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Topic: [Mathqa]prime numbers
Replies: 0

[Mathqa]prime numbers
Posted: Feb 19, 2001 7:30 PM

I don't know how is the state of art about prime numbers, I write what I
found, and sorry if this relationships were already shown.
Let the following series (of integers) so formed:
F(n) = F(n-1)+F(n-2)-F(n-3)
Sequence of integers {1,2,3,4,5,6,7,8,9,10,.....}can be written as follow,
giving initial conditions
F(0)=1; F(1)=2; F(2)=3
so F(3)=3+2-1=4;
F(4)=4+3-2=5;
F(5)=5+4-3=6;
........................and so on.
Let now
initial conditions F(0)=1; F(1)=5; F(2)=7; [let 2 and 3 prime numbers
at priori fixed]

and the following one so formed:
H(n,n) = [F(n)]*[F(n)]
then
P(n) = {{F(n)}-{H(n,n)}}= {(F(n)}- {{F(n)}*{F(n)}}
{P(n)} is the Prime Numbers series [escluding 2 and 3 fixed as prime
numbers at priori].

{F(n)}=
{1,5,7,11,13,17,19,23,25,29,31,35,37,41,43,47,49,53,55,59,61,65,67,71,
73,77,79,83,85,89,91,95,97,101,103,107,109,113,115,119,121,125,127,131,
133, 137, 139,143,145,149,151,155,157,161,163,...}

{H(n,n)}={{F(n)}*{F(n)}=
{(5*5),(5*7),(7*7),(5*11),(5*13),(7*11),(5*17),(7*13),(5*19),(5*23),
(7*17),(11*11),(25*5),(7*19),(11*13),(5*29),(5*31),(7*23),...}=
={25,35,49,55,65,77,85,91,95,115,119,121,125,133,143,145,155,161,...}

[I haven't though till now how to esclude the doubles i.e. among (5*7) and
(7*5) I need just one, anyway is simply to formalize I mean because with
H(n,n) we
have a simmetric matrix and we need only of the elements over the main
diagonal, diagonal included, and multiple of 5 escluded]

1 5 7 11 13 17 19 ...
5 25 35 55 65 85 95 ...
7 35 49 77 91 119 133 ...
11 55 77 121 143 187 209 ...
13 65 91 143 169 221 247 ...
17 85 119 187 221 289 323 ...
19 95 133 209 247 323 361 ...
... .. .. .. .. .. ..
...

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

so

P(n)={P(n)}={{F(n)}-{{F(n)}*{F(n)}}={1,5,7,11,13,19,23,29,31,37,41,43,47,
53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,
151,157,163,...}
prime numbers series [escluding 2 and 3 supposed prime numbers at priori
fixed]

{F(n)}-{H(n,n)} is possible because H(n,n) is included in F(n)
and all its elements are contained in F(n).
I think that's easy to demostrate.
F(n) you can also write as shown:
F(n)=((6*n)-+1)={1,5,7,11,13,17,19,23,25,29,31,35,37,39,41,43,47,49,
53,55,59,61,65,67,71,73,77,79,83,85,89,91,95,97,...}
n=0,1,2,3,....infinite
and it contains all prime numbers escluding 2 and 3.
What I am tring to tell is that there is a way to 'clean' it obtaing
the prime number series using an analitic formulation and not using
some software or complex alghoritms or attempting.
This sequence includes all prime numbers [excluding 2 and 3],
but as already expained we need to filter out
all elements which are not primes.
Therefore H( ) is included in F(n) and not viceversa.
H( ) can be though as a matrix (boundary lines escluded) so bulit:
1 5 7 11 13 17 19 23 25 29 31...
5 25 35 55 65 ................
7 35 49 77 91 ................
11 55 77 121 143 .................
13 65 91 143 169 ..................
17 85 ...........................................
19 ....
23 ....
25 ...
29 ...
31 ...

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

H(n,n) is formed by only all internal numbers of this matrix, because the 2
semi-infinite boundary lines which permit to build this matrix (all
multiplications of each elements of F(n)) are the sequences of F(n) 2 times
written.
P(n) = F(n) - [F(n)]^2
or
{P(n)}= {F(n)} - {{F(n)}*{F(n)}}
Thinking of the implementation of this idea, F(n) is easy
to think and generate, but H(n,n) seen that is a matrix, is not a ordinate
sequence, in fact it can be generated
thinking of a cicle for ...next with index i,j = 1,5,7,.......N with N big
at pleasure.
After that it needs to order the sequence with a alghoritm, not so
difficult
I mean to do.
The main problem is to compare the 2 sequences so formed (F(n) and H(n,n)):
the only solution I can imagine at the moment is to take each element of
F(n),
big at pleasure,
and compare it with H(n,n) sequence: if it is not included in H(n,n), then
this
element
is a prime number.
Another problem to solve is how big at pleasure must be the maximal N of
index j,j
to be sure that the generated sequence of H(n,n) ordered 'covers' all
possibility.
I explain myself with an easy example:
if we take the number 65 or 85 or 91 belonging to F(n) sequence
and we compare these numbers with H(n,n) 'ordered' (just till 11; or
N=11),
we
make a mistake because we get:
F(n) = {1,5,7,11,13,.....,65, ...., 85,..........91,.....}
and H (n,n) = {25,35,49,55,77,121}
[ i.e H(n,n) for N=11 mean all multiplications of each element of F(n) for n
= 1
to N, or for n=1 to 11]

because N=11 doesn't cover all possibilities.
To cover all possibilities till 91 we need to take at least N=17. (see the
matrix)
So I mean we don't need to take all multiplications till 91 (i.e. N=91) but
a smaller number.
I haven't yet found the relationship which permits to be sure to 'cover'
all possibilities,
anyway called 'nf' each element of F(n) examinated and N the maximal number
of
indexes i,j
necessary to build the matrix H( ) to compare with 'nf' examinated, then:
N<nf
and for bigger numbers should be always more the difference between N and
nf
(Eulero studies about primes could help in this matter).
Concluding I don't know an efficient method how to decide whether an
integer H( ) belongs to the set F(n) or not, I mean it can be found at the
moment just comparing the 2 sequences.
Thinking of Fibonacci sequence the problem should be the same:
how can we say if a number big at pleasure belongs to Fibonacci sequence or
not?
Note that (obviously for properties of multiplications) having 3 initial
conditions in any lines of the matrix the next element follow always the
rule F(n)=F(n-1)+F(n-2)-F(n-3)
i.e. element 169=13*13 = F(n-1)+F(n-2)-F(n-3)= 143+91-65
143=11*13= 121+77-55 =13*11=91+65-13
I think this can be a nice property for alghoritmic implementations.
Another similar way to build the series H(n) is as
follow:
H(n,n)=H(k,nf)= (6k*(nf)-+(nf)) = {(6k*(nf)-+(nf))}={(6*n-+1)(6*k-+1)}
where n,k=1,2,3,...... and nf is each element of F(n)
i.e.
k=1 nf=5 H(1,5)= {25;35}
k=2 nf=5 H(2,5)= {55;65}
k=3 nf=5 H(3,5)= {85;95}
...........................
k=1 nf=7 H(1,7)= {35;49}
k=2 nf=7 H(2,7)= {77;91}
...........................
k=1 nf=11 H(1,11)= {55;77}
...........................

H H H H H H
F(n) k*6nf-nf k*6nf+nf
n 6*n-+1 k=1 k=1 k=2 k=2 k=3 k=3 ...
6*n-+1 1 5 7 11 13 17 19 ...
1 5 25 35 55 65 85 95 ...
1 7 35 49 77 91 119 133 ...
2 11 55 77 121 143 187 209 ...
2 13 65 91 143 169 221 247 ...
3 17 85 119 187 221 289 323 ...
3 19 95 133 209 247 323 361 ...
... .. .. .. .. .. .. ..
...............................................................

which can be written also as a triangle:

1
5 5
7 25 7
11 35 35 11
13 55 49 55 13
17 65 77 77 65 17
19 85 91 121 91 85 19

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

I am examinating this relationship for a large number (5000) of odd numbers
of the
kind 6*n-+1 (the sequence F(n)), and till now I haven't found any exception
to this
'rule' supposed. I mean for exception some odd prime number (escluding 3)
not
>included in F(n), and also some element of F(n) not filtered with H( ), or
to be clear
some 'nf' which is not filtered out by H( ), taking N of index i,j of the
matrix H( )
sufficiently big at pleasure.

Best regards,

Simone Caramel

p.s. sorry for english mistakes :-)

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