On 1/1/04 Andy wrote:

One of these days I'll learn to ignore these posts, but until then...

It's bad enough working out what the mirror sequence is. It's

impossible

from the way you've defined it, I had to look at the examples instead.

So you take all the odd numbers between (2n)^2 and (2n+1)^2, of which

there are of course 2n. You then write a sequence of pairs:

{(2n)^2+1, 2n-1},

{(2n)^2+3, 2n-3},

{(2n)^2+5, 2n-5},

...

{(2n)^2+2n-3, 3},

{(2n)^2+2n-1, 1},

{(2n+1)^2-2n, 1},

{(2n+1)^2-2n+2,3},

...

{(2n+1)^2-4, 2n-3}

{(2n+1)^2-2, 2n-1}

You then claim that, of the 2n terms in this sequence, 2n-2 of them

consist of 'actual divisors'? Meaning what? What divides what? It's

not

clear.

Andy

Hi Andy and all:

Hang in there Andy. You are almost at the end of this

apparently tortuous path I have imposed upon you, and my apologies

for that.

Let me correct a few of your comments. I am studying the odd

numbers between N^2 and (N+1)^2, not between (2N)^2 and (2N+1)^2.

You also have more terms in the equations than I do. I don't know

why, but I hope it will sort itself out from the comments below.

The first pair should read N^2+1 as the number to be divided, and

N-1 as the number to divide into it. Don't worry yet that N-1 likely

won't really divide into N^2+1. Read on to find out why.

Glad to hear that you got through the definition of "mirror

sequence" OK. I'll expand and illustrate that concept now for others

who might not understand what I said, which will lead right into your

question about what divides what. You can skip to the heading WHAT

DIVIDES WHAT if you have the mirror sequence under your belt.

For those of you who are catching up (if anybody), refer to

message # 14289 for my initial submittal, and message #14307 for the

clarification of the first part of my arguments, where I left off

with values for KE and LE for the case N = 12.

MIRROR SEQUENCE clarification

As I stated in my original submittal, the mirror sequence is a

sequence of primary odd divisors that is centered at 2*the product of

all odd integers to N and spans +/- N from that center. Continuing

with N=12 as the example, the mirror sequence set is centered at

2*3*5*7*9*11= 20790. The span + or N from 20790 is 20778 through

20802. Using odd numbers only, we end up with this series of odd

numbers in that interval:

20779 20781 20783 20785 20787 20789 20791 20793 20795

20797 20799 20801

Let's call this subset (20779 through 20801) of the complete positive

integer set the reference set (REF set) just to give it a name.

Continuing on, we know that 20790 is divisible by all the numbers we

multiplied to get that product, so we then know that a number 3 away

from the center of 20790 is divisible by 3 (i.e. 20787 and 20793), a

number 5 away is divisible by 5, etc. We will want to use only the

divisors in the LE set, (3,5,7,9,11) since we are still talking about

N=12. Now lets match up the REF set with it's divisors from the LE

set.

REF set

20779 20781 20783 20785 20787 20789 20791 20793 20795 20797

20799 20801

LE set values

11 9 7 5 3 X X 3 5 7

9 11

Note that we have used the full LE set of values twice. Note too

that 11 actually divides 20779 and 20801, 9 actually divides

20781and 20799, 7 actually divides 20783 and 20797, etc. The two

X's in the middle indicate there are no odd values in the LE set that

divide into them. Of course, there are odd divisors higher than 11

that may divide 20789 and/or 20791, but let's not bother to check,

since it is not germane to this argument - for N=12, we are limited

to divisors up to 11.

MS, the number of actual divisors in the mirror sequence is:

MS =2*LE, per equation 5 in my initial post.

WHAT DIVIDES WHAT clarification

Why have we gone through the above exercise to find divisors of a

set of numbers from 20779 to 20801 (the REF set), when we are

analyzing numbers between 145 and 167, and to boot, have chosen to

not investigate some possible divisors in the REF set? Because we

are after an upper bound case of divisors in an interval 2N. The

mirror sequence is the result. The upper bound of divisors in the

interval will allow us to calculate the lower bound of primes in the

interval, which is what this argument is all about.

Now that we have the above sequence of divisors containing 10

divisors and two spaces (X's), we will use it in the range from 145

to 167 as the most dense distribution of divisors that could be

amassed in that range, whether they properly divide numbers in the KE

set OR NOT. But that's OK. Here's how that looks:

KE set 145 147 149 151 153 155 157 159 161 163 165

167 (the set of numbers under study)

MS set 11 9 7 5 3 X X 3 5 7 9

11 (the mirror sequence divisors)

Of course we know that 11 does not divide 145, 9 does not divide

147, etc (3 happens to divide 153 and 159,and that's OK too, but not

significant). The point is that we have applied the most dense set

of divisors to the 2N interval 145 thru 167, yet there are still two

more numbers to divide than we possibly could have divisors for.

Therefore we must have at least two primes between these squares.

This argument holds for all squares.

Regards, Bill- The best way to avoid torturing a mathematician is to say nothing, or

maybe talk about the weather. Barring that, clarity and conciseness

expressed in terms familiar to them is the only medicine that eases

the pain. Remember, short does not guarantee concise. Even us

amateur geniuses aren't smart enough to always know (or even

discover) the correct mathematical term to use in a given instance.

But, even though we might not have the correct terminology at our

disposal, making up new terminology only makes matters worse and

should be avoided if possible. The statement "odd divisors not

exceeding n" would be understood, whereas a new term "primary

divisors" requires a definition and increases the

reader's "conceptual burden" <--Do you know what that term means?,

neither do I.

The mirror sequence you speak of is well known and plays a role in

the foundations of many theorems and allows prime hunters to save

tons of computer cycles, which must be a lot since they don't have

mass.

Suggested reading - Wilson's Theorem, Quadratic Reciprocity,

LeGendre's Prime Counting function, the theories of linear and

quadratic congruences, theory of quadratic residues...

Keywords - primorial, factorial, sigma divisor function, Euler

Totient function, Jacobi symbol...

...anybody else have suggestions?

> MIRROR SEQUENCE clarification

of

> As I stated in my original submittal, the mirror sequence is a

> sequence of primary odd divisors that is centered at 2*the product

> all odd integers to N and spans +/- N from that center.

The divisor structure you describe centered on ...20789, 20791... ,

is exactly the same as that of ...-1, 1... , there is no need to look

at such a large number only for this information. That number, by

the way, = n!! for odd n, where !! is the double factorial function.

The density distribution of all odd primes <= sqrt(n) as relates to

their occurence as a factor in a sequence of n consecutive odd

integers, is neither maximum nor a minimum when centered over {-

n,...,n} and cannot be used as a bound.

Here's an example

For, {1,2,3,4,5,6,7,8,9}, 3 divides elements {3,6,9}, 3 out 9

elements, density=1/3

For, {1,3,5,7,9,11,13,15}, same result, 3 divides 3 out of 9 elements

For, {15,17,19,21,23,25,27}, 3 divides 3 out of 7 elements > 1/3

For, {1,3,5,7,9,11,13,15,17,19}, 3 divides 3 out of 10 elements < 1/3

The pattern within n odd numbers centered on n!! is the exact same

pattern as the n odd numbers centered on zero. All such patterns for

any given span of n consecutive odd numbers, transposed say, a

distance x from the center point at 0, will also occur as a mirror

image at n!!-x, this is true. In fact, all unique, possible

combinations are exhausted as the center point moves from 0 to 2 to

4...to (n!!-2) to n!! Despite being "equivalent", the mirror image

pairs are unique from each other within our frame of reference of

increasing odd integers. Except that the sequences centered on 0 and

n!! are not only equivalent, the are equal and identical. The

sequence centered on 3 is identical to that centered on n!! + 3,

while it's mirror image occurs at n!!-3 and repeats every k*n!!

transposition of it's center, k = -inf, +inf. Thus there are n!!

unique divisor distribution patterns for a sequence of n consecutive

odd integers when only the odd divisors not exceeding n are

considered in the make-up of the pattern. If mirror images are not

considered unique, there are (n!!+1)/2 unique distributions.

> Now lets match up the REF set with it's divisors from the LE

20797

> set.

>

> REF set

> 20779 20781 20783 20785 20787 20789 20791 20793 20795

> 20799 20801

7

>

> LE set values

> 11 9 7 5 3 X X 3 5

> 9 11

Au contraire.. there must be and there are odd divisors higher than

>

> Of course, there are odd divisors higher than 11

> that may divide 20789 and/or 20791, but let's not bother to check,

> since it is not germane to this argument

11 that do divide those numbers. The same is true for most all odd

numbers greater than 11. And one must bother to check them, or at

least account for them, before a conclusion can be drawn.

In a nutshell, I think you are focusing on numbers relatively prime

to n as if all of them were primes, but only a subset of the integers

relatively prime to n are actually primes. It almost looks that way

for small n, but in reality, the ratio of primes < n over integers

less than and relatively prime to n, tends to zero with increasing n.

-Dick