sw00888math--DFnote-and-squares-da-patterns

32


 DF = Descension Family          da = digit addition
 nnss = natural number summation sequence

 DF note:  As God would have it, when I began unstacking the high
pile of manilla folders holding writings of mine from my computer stool,
the first group did not separate as normal, revealing a sheet of lined
paper on which I had penned some math thoughts on 7-31-06.  I let
it be and moved what I held/ over to its usual location on my bed.
Later, I read what I'd noted, and found it of interest in one particular
way: an nDF descends to a value which is n - 1.  If n = 10 (5 x 5),
then (5 + 5) - 1 = the final resultant for 10DF.  This formula is not in
the note, but the note's examples imply it.

My project today is a continuation of the da patterns for the decades
pertaining to odd squares, and possibly to even squares.  Want to be
more explicit, and want to investigate if and where intersectings occur.
-
First, a brief review:  The 2 5 8 pattern for the only group in which
paired twin primes can be found looks like this:
[ Insert:  Decade # equals (decade's f-d-5 integer plus 5) divided by 10. 
   Example: (15 + 5)/10 = 2 ]
2  5  8  -  15/3 = 5 (5)          45/3 = 15 (6)          75/3 = 25 (7)                     
2  5  8  -  105/3 = 35 (8)          135/3 = 45 (9)          165/3 = 55 (1)               
[ Insert:  The 2nd row decades are 11 (2), 14 (5), 17 (8). ]
2  5  8  -  195/3 = 65 (2)          225/3 = 75 (3)          255/3 = 85 (4)               
2  5  8  -  285/3 = 95 (5)          315/3 = 105 (6)          345/3 = 115 (7)           
2  5  8  -  375/3 = 125 (8)          405/3 = 135 (9)          435/3 = 145 (1)          
2  5  8  -  465/3 = 155 (2)          495/3 = 165 (3)          525/3 = 175 (4)          
:
[ Insert:  Have decided to use the f-d-5 integers for the f-d-1 and f-d-9 groups. ]
1  9  4  -  [1] 5/3 = 1.6 (1)    [81] 85/3 = 28.3 (9)    [121] 125/3 = 41.6 (4)    
1  9  4  -  [361] 121.6 (1)    [441] 148.3 (9)    [841] 281.6 (4)                      
7  9  7  -  [961] 321.6 (7)    [1521] 508.3 (9)    [1681] 561.6 (7)                   
7  9  7  -  [2401] 801.6 (7)    [2601] 868.3 (9)    [3481] 1161.6 (7)                
4  9  1  -  [3721] 1241.6 (4)    [4761] 1588.3 (9)    [5041] 1681.6 (1)             
4  9  1  -  [6241] 2081.6 (4)    [6561] 2188.3 (9)    [7921] 2641.6 (1)             
:
1  5  8  -  [9] 1.6 (9)          [49] 15 (4)          [169] 55 (7)                           
2  8  1  -  [289] 95 (1)          [529] 175 (7)          [729] 241.6 (9)                 
1  2  5  -  [1089] 361.6 (9)          [1369]455 (1)          [1849] 615 (4)
5  2  1  -  [2209] 735 (4)          [2809] 935 (1)          [3249] 1081.6 (9)         
1  8  2  -  [3969] 1231.6 (9)      [4489] 1495 (7)      [5329] 1775 (1)              
8  5  1  -  [5929] 1975 (7)      [6889] 2295 (4)      [7569] 2521.6 (9)               

In the first group's first column is a 95.  In the third group's first column is a 95.
That says the same decade is being referenced in both.  That decade is 29.
That says 289 is a square and therefore decade 29 is eliminated.  This is what
I mean by intersectings.

More to come.

PM 2:24 - I don't do math the way professional mathematicians do;
yet I strive to be rigorous.  Ideally, I would like to be able to ascertain
all the essential qualities of an integer just by closely observing it.  I am
seeking methods which will apply into infinity, but I prefer methods 
which do not require intricate formulas.
Of the 18 terms in the first cycle of the f-d-9 group, 12 intersect with
terms in group 1.  Of the 18 terms in the first cycle of the f-d-1 group,
0 instersect with terms in group 1.  Of those that interact, four are of
the da2 kind, four of the da5, and four of the da8.  The square in the
first da5 is 49, and the square in the first da8 is 169.  In groups 2 and
3 the order is irregular within each cycle, but does have a pattern.  In
group 1 the order is regular, but group 1's order does not depend on
where square's occur but rather on where "3" divides evenly into an
f-d-5 integer.  In group 2 "3" does not divide evenly into any f-d-5
integer.  In group 3 "3" divides evenly into an f-d-5 integer 12 times
in each 18-term cycle.  In group 3 the first da8 is in R1C3, but in the
first group it is in R2C3.  What does this tell about what is in R1C3
in group 1?  That it does not have an f-d-9 square in it, but that some
other eliminator of a form not here being tested might be lurking there
remains a possibility.  I believe it's time for a different chart.

These are decade # / da charts:
-
Group 1
2 / 2          5 / 5          8 / 8
11 / 2          14 / 5          17 / 8
20 / 2          23 / 5          26 / 8
29 / 2          32 / 5          35 / 8
38 / 2          41 / 2          44 / 8
47 / 2          50 / 5          53 / 8
:
Group 3
1 / 1          5 / 5          17 / 8
29 / 2          53 / 8          73 / 1
109 / 1          137 / 2          185 / 5
221 /5281 / 2          325 / 1
397 / 1    449 / 8          533 / 2
593 / 8          689 / 5          757 / 1

After some figuring, figured out 689 / 5 in group 3 is in
R5C2 of cycle 13 in group 1, which cycle's 18th term is
701, there being a 54 spread in each cycle after cycle 1.
12 x 54 = 648; 648 + 53 = 701.  701, 698, 695, 692, 689.
(689 x 10) - 1 = 6889.

More to come.

05/16/08  AM 7:32  -  A little something that provides both an 
nnss magic number and the number to minus from and plus to it:
2 x 0 x 0 = 0    
2 x 1 x 1 = 2    2 x 2 x 2 = 8    2 x 3 x 3 = 18    2 x 4 x 4 = 32    2 x 5 x 5 = 50
2 x 6 x 6 = 72
for generating the natural number summation sequence:
0          
1    3          6    10          15    21          28    36          45    55          66    78
and dividing one of these magic numbers by "2" begets a square as with 50/2
-
a different view of these nnss magic numbers:
0          2          8          18          32          50          72
     2          6          10          14          18          22 
so
(4 x 0) + 2    (4 x 1) + 2    (4 x 2) + 2    (4 x 3) + 2    (4 x 4) + 2    (4 x 5) + 2
-
Also, as can be seen, I have separated the nnss into sets.
Adding the integers in a set gets an even square,
 and subtracting a set's lesser # from its greater # 
 gets the even square's square root.
Adding the integers between sets gets an odd square,
 and subtracting the lesser # from the greater # 
 gets the odd square's square root.
-
a way to allow the term position (tpo) of an nnss magic number 
 to equal the second and third square root associated with that number:
0's tpo is 0    2's tpo is 1    8's tpo is 2    18's tpo is 3    32's tpo is 4
Thus, for this sequence, tpo 3 = 2 x 3 x 3.    18 - 3 = 15    18 + 3 = 21
-
Another view of the squares:
0 1 2    3 4 5    8 9 10    15 16 17    24 25 26    35 36 37    48 49 50

During lunch break I uncovered digit addition relationships 
between square roots and their squares.  Examples follow:
1    1 = 1  1 = 1    19 = 1  361 = 1    28 = 1  784 = 1    73 = 1  5329 = 1
2    2 = 2  4 = 4    11 = 2  121 = 4    38 = 2  1444 = 4    47 = 2  2209 = 4
3    3 = 3  9 = 9    |    6 = 6  36 = 9    |    9 = 9  81 = 9
4    4 = 4  16 = 7    13 = 4  169 = 7    22 = 4  484 = 7    103 = 4  10609 = 7
5    5 = 5  25 = 7    14 = 5  196 = 7    23 = 5  529 = 7    41 = 5    1681 = 7
7    7 = 7  49 = 4    16 = 7  256 = 4    25 = 7  625 = 4    61 = 7  3721 = 4
8    8 = 8  64 = 1    17 = 8  289 = 4    26 = 8  676 = 4    71 = 8  5041 = 4
Note the 1 4 9 7 7 9 4 1 9 routine.  To start:  If a square root is of the 3, 6, or 9
variety, its square will be of the 9 variety as in 21 = 3  441 = 9.  Sqrts of the
4 and 5 variety net squares of the 7 variety.  Sqrts of the 2 and 7 variety net
squares of the 4 variety.  Sqrts of the 1 and 8 variety net squares of the 1 variety.
All this helps in that knowing the da of a square limits its possible square roots.
The square "6889" = 31 = 4 and so signals that its sqrt must be a da2 or a da7.
One is still left guessing, or trying other methods.  6888/4 = 1722    1722/3 = 574
574/7 = 82    3 x 4 x 7 = 84    82 x 84 = 6888    83 x 83 = 6889    83 = 11 = 2


May 17, 2008 note - Stating the implicit: Every nine integers.       






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Brian A. J. Salchert