ado
Posts:
12
Registered:
12/8/04


[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(n1)+F(n2)F(n3) 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+21=4; F(4)=4+32=5; F(5)=5+43=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 semiinfinite 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(n1)+F(n2)F(n3) i.e. element 169=13*13 = F(n1)+F(n2)F(n3)= 143+9165 143=11*13= 121+7755 =13*11=91+6513 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} ...........................
If you follow this procedure you obtain the matrix already shown:
H H H H H H F(n) k*6nfnf 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
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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.
Waiting for some critical comments.
Best regards,
Simone Caramel
p.s. sorry for english mistakes :)
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