sw00289math-numbertheory

Math   Twentieths Funky Bar Graph              

 -          and some nnss-related findings
                                   [ last modified: 2008-10-27 ]

 kestroke formula for bottom line:    n + (n + 1)
   If you want to make a graph for twenty twentieths,
   you will need 20 + (20 + 1) bottom-line keystrokes.

 #

 033107 insight which uses nnss terms:  see at *** below
          and also an insight which uses non-nnss integers

 032807
 Note:   n + (n + 1)    
            is a natural number summation sequence
            formula
            when a term in the nnss is an odd integer.
            There are therefore only certain numbers
            which can be "n" numbers, and there is a
            pattern inherent in this set of numbers.
            0 and 1 are the first pair of "n" numbers.
            7 and 10 are the second pair of "n" numbers.
            22 and 27 are the third pair of "n" numbers.
            45 and 52 are the fourth pair of "n" numbers.
            What can be determined about these numbers?
            One: the difference between the numbers
            in each succeeding pair increases by a factor of 2.
            Two: the difference between the greater number
            in a given pair and the lesser number in the 
            following pair increases by a multiple of 6. 
            Three: notice how multiples of 8 relate to these
            "n" numbers.  (8 x 0) - 0 = 0  /  (8 x 1) - 7 = 1
            (8 x 3) - 22 = 2  /  (8 x 6) - 45 = 3
            (8 x 10) - 76 = 4  /  (8 x 15) - 115 = 5
            -
            Stop.  22 - 7 = 15.  (7 x 2) + 1 = 15.
                     45 - 22 = 23.  (7 x 3) + 2 = 23.
                     76 - 45 = 31.  (7 x 4) + 3 = 31.
                     115 - 76 = 39.  (7 x 5) + 4 = 39.
                     So, this difference group of numbers
                     wherein the difference between terms
                     is 8: 
                     7, 15, 23, 31, 39, 47, 55, 63, 71, 79, n.
            -
            Now, (7 x 6) + 5 = 47.  47 + 115 = 162.
                      (8 x 21) = 168.  168 - 162 = 6.
                   -
  (7 x 7) + 6 = 55.  55 + 162 = 217.
                      (8 x 28) = 224.  224 - 217 = 7.
 -
                    (7 x 8) + 7 = 63.  63 + 217 = 280.                   
                      (8 x 36) = 288.  288 - 280 = 8.                 
                   -            
                    (7 x 9) + 8 = 71.  71 + 280 = 351.
                      (8 x 45) = 360.  360 -351 = 9.             
                   -                    
                    (7 x 10) + 9 = 79.  79 +351 = 430.
                      (8 x 55) = 440.  440 -430 = 10.
            -
            45 is both in the "n" number group
                                   and is a term in the nnss
            because 22 is in the "n" group
                                   and 22 + (22 + 1) = 45.
 -

 ***
        (7 x 0) + 0 = 0            (7 x 0) + 1 = 1
        (7 x 1) + 0 = 7            (7 x 1) + 3 = 10
        (7 x 3) + 1 =22            (7 x 3) + 6 = 27
        (7 x 6) + 3 = 45           (7 x 6) + 10 = 52
        (7 x 10) + 6 = 76          (7 x 10) + 15 = 85
        (7 x 15) + 10 = 115       (7 x 15) + 21 = 126
 #
        0 + (0 + 1) = 1       |       1 + (1 + 1) = 3
        7 + (7 + 1) = 15       |       10 + (10 + 1) = 21
        22 + (22 + 1) = 45       |       27 + (27 + 1) = 55
        45 + (45 + 1) = 91       |       52 + (52 + 1) = 105
 -
 difference notes:  
   pairs: 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50
   < run: 14, 30, 46, n   (beinning with 1): 1, 15, 45, 91, n
   > run: 18, 34, 50, n   (beginning wirh 3): 3, 21, 55, 105, n 
 ?: How can one tell from the "pairs" differences/ what the
     "< run" and "> run" integers are?
 Answer:  Half of a pair difference times the pair difference
              equals an integer from which when the half of its
              pair difference is subtracted out = its "< run" #
              and from which when the half of its pair difference 
              is added in = its "> run" #.  Example:
              14/2 = 7.  7 x 14 = 98.  98 - 7 = 91.  98 + 7 = 105.
 -
 12:39 PM  March 31, 2007
 -
 This means that/ beginning with a positive "2" is a set of
 positive integers controlled by the positive even-number
 multiples of positive "8" from which the terms in the nnss
 (natural number summation sequense) can be derived.
 "98" is the fourth term in that set.  Here is an initial look:
 2       18        50        98        162        242        338        450
    16      32       48       64        80  96        112
 -
 How can one know that "7" is the essential positive integer
 associatedwith "98"?Enter termposition.  2 x 4 = 8, and
 and 8 - 1 = 7.  How can one know that at term-position 4
 the positive integer associated with it is "98"?  This is a bit
 more difficult, and also contains a circular trap.  This is the
 trap:   98 -2 = 96.  96/16 = 6.  "6" is--if one accepts "0"
 as term 1 in the nnss--at term-position 4, with "1" being at
 term-position 2 and "3" being at term-position 3.  So: time-
 out.  [ Used my in-bold numbers in an advanced search and
 got a single result which is attributed to Joshua Zucker and 
 the Castilleja School MathCounts club (joshua.zucher (AT)
 stanford alumni.org), Nov 07 2002, and has to do with:
 Maximum number of regions the plane can be divided into
 using n (concave) quadrilaterals.  Do what I did to learn more. ]
 This doesn't settle my quandary, which is quite different.  It 
 may be that the trap I exposed is actually a proof, but I desire  
 a less circuitous answer.  98 - 4 = 94.  50 - 3 = 47.  18 - 2 =
 16.  2 - 1 = 1.  15, 31, 47, n which suggests "n" will be 63.
 63 + 94 = 157, and 162 - 5 = 157.  So now yet another set
 of differences begins, but to what purpose?  Obviously, 16
 or (8 x 2) controls this set, and (63 + 1)/16 = 4.  Perhaps I
 will return here tomorrow.  Can't let this go.  2 + (16 x 1) =
 18; 18 + (16 x 2) = 50; 50 + (16 x 3) = 98; 98 + (16 x 4) =
 162.  The only answer I can come up with is the term-position
 match answer, aided by knowing that if 2 less than a number
 in my in-bold set is evenly divisble by 16, that number may
 be correct.  I must go again.  4:53 PM - Who stole the magic?
 3 - 1 = 2.  10 - 6 = 4.  21 - 15 = 6.  36 - 28 = 8.  55 - 45 = 10.
 I am trying to see something.  Toward that end I have left out
 "0".  50 is at term-position (t-p) 3, and (50 - 2)/16 = 3.  If I let
 50 be at t-p 5, then its position would match its -/+ difference.
 My in-bold numbers do pertain only to the odd nnss terms, but
 it appears I simply ought to make believe the even nnss terms
 still matter/ and assign only odd term positions to my in-bold 
 numbers.  What thinkest thou?  Doing so woud certainly simplify
 matters.  I am going to do it.  So: How can one know that at
 t-p 7 the positive integer associated with it is "98"?  98/7 = 14,
 which appears to be at t-p 4 in the "pairs" run, but from here
 on will be at t-p 7 in that run.  There are times when making
 believe has its advantages.  


 - 
 see my nnss entry
 -

 ~


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twentieths funky bargraph

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Brian A. J. Salchert