Problems & Puzzles: Puzzles

 Puzzle 827. A sieve for twin primes John Arioni sent the following nice puzzle closely related to Sunduram's sieve, Puzzle 821. This is a sketch of the matrix M you can construct using rules 1) and 2) 4 5 7 8 10 11 13 14 16 17... 7 8 12 13 17 18 22 23 27 28... 10 11 17 18 24 25 31 32 38 39... 13 14 22 23 31 32 40 41 49 50... 16 17 27 28 38 39 49 50 60 61... ... Rule 1) columns 1, 3, 5, ….... 2n-1....... M[r,c] = 2rc' + r + c' where c' = (c+1)/2 Rule 2) columns 2, 4, 6, ….... 2n ….... M[r,c] = M[r,c-1] +1 Q1. Show that any number x not present in the matrix produces the couple 2x-1,2x+1 of twin primes. Q2 How can you construct a matrix where any integer x not present in the matrix produces pairs of primes p, p+d ? ( d is any even integer ). Q3 Construct a simple algorithm to sieve pairs of primes p, p+d , given an integer d = 2n .

Contribution came from Emmanuel Vantighem

***

Emmanuel wrote:

Q1. The question should be : show that any number  x > 1 not present in the matrix produces the couple  2x-1, 2x+1  of twin primes.
This can then be proved as follows :
we can write :
M[r,c] = (2r+1)(c+1)/2 + r   if  c  is odd
M[r,c] = (2r+1)c/2 + r + 1   if  c  is even.
Thus :
2 M[r,c] + 1 = (2r+1)(c+2)  if  c  is odd
2 M[r,c] - 1 = (2r+1)(c+1)   if  c  is even.
Hence, if an element  x  is in the matrix, either  2x+1  or  2x-1  is composite.
Suppose  x  is a number > 1  such that  y = 2x-1  is composite.
Then y is odd and  > 1  and thus  y  can be written as a product of two odd numbers, say :
y = (2r+1)(2u+1), which is easily identified with  2M[r,2u] - 1.
Thus, x = M[r,2u]  is in the matrix.
Suppose  x  is a number > 1  such that  y = 2x+1  is composite.
Then we also can write  y  as a product of two odd numbers, i.e. :
y = (2r+1)(2u+1)  which is easy to identify with  2M[r,2u-1] + 1.
Hence, x = M[r,2u-1]  is an element of the matrix, QED.

Q2.
Given a number  k > 0, I define a matrix  M  which has the following property :
Every number  x > 1  not in the matrix has the property that  2x-1  and  2x+2k+1  are both prime.
I simply define the entries  M[r,c] by :
M[r,c] = (2r+1)(c+1)/2+r+1  if  c  is odd
M[r,c] = (2r+1)c/2+r-k    if  c  is even.
Example, when  k = 2, the matrix is :
5   2   8    5   11   8   14  11   ...
8   5  13  10  18  15  23  20   ...
11  8  18  15  25  22  32  29   ...
...     ...
Here,  4  is not in the matrix : 2*4-1 = 7  and  2*4+5 = 13  are both prime.

The proof is - mutatis mutandis - the same as in Q2.

Remark : you may take  k = 0  in  Q2.  Then  M  will not be the same as in Q1 : the columns with numbers  2i+1  and  2i  are simply interchanged.  Thus : the numbers  x  are the same in Q1 as in Q2.

***

ON April 30, John Arioni wrote:

Let q = (2n+1) be the generic odd integer (n =1,2,3,....).

The notations :

h(q) = n = (q-1)/2

H(q)= (n+1) = (q+1)/2

will be helpful in the following – trivially H(q)= h(q) +1 = h(q+2).

Q1

The matrix M can be split into the following two:

___Matrix A ________________ Matrix B_______

4 7 10 13 16 …..                   5 8 11 14 17 …...

7 12 17 22 27 ….                 . 8 13 18 23 28 ….

10 17 24 31 38 ….                11 18 25 32 39 …..

.......................... ...............................

Columns 1, 2, 3, ….c in A are columns 1, 3, 5, …...2c-1 of M

Columns 1, 2, 3, ….c in B are columns 2, 4, 6, .….. 2c of M

A[r,c] = 2rc+r+c B[r,c] = 2rc+r+c+1

But what are 2rc+r+c and 2rc+r+c+1 ?

Well, the product (2r+1)*(2c+1) = 4rc+2r+2c+1 with r and c varying

from 1 to infinite produces all of the odd composite integers, and:

h(4rc+2r +2c+1) = 2rc+r+c = A[r,c]

H(4rc+2r +2c+1) = 2rc+r+c+1 =B[r,c]

then, if q is any odd composite integer h(q) is an entry in A and H(q) is

an entry in B, while, if p is any prime number, h(p) is not an entry in A and

H(p) is not an entry in B.

Now, if p is a prime number then H(p) is not present in matrix B, but in the

case that p+2 is composite h(p+2) = H(p) is present in matrix A and then

H(p) is present in matrix M.

Differently if p+2 is prime then h(p+2) = H(p) is not present in matrix A as well

as it is not present in matrix B, which means that H(p) is not an entry of matrix M.

As: 2*H( p)–1 = p and 2*H( p)+1 = 2* h(p+2)+1 = p+2 , we can state that:

every x not present in the matrix produces a pair (2x-1),(2x+1) of twin primes.

Q2

Let's consider the following matrices A and C:

___Matrix A ________________ Matrix C_______

4 7 10 13 16 ….. 6 9 12 15 18 …...

7 12 17 22 27 ….. 9 14 19 24 29 ….

10 17 24 31 38 …. 12 19 26 33 40 …..

.......................... .................................

Entries in C are the corresponding in B increased by 1, then if q is the generic

odd composite integer q = (2r+1)*(2c+1), the entries of C are :

C[r,c] = B[r,c] +1 = (2rc+r+c+1)+1 = H(q)+1 = H(q+2) = h(q+4) = h(q)+2

Let's define AUC :

________Matrix AUC ________

4 6 7 9 10 12 13 15 16 18 …..

7 9 12 14 17 19 22 24 27 29 …..

10 12 17 19 24 26 31 33 38 40 …..

...........................................

Columns 1, 3, 5, …...2c-1 in AUC are columns 1, 2, 3, ….c of A

Columns 2, 4, 6, .….. 2c in AUC are columns 1, 2, 3, ….c of C

If p is any prime, then h(p)+2 is not present in C, but, if p+4 is composite, then

h(p+4) = h(p)+2 is present in matrix A: in this case h(p)+2 is present in AUC.

Differently if p+4 is prime then h(p+4) = h(p)+2 is not present in A as well as it

is not present in C, then h(p+4) is not present in AUC.

As p+4 = 2* h(p+4)+1 and p = 2* h(p+4)+1- 4, we can state that: every number

x not present in AUC produces a pair (2x-3),(2x+1) of primes ( p, p+4 ):

e.g. 3 produces 3,7 , 5 produces 7,11, 8 produces 13,17, and so on.

Now we are ready to generalize:

let d be any even integer and and q the generic odd composite integer:

we can construct a matrix (say Y) with entries Y[r,c] such that

Y[r,c] = h(q+d) = h(q)+d/2= 2rc+r+c+d/2

Let AUY be a matrix such that:

Columns 1, 3, 5, …...2c-1 are the columns 1, 2, 3, ….c of A

Columns 2, 4, 6, .….. 2c are the columns 1, 2, 3, ….c of Y

Well, if p is prime then h(p)+d/2 is not present in Y, but, if p+d is composite, then

h(p+d) = h(p)+d/2 is present in matrix A: in this case h(p)+d/2 is present in AUY.

Differently if p+d is prime then h(p+d) =h(p)+d/2 is not present in A as well as it is

not present in Y so h(p)+ d/2 is not present in AUY

As p+d = 2*h(p+d) +1 and p = 2* h(p+d)+1- d we can state that: every number x

not present in AUY produces a pair (2x+1-d),(2x+1) of primes ( p, p+d)

note: generally p+d is not the prime successive to p,

but it is simply a prime at a distance d from p.
i.e. d is not generally the same as gap.

for d=2 and d=4, d is is the same as gap

Q3

A simple algorithm to sieve pairs of primes p,p+d makes use of an array (arr( ) ) to

record all of the entries of matrix AUY above while computing A[r,c] = 2rc+r+ c and

Y[r,c] = 2rc+r+ c+d/2. Once all steps of calculus have been completed arr(i) =1 if i

is an entry of AUY, differently arr(i) = 0.

The algorithm operates on a list of integers whith upper limit = Sup.

d and Sup are parameters that are acquired in the input section

```A premise in order to better understand the algorithm:

As A[r,c] = 2rc+r+c = (2r+1)c+r , the entries, row by row, of A are:

A[1,c] = 3c+1 = h(6c+3)        (6c+3) is the generic odd multiple of 3
A[2,c] = 5c+2 = h(10c+5)    (10c+5) is the generic odd multiple of 5
A[3,c] = 7c+3 = h(14c+7)    (14c+7) is the generic odd multiple of 7
…..
A[r,c] = (2r+1)c+r
…..```

Here below the code in Visual Basic for the section of calculus.

row = 1

``` col = 1
While (2 * row + 1) * col + row < Int(Sup / 2)
oddinteger = (2 * row + 1)
littleh = oddinteger * col + row
While littleh - d /2 <= Int(Sup / 2)
arr(littleh) = 1
arr(littleh + d /2) = 1
littleh = littleh + oddinteger
Wend
row =row + 1
col = row                 '***
Wend

All we need, now, is to test arr(i) for i=1,2,3,....and print (2i+1-d),(2i+1)
every time we encounter a zero in arr(i).

' ***   col = row instead of col = 1
because  A and Y are symmetric matrices```

***

 Records   |  Conjectures  |  Problems  |  Puzzles