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Curious about a pattern to prime products.

  1. Jun 13, 2005 #1
    Curious about this pattern.

    list(6N,-1,+1)less list((6n,-1,+1)*(6N,-1,+1)) produces 100% primes =>5 for as far as I am able to take it.

    Sorry the dwgs would not copy so have added attachments.

    This pattern of products repeats by value (first digit set, second digit set, etc) and by position so as I am humbled by not knowing enough to take it much farther on my own. just an average guy looking at some patterns and do not know just how to present them. I try but the brain ain't what it was.

    A=6N,-1,+1 gives a list 1,5,7,11,13… for values of N=0:step 1 and for A being a whole real number. This list A contains Primes and Prime Products >=5. The first Prime Product in the list A is 25. (5*5) and so on as you build an array using (list A)*(list A) to give a complete list of Prime Products in list B. (within the array are duplicates as a*b=c and b*a=c and these duplicates are separated or isolated as values of n2 ) Then if you remove list B from list A you have 100% primes in sequence and without any exceptions that I have found. Exploring this farther I find that the individual product list for value A is 6NA,-A,+A. example would be A=5: N=0:step1 begins with value A step 1,+1,..gives 25,35,55,65,..


    Please note that the products of 5 are underlined and form a grid that is consistent throughout the lists and arrays. As is true for all values A with horizontal and vertical columns being first and second products etc. Although not shown here the positional
    Sequence is identical.

    Question: Could these patterns of the list A, of the product list B and the positional of each be a basis for a derivable pattern to the Prime. All products sequence both in value and position and the positional placements are fixed and all repeat in finite sets as 2,102,1002…5,155,1505…1,31,301,3001. For the 1 digit set you need only look at the array above and note the value of the first digit repeats in sets of 10 out of 30 as (1,5,7,1,3,7,9,3,5,9) for the two digit set you must go to 100 out of 300, etc. As all appear to be finite repeating sets and if you only consider the positional array then it would seem that the positional array would also repeat and allow the removal of products not by value but by positions that repeat.

    I simply do not know and would like an opinion please. I am just an average Bob that was curious and without any math or programming background needed to test this.
    Thank you in advance
    Bob.
     

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    Last edited: Jun 13, 2005
  2. jcsd
  3. Jun 13, 2005 #2

    shmoe

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    All primes greater than five will be of the form 6n+1 or 6n-1, since every other option will be divisible by either 2 or 3. If you have a composite of this form, then it's factors must also be of this same form, other wise the factors would be divisible by 2 or 3. So if you take your list A and remove everything on it thats a product of two elements on this list (except where 1 is a factor), you will then have all the primes and only primes left. I don't know if this answers your question as I don't really understand the latter half of your post. Your attatched document is garbled for me, so that's not really helping.
     
  4. Jun 14, 2005 #3
    Brain dead tonight....messed up the post so the attachments hopefully will expand a little. 6N-1,+1 is the list A. A list * A list= B list. Remove all values of B list from A list and you have a sequence of Primes =>5. Products of values of A list can be generated using 6NA,-A,+A. The products of A increment in sets of 10*10k out of 30A*10k as for 5 having sets of 10 products out of 150 and for 7 it would be 210. Also the products of A are A(+4A,+2A) which for 5 would be (+20 and +10). This last is where I am going. 5 is position 2 on list A and 25 ( the first product of 5) is position 9 and then 12 which is then 2(+7,+3) positionally. for 7 the position is 3 (+9,+5) Even positions are converted as v5,p2 gives (5+2) step 7 and (5-2) step 3. and odd is converted as v7,p3 gives (7+3)-1 step 9 and (7-3)+1 is step 5. This positional sequence is independent and sets up the condition whereby a "hit" is a product and a miss is a prime in list A for values =>5. I believe that the repeating patterns support a value based repeat but a positional pattern base should develope a harmonic of repeatability. ie the position for v7 is p3 and the 103p position is 307v and for v11 you have p4 and the 104p is 311. so what happens at p1003 or p1004 and do they need to be calculated. Hope this helps explain a little better.
     
  5. Jun 14, 2005 #4
    you gott to explain your list better of A or B...btw word of advice i wouldn't post such a thing in a public place because if one day you are right some one may steal the idea from you. Go to a close university and ask the number theorists in there.
     
  6. Jun 14, 2005 #5

    shmoe

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    This is fine, though the notation your using for these lists is confusing in places. The rest of what you're saying is also confusing, you're introducing many symbols (v,p, etc.) and not explaining what they are.

    However, what you're saying looks vaguely like something that's recently been beaten to death here. Take a look at https://www.physicsforums.com/showthread.php?t=66465 some of it's loopy but does anything look familiar?
     
  7. Jun 19, 2005 #6

    HAP

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    Well, you have a listA that contains all numbers that can be descriped by (6N, +/- 1). Then you make another listB that contains all numbers from listA * listA or in other words (6N, +/- 1)*(6M, +/- 1) = (36*N*M, -/+ 6N, +/- 6M, +/- 1)... And then you claim that a primelist (listP) can be descriped by: listA without listB.

    But your generated listP includes an unlimited number of non-primes!

    Let z_n be (6N, +/- 1)^n. Then you have all z_1 in your listA and remove all z_2 from it. But you have to be aware that all z_3 (plus many more) are still in listA and do in fact include non-primes. The first example of this is (6*21-1)=125=5^3.
     
  8. Jun 19, 2005 #7

    shmoe

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    Assuming I understand what he's saying, then no, you will have only primes left (and all primes >3).

    "List A" is the set [tex]A=\{a|\ a\in\mathbb{N},a>1,a\equiv\pm 1\text{ mod 6}\}[/tex]. Note A contains the set of primes greater than 3 as it's the set of naturals not divisible by 2 or 3, and I've removed 1 (which is what he seems to want).

    "List B" is the set [tex]B=\{a\times b|\ a,b\in A\}[/tex].

    Then A-B (everything in A not in B, aka "listP") is exactly the set of primes greater than 3. 5 and 25 are in A, so 5*25 is in B and hence not in A-B. Let a=bc be a nontrivial factorization of any composite in A. Then b and c are in A as well otherwise 2 or 3 divides a, so a is in B and hence not in A-B.

    Your z_2 seems to be the set [tex]\{a^2 |\ a\in A\}[/tex] which is different from how I've interpreted "list B" (and indeed how you seem to have interpreted it in your first paragraph)?
     
  9. Jun 19, 2005 #8

    HAP

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    Ahh, my mistake..

    But then I think I can explain this pattern a little more general so it will work for other base-values than 6. Look at the base-6 (2*3) number system, if that is what it's called in english.. It will include some symbols s_0, s_1, ..., s_5. If you're looking for primes here, you can look away from s_0, s_2, s_3 and s_4 since they are all divisible to either s_2 or s_3; Obvious you can now concentrade on s_1 and s_5.
    If you are to write a base-10 number(n) as a base-6 number, the last symbol would be (n mod 6). And the opposite way would be n+6x.. Since we know that only n=1 and n=5 can descripe primes, we can conclude that all primes (exept 2 and 3) are on one of the formulars 6x+1 and 6x+5 (which is equal to 6x-1 because of the modulation).

    Lets try this on our base-10 (2*5) number system:
    We can concentrade on a_1, a_3, a_7 and a_9. Then all primes (except 2 and 5), can be written as (10x +/- 1) or (10x +/- 3).

    Asuming I haven't made more mistakes, I have now showed a simple method to apply rshriner's observations on different base numbers..
     
  10. Jun 19, 2005 #9

    shmoe

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    "base" isn't the word you're looking for, that refers to different ways to represent your numbers, binary, hexadecimal, the usual base 10 etc.

    You could call it modulus if you like. Mod 6 all primes except 2 and 3 are congruent to 1 or 5. Mod 10 all primes except 2 and 5 are congruent to 1, 3, 7, or 9. Mod n all primes that don't divide n are congruent to a member of [tex]\{r|\text{ gcd(r,n)=1 and }1\leq r\leq n\}[/tex], the standard reduced residue system mod n. This is of course just different terminology for arithmetic progressions.
     
  11. Jun 20, 2005 #10

    HAP

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    I'm afraid you have to explain what gcd(r,n) is... It must be some standard function or technical term I ain't familiar with!
     
  12. Jun 20, 2005 #11

    matt grime

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    greatest common divisor (aka least common multiple); if you want to learn about number theory then you need to buy a book (if yuo can'ta take a course at university). gcd (or hcf) is the most fundamental thing in it.
     
  13. Jun 20, 2005 #12

    HAP

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    Just confusing learning math in two languages:

    "least common multiple" (lcm) is "mindste fælles multiplum" (mfm) in danish..
    "greatest common divisor" (gcd) is "største fælles divisor" (sfd) in danish..

    But they are two different things matt grime; greatest common divisor isn't also known as least common multiple! But I guess you just wrote to quick.. :)
     
  14. Jun 20, 2005 #13

    matt grime

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    yes, sorry, meant highest commion factor, hcf.
     
  15. Jun 20, 2005 #14
    If you haven't already, I suggest you get Anders Thorup's "Algebra" from Naturfagsbogladen which lies in August Krogh Institutet on Jagtvej. It's cheap (roughly 120 kr. when I bought it) and is used as a second year introduction to algebra on Københavns Universitet.

    It's an excellent introduction to number theory.
     
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