An interesting pyramid sequence.

From: Dan (30pack_at_sbcglobal.net)
Date: 06/27/04 

Date: 26 Jun 2004 23:13:15 -0700

I can print this out in both formats, as a continuing
sequence or starting with left diagonal it places all
diagonals in column form.
I set it up this way here for an easy explanation on how
numbers are generated.

                      1
                    2 3
                  4 5 6
                7 9 11 12
              13 16 20 23 24
            25 29 36 43 47 48
         49 54 65 79 90 95 96
       97 103 119 144 169 185 191 192
    193 200 222 263 313 354 376 383 384
  385 393 422 485 576 667 730 759 767 768
769 778 815 907 1061 1243 1397 1489 1526 1535 1536
... etc.
Where starting with the 3rd row all the inner triangles are
created by summing the two diagonal integers above it and
placing this sum beneath and in the center of the two integers
above.
With the help of two arrays the algorithm works this way--
If there is not two integers diagonally above the spot,
it just terminates the row by incrementing by (1) from the previous
sum and starts a new row by incrementing by (1) more.

Where the Pascal triangle has a ratio of 2 between the sum
of each row, the ratio in this sequence converges (2)<---(5).
The last # in each row divided by the first # of the same row
converges (1.5)---->(2). So you have a convergence on (2) working
both ways!
Also the right diagonal starting with (3) doubles. Each of
their squares falls on their downward left diagonal that ends in a
column left of center.
Other than the right diagonal, I believe the highest integer that
has a square later in the sequence is (13). This is hard to prove!
 
I am not sure but believe there is only (5) triangle numbers,
[1,2,3,6,36], in this sequence--->oo. Also hard to prove!

Into row 54 it goes into scientific notation because each integer
becomes >16 digits in length.

There's other interesting stuff but I will end it here.
 
Sequence not in OEIS.

Dan

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