## Number crunch

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• We hear a lot about n^2+1 containing a infinite number of primes. A similar function may well be: floor(n^(3/2))+/-k (and obviously 3/2 is a variable too) Does
Message 1 of 6 , May 1, 2002
We hear a lot about n^2+1 containing a infinite number of primes.

A similar function may well be:

floor(n^(3/2))+/-k

(and obviously 3/2 is a variable too)

Does this function, or any variants, contain an infinite nuber of primes?

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• ... Good question. The forms grow slowly enough for them to statistically be expected to contain an infinite number of primes (anything sub-exponential has
Message 2 of 6 , May 1, 2002
--- Jon Perry <perry@...> wrote:
> We hear a lot about n^2+1 containing a infinite number of primes.
>
> A similar function may well be:
>
> floor(n^(3/2))+/-k
>
> (and obviously 3/2 is a variable too)
>
> Does this function, or any variants, contain an infinite nuber of
> primes?

Good question. The forms grow slowly enough for them to statistically
be expected to contain an infinite number of primes (anything
sub-exponential has this property).

Thay also probably have no divisor properties which would cause them
to deviate from a simple statistically predicted density.
(which isn't true about a lot of forms).

Phil

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• I came up with this: integral(x=1,infinity of sigma(n=1,2 of x^n)) contains all the primes, and further more: integral(x=1,infinity of sigma(n=1,2 of
Message 3 of 6 , May 2, 2002
I came up with this:

integral(x=1,infinity of sigma(n=1,2 of x^n)) contains all the primes, and
further more:

integral(x=1,infinity of sigma(n=1,2 of x^n))==sigma(n=1,2 of
integral(x=1,infinity of x^n)

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: Phil Carmody [mailto:thefatphil@...]
Sent: 01 May 2002 23:08

--- Jon Perry <perry@...> wrote:
> We hear a lot about n^2+1 containing a infinite number of primes.
>
> A similar function may well be:
>
> floor(n^(3/2))+/-k
>
> (and obviously 3/2 is a variable too)
>
> Does this function, or any variants, contain an infinite nuber of
> primes?

Good question. The forms grow slowly enough for them to statistically
be expected to contain an infinite number of primes (anything
sub-exponential has this property).

Thay also probably have no divisor properties which would cause them
to deviate from a simple statistically predicted density.
(which isn't true about a lot of forms).

Phil

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• ... I m fighting your notation somewhat. I assume sigma is summation? Is sigma(n=1,2 of x^n) just x+x^2 = x(x+1) ? The integral from 1..+oo of x(x+1) diverges.
Message 4 of 6 , May 2, 2002
--- Jon Perry <perry@...> wrote:
> I came up with this:
>
> integral(x=1,infinity of sigma(n=1,2 of x^n)) contains all the
> primes, and
> further more:
>
> integral(x=1,infinity of sigma(n=1,2 of x^n))==sigma(n=1,2 of
> integral(x=1,infinity of x^n)

I'm fighting your notation somewhat. I assume sigma is summation?
Is sigma(n=1,2 of x^n) just x+x^2 = x(x+1) ?
The integral from 1..+oo of x(x+1) diverges.
What is "contains" supposed to mean in the context "+oo contains all
the primes"?

Phil

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• First, a couple of typos: Should be intergral(n=1+e,...) as the integral from 1 includes Z, and is trivial. Also the function should be floor(x^n), not x^n,
Message 5 of 6 , May 3, 2002
First, a couple of typos:

Should be intergral(n=1+e,...)

as the integral from 1 includes Z, and is trivial.

Also the function should be floor(x^n), not x^n, although it could equally
be any variety.

Second, the integral can be refined:

As n^2>=2n for n>=2, (hence by Bertrand there is a prime), we can use the
identity:

n^x=2n

Taking logs

xlogn=log2+logn

x=log2/logn + 1

Hence we can use the equation:

sigma(n=1,infinity, of integral(x=1+e,log2/logn + 1, of floor(n^x))

And this contains all the primes.

What can the bounds be reduced to?

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• I got a bit twisted. integral(x=1+e,2, of sigma(n=1,infinity, of floor(n^x))) i.e. the numbers: 1^(1+e) through to 1^2 + 2^(1+e) through to 2^2 + ... The point
Message 6 of 6 , May 3, 2002
I got a bit twisted.

integral(x=1+e,2, of sigma(n=1,infinity, of floor(n^x)))

i.e. the numbers:

1^(1+e) through to 1^2
+
2^(1+e) through to 2^2
+
...

The point being that by Bertands, every prime is represented by the
integral, (and the sum is infinite, but I am not currently concerned with
this).

The second statement is the identity that the sigma and integral are
interchangeable:

sigma(n=1,infinity, of integral(x=1+e,2, of floor(n^x)))

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: Phil Carmody [mailto:thefatphil@...]
Sent: 02 May 2002 20:05

--- Jon Perry <perry@...> wrote:
> I came up with this:
>
> integral(x=1,infinity of sigma(n=1,2 of x^n)) contains all the
> primes, and
> further more:
>
> integral(x=1,infinity of sigma(n=1,2 of x^n))==sigma(n=1,2 of
> integral(x=1,infinity of x^n)

I'm fighting your notation somewhat. I assume sigma is summation?
Is sigma(n=1,2 of x^n) just x+x^2 = x(x+1) ?
The integral from 1..+oo of x(x+1) diverges.
What is "contains" supposed to mean in the context "+oo contains all
the primes"?

Phil

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