sw00883math--my-paired-twin-primes-conjecture

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 Yesterday (Friday, May 09, 2008) I concluded my sw00882math/

descension-by-squares-groups/2841 with this conjecture:  Regarding

paired twin primes, the even integer between one of the twin primes is

always divisible by "4" and the even integer between the other of the

twin primes is always divisible by "6".  Why did I posit this seemingly

senseless conjecture?  Because I wanted to use even integers since I

had set up a descension-by-squares grouping controlled by "4".  As

it happens, the only way my groupings matter is that it was through

them that I uncovered the "4" and "6" phenomenon.  Early this after-

noon, I was trying to find out if I was wrong, an effort which led to:

"3" divides into every six integers, the final digits of which (excluding

the special case: 3 x 3) are 5, 1, 7, 3, 9 as in 15, 21, 27, 33, 39.

Here is a run with the above numbers in italics:

11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43

If the final digit is "1" or "7" or "3" or "9", there cannot be any

paired twins primes because 1 + 4 = 5, 7 - 2 = 5, 3 + 2 = 5, 9 - 4 = 5.

Therefore, the final digit must be "5" and the integer must be divisible

by "3".  5 - 3 = 2 and 5 + 3 = 8.  This then is my proof, such as it is.

Also, where the integer whose final digit is "5" is not divisible by "3"/

neither the integer whose final digit is "2" nor the integer whose

final digit is "8" is evenly divisible by "6". 



My effort began with this:

6         102          108          114          120          126          132

4    100     104     108     112     116     120     124     128     132      136

- - -

  99   101   103   105   107   109   111   113   115   117   119   121   123   125   127

   3                  3x5x7                   3             5      3      7      11      3      5



5:01 PM - Have uncovered a flaw.  At the early going it involves "7".  It
is quite obvious, but I failed to check.  "3" divides evenly into an integer
whose final digit is "5" every thirty integers greater than "15" but it isn't
until ninety above "15" that paired twin primes are encountered.  [ This
doesn't invalidate my conjecture as stated, but it certainly weakens it. ]
One obstacle is the odd square.  However, the (8 x an nnss term) + 1
method isn't the only way to find an odd square.  These mulitplications:
2 x 4, 4 x 6, 6 x 8, 8 x 10, 10 x 20, and so on + 1 is another method.
But I had mentioned "7".  Every fourteen from forty-nine it's out to clip
ya, and it doesn't follow an easy-to-detect routine the way "5" and "3"
do.  Perhaps "3" can help, or "5", or both.  Some integers "7" divides
evenly into: one above 55, two below 65, two above 75, one below
85, one above 90, two below 100, at 105, two above 110, one below
120, one above 125, two below 135, at 140, two above 145.  Dizzy
yet?  Which one did I miss?  Correct: at 70.  Routine?  Going back to
7 x 1, the routine begins at two above.  So: two above, one below, one
above, two below, at  |  two above, one below, one above, two below,
at.  I hate "7", which is to say nothing about the greater more difficult
integers.  What about how it relates to mulitiples of "3"?  Which ones?
How about just those that add to 3.  four above 3, two above 12, at 21, 
two below 30, four below 39, three above 39, one above 48, one below 
57, three below 66  |  four above 66, two above 75, at 84, two below
93, four below 102, three above 102, one above 111, one below 120,
three below 129.  All this reminds of a poem I wrote in 2006.  After I
find it I'll place a link to it.  It is 9:35 PM.  Earlier a storm passed.  Need
to see if another one's coming.  Gusts just thrust us.  Satellite's showing
an enlarging green patch over this city.  Dost one blame it on the coal-
fired power plant?  Or is it merely a reflection of the city's lights?  "Bone
Ache" is the poem.  It is the fourth one on its page.  A quick six lines.
It was written September 29, 2006.
It fits this occasion.     





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Brian A. J. Salchert