- Hi All: Here's my take on showing that there are at least two primes

between adjacent squares. Let me know what you think. My apologies

in advance if the tabular examples are scatterred. Has anyone else

been looking at squares and primes recently?

ARGUMENTS TO DETERMINE THE MINIMUM NUMBER OF PRIMES BETWEEN TWO

ADJACENT PERFECT SQUARES

This article presents a methodology leading to a proof that at

least two prime numbers exist between any adjacent perfect squares.

The basic part of this calculation is elementary. The keys to the

solution are in the observations and assumptions made along the way.

METHODOLOGY

First calculate the number of possible primes between any two

adjacent perfect squares, then calculate the number of possible

primary divisors for that set. Next use the most dense distribution

of divisors occurring anywhere in the complete number field, which is

the "mirror sequence" (ref note 1). Examples are given at the end

of this article to illustrate the methodology presented here.

DEFINITIONS

N^2 = the value of any perfect square, where N is an integer

from 2 to any N.

Squares interval = numbers between but not including N^2 and

(N+1)^2

Possible primes = all the odd numbers in the squares interval

Divisor = the denominator in any division operation

Primary divisors = unique divisor values that are less than or

equal to N

KE = the number of possible primes in the interval range if N

is even

KD = the number of possible primes in the interval range if N

is odd

LE = the number of primary odd divisors less than N, not

including 1, if N is even

LD = the number of primary odd divisors less than and including

N, not including 1, if N is odd

Mirror sequence = that sequence of primary odd divisors that is

centered at 2*the product of all

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

MS = the number of actual divisors in the mirror sequence

J = the minimum number of primes between any two adjacent

squares

CALCULATIONS equation number

KE =

2*N/2=N

(1)

KD = KE-1 =N-1 (ref Note 2)

(2)

LE = KE/2 -1 = N/2-1

(3)

LD = KD/2 -1 =(N-1)/2-1 (ref Note 2)

(4)

For N even:

MS = 2*LE = 2*(N/2-1) = N-2 (5)

J = KE MS = N (N-2) = 2

(6) answer

For N odd:

MS = 2*LD = 2*[(N-1)/2-1] (7)

J = KD MS = N-1 2*[(N-1)/2-1] = N-1 (N-1-2) = 2

(8) answer

NOTES:

1. The most dense distribution of divisors is determined by

examining randomly generated sequences, perturbations of the mirror

sequence, and actual divisor sets. For all variances, the divisor

sequence must be internally mathematically consistent, i.e. every

third odd number must be divisible by 3, every fifth odd number by 5,

etc. For the purposes of this argument, it is not necessary that a

divisor in the mirror sequence be a true divisor of the possible

prime with which it is paired. Using these rules, random sequences

create spaces when the mirror image set is generated if the original

set is not in numerical sequence. Further, perturbations of the

mirror sequence show that any change in the original sequence will

create one more space. It is possible to create a sequence N-4 long

when the divisors are not in numerical order. However the mirror

sequence is N-2 long over the interval N, and thus more dense. These

findings need to be formally presented before this 2 prime proof is

complete. Mirror sequences are illustrated in the examples.

2. KD and LD must be formulated differently than KE and LE to avoid

round-off inaccuracies

when calculating J. The reformulation can be made by

observing that, for odd N's, N^2+2N

is the last odd number in the squares interval, and is

divisible by N. Assigning N as divisor

of N^2+2N then diminishes the available possible primes by 1,

and also the available primary

divisors by 1.

EXAMPLES

Examples of possible primes (KE or KD) and their mirror sequence

(MS).

EVEN N ODD N

Code: X = no value

N = 12 N = 13

KE = 12 MS = 10 KD = 13 MS = 10

145 11 171 11

147 9 173 9

149 7 175 7

151 5 177 5

153 3 179 3

155 X 181 X

157 X 183 X

159 3 185 3

161 5 187 5

163 7 189 7

165 9 191 9

167 11 193 11

N^2+2N = 195 13*15

N = 2 N = 3

KE = 2 MS = 0 KD = 2 MS = 0

5 X 11 X

7 X 13 X

N^2+2N = 15 3*5

N = 4 N = 5

KE = 4 MS = 2 KD = 4 MS = 2

17 3 27 3

19 X 29 X

21 X 31 X

23 3 33 3

N^2+2N = 35 5*7

N = 1234E10000, or any even N

KE = N MS = N-2

N^2+1 N-1

N^2+3 N-3

N^2+5 N-5

N^2+7 N-7

* *

* *

* *

X X

X X

* *

* *

* *

(N+1)^2-8 N-7

(N+1)^2-6 N-5

(N+1)^2-4 N-3

(N+1)^2-2 N-1

Examples of calculations

Even N's

N 12 90 4566 1234E10000

KE 12 90 4566 1234E10000

LE 5 44 2282 617E10000

MS 10 88 4564 (1234E10000)-2

J 2 2 2 2

Odd N's

N 13 111 9979 123456789123

KD 12 110 9978 123456789122

LD 5 54 4988 61728394560

MS 10 108 9976 123456789120

J 2 2 2 2

END - Could you re-explain what divisors are? And hence what primary divisors

are? It's not clear from what you write, and (3) and (4).

I mean a divisor is defined as "denominator in any division operation".

That could be anything at all, couldn't it?

Presumably you're studying the set J of integers between n^2 and (n+1)^2,

for n even (there's no need to mess things up by studying even and odd

separately throughout - just look at the even). By divisors, I assume

(hope) you mean those integers which divide at least one member of J. Is

that right? Then the primary divisors are just a certain subset of the

divisors.

Please confirm that, or correct me if Im wrong.

Andy - Hi Andy ; Thanks for taking the time to review my post of 12/30.

Here are the answers to your questions.

Yes, I am studying the set of integers between N^2 and (N+1)^2,

more specifically the odd integers contained therein, as they are the

possible prime numbers in that set.

I agree that we can just look at the evens for now to study the

technique I am presenting.

I would call the KE values (for even N) the set of numbers I am

studying, not J. J is just one number, being equal to the minimum

number of primes between two squares.

The fancy footwork associated with the definition of divisors and

primary divisors is for the purpose of distinguishing between those

divisors of a number in the KE set that are less than or equal to the

square root of that number. For example, if a number in the KE set

were 255, then 3,5 and 17 would be divisors, but only 3 and 5 would

be primary divisors. I hope this helps. Thanks again for your

inputs. - Hi Bill,

I had the same confusion about your intention on divisors and now

that I see your reply, another question comes up. If by primary

divisors, you mean those that are < sqrt(x), you may have a problem

with the reasoning because, in the example of x=255, not only are 3 &

5 primary divisors, but also the trivial 1 & the not so trivial 15

would be defined as primary divisors. If by primary divisors you

mean only those divisors that are < sqrt(x) and prime, then the

definition excludes 15, but what then is the impact on your

argument?

-Dick

--- In primenumbers@yahoogroups.com, "billoscarson"

<billroscarson@p...> wrote:>

the

> Hi Andy ; Thanks for taking the time to review my post of 12/30.

>

> Here are the answers to your questions.

>

> Yes, I am studying the set of integers between N^2 and (N+1)^2,

> more specifically the odd integers contained therein, as they are

> possible prime numbers in that set.

and

> I agree that we can just look at the evens for now to study the

> technique I am presenting.

> I would call the KE values (for even N) the set of numbers I am

> studying, not J. J is just one number, being equal to the minimum

> number of primes between two squares.

> The fancy footwork associated with the definition of divisors

> primary divisors is for the purpose of distinguishing between those

the

> divisors of a number in the KE set that are less than or equal to

> square root of that number. For example, if a number in the KE set

> were 255, then 3,5 and 17 would be divisors, but only 3 and 5 would

> be primary divisors. I hope this helps. Thanks again for your

> inputs. - 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