sw00875math--even-squares-to-odd-squares

  This is another revisit to the natural number summation sequence
and its relation to "8".

Excluding "0", the first twelve nnss terms are: 
1 3, 6 10, 15 21, 28 36, 45 55, 66 78

   These can be approached variously; but how even squares are
progenitors of them, and how odd squares are derived from them
is valuable knowledge.

As can be seen, this sequence can be shown in 2-term groups
because the first two are odd integers and the second two are
even integers, and so on in that alternating manner.

Things to note:
1 + 3 = 4; 6 + 10 = 16; 15 + 21 = 36; 45 + 55 = 100; and
so on in that manner.  Obviously, the sums are even squares,
meaning that hidden in any even square are two terms of the
natural number summation sequence.  Therefore, if 12 x 12
(or 144), is chosen for inspection, how are the nnss terms
in it exposed?  If 144 is divided by 2, the quotient is 72.  If
12 is divided by 2, the quotient is 6.  72 - 6 = 66, and 72
+ 6 = 78.  As per above, 66 and 78 constitute a group of
two even terms in the natural number summation sequence
(nnss).
-
How do odd squares relate to this?  Every odd square is
1 > any nnss term (including "0") multiplied by "8".  So,
say the target is 13 x 13 (or 169).  How is it discovered?
One way is to subtract "1" from "169" and then divide
"168" by "8".  Having done this, the next question is: 
What is the proof that "21" is an nnss term?  There are
several ways, but I want to try something.  In the nnss
there are odd groups and even groups.  Since the first
group is an odd group, I'm calling the odd groups the
minor sequence, making the even groups the major
sequence.  Now 1 + 3 = 4, so "4" is a number which
can be used as a divisor.  Here's a run:
.25, .75; 3.75, 5.25; 11.25, 13.75; 22.75, 26.25
Here's a run using even nnss terms:
1.5, 2.5; 7, 9; 16.5, 19.5; 28, 36
I'm sure you see the difference and also the nuances.
4 into 21 = 5.25, but--.  .25 represents "1"; therefore,
if 5.25 is divided by .25, the quotient should be 21,
and it is.  "21" is the sixth term in the nnss, which means
that it is the sum of 1 + 2 + 3 + 4 + 5 + 6.  If term position
(tpo) is used as a guide, the results of dividing each term by
its position creates an incontrovertible sequence:
1, 1.5, 2, 2.5, 3, 3.5, n
Each of these numbers can also be obtained by adding "1"
to an nnss tpo and then dividing by 2.  From this it is also
possible to get back to an nnss term's value by subtracting
.5 from the (tpo + 1)/2 quotient and then multiplying that 
by the (tpo + 1).  3.5 - .5 = 3, and 3 x 7 = 21.
[ By these methods any unknown nnss term can be found.
   Example:  Let 63 be the tpo for an nnss term.  63 + 1 = 64.
   64/2 = 32.  64 x 31.5 = 2016. ]  So sayeth my calculator.
What then is this?: (8 x 2016) + 1.  16129.  What is the sqrt
of that?  Three calculator tries got me "127".
-
3² 5²   7² 9²   11² 13²   15² 17²   19² 21²   23² 25²
Now, the square root (sqrt) 
of 4 is 2; of 16 is 4; of 36 is 6; of 64 is 8; of 100 is 10; of 144 is 12
and
2 x 2 = 4, 2 x 4 = 8, 2 x 6 = 12, 2 x 8 = 16, 2 x 10 = 20; 2 x 12 = 24
Remember
1 + 3 = 4, 6 + 10 = 16, 15 + 21 = 36, 28 + 36 = 64, 45 + 55 = 100; 66 + 78 = 144
and
4 x 1 = 4, 4 x 2 = 8, 4 x 3 = 12, 4 x 4 = 16, 4 x 5 = 20; 4 x 6 = 24
So, if the 2-term group position is 32, the integer between the odd square roots
 of each term in that group must be 4 x 32, which is 128.  See "127" above.
If the spread between the nnss terms in a group is known,
 then that value divided by 2 will equal that group's position (gpo). 
Therefore, (32 x 2) + 2016 must equal a term whose value is 2080;
 and (8 x 2080) + 1 must equal an odd integer which is 129 x 129.
 According to my calculator/ both the former and the latter is 16641.
Of course, then, the difference between the odd squares 
 divided by the spread between the related nnss terms
 will always equal "8".  




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Brian A. J. Salchert