2k-1)<=N, so k = [(N+j)/2j]

and then the formula becomes

[N/j] - [N/2j] = [(N+j)/2j]

which was actually guessed by an anonymous poster a while back when I was talking about my "tidbits".

Some poster on sci.math made a post like that was a big deal, but in mathematics guessing is one thing, proving is another.

I've seen a lot of talk about these simple relations already, as if…as if mathematicians knew them before my work, and the discussions I started.

But that's how today's mathematicians operate.

They don't care about the math, only about what's said about the math.

They are all about prestige and holding on to their social position and the math be damned.

Go out, look in number theory texts or other math books and find that relation.

I hope you do find it, as if this is new then mathematicians are worse than pathetic, because if the people of sci.math who are still trying to push the idea that my work is useless or old never learned that relation but STILL are trying to downgrade it, then there's just no doubt that they hate mathematics.

Math is not a social game. People who truly love mathematics don't stop to first decide if they like someone before they accept a result.

That's the beauty of mathematics, but there are these people who call themselves mathematicians who other people call mathematicians who get paid, and they probably figure that if they don't play these social games then maybe they can't get their wife that car, or maybe they can't put their kid through school.

But they are willing to sacrifice you for their needs.

To them amateur mathematicians are worse than scum, and scarier than nuclear bombs.

But the math doesn't care about their mortgages. It doesn't care about their political needs. And it doesn't care about them.

Today's mathematicians have to hate mathematics because mathematics doesn't look out for them. It doesn't pay attention to their needs. It doesn't worry about their bills.

Mathematics just doesn't care, so they don't care about mathematics.

They may SAY they care, but that's just to get money and prestige.

Don't believe me? Then find

[N/j] - [N/2j] - 1 = [(N-j)/2j]

where j is a natural greater than 1, in a math textbook.

When you don't find it, understand the people you're dealing with, no matter what label they have on themselves.

They may call themselves mathematicians, but they hate mathematics.