With
    1  2  3  4  5  6 ...
I obtained
            1
            2    1
            3    4    1
            4   10    6    1
            5   20   21    8    1
            6   35   56   36   10    1
            7   56  126  120   55   12    1
whose row sums correspond to the first bissection of Fibonacci numbers.
A few days after, I tried
(1)            1   1   2   3   4   5   6  ...
Hence
              1  
              1   1
              2   2   1
              3   5   3   1
              4  10   9   4   1
              5  18  22  14   5   1
              6  30  48  40  20   6   1
(of whom the 45 degrees diagonal sums
(2)        1   1   3   5   10   18   34   63
are of an unknown sequence).
The row sums and their successive differences 
(3)    1   2   5   12    28    65    151   351
         1   3   7    16    37    86    200
           2   4    9    21    49    114
             2   5    12    28    65
consituted my first    "suite en trios" .
After, I chose arbitrarly
(4)    1   2   3   5   10    21    43    86    171
         1   1   2   5    11    22    43    85
           0   1   3    6    11    21    42
             1   2   3     5    10    21
I integrated the first line of this sequence in the    "tableau en trios"
(5)  1   1   1    0    0     0      1      (6)  1   0   0    0    1     1
     1   2   3    5   10    21     43           1   2   6   20   61   183     
     1   3   7   20   60   182    547           1   3  12   51  205   820
     1   4  13   51  204   819   3277           1   4  20  104  520  2605
     1   5  21  104  520  2604  13021           
The third one is formed with the third line of (4)  0 1 3 6

The first line of the (5) array generates the present sequence by the principal diagonal :
(7)   1   2   4   7   11   16   23  37  74   175   431   1024    2291
        1   2   3   4    5    7   14  37  101   256   593    1267
          1   1   1    1    2   7   23  64   155   337   674
            0   0   0    1    5   16  41   91   182   337
              0   0    1    4   11  25  50    91   155
                0   1    3    7   14  25   41    64
                  1    2    4   7   11  16    23
                    1    2    3   4    5    7
                       1    1   1    1   2
                         0    0   0    1 
                            0   0    1
                              0    1
                                1
The recurrence is, from the sixth term, 
     u(n)=6*u(n-1)-15*u(n-2)+20*u(n-3)-15*u(n-4)+6*u(n-5)
The array contains many 2^n like the other sequences I presented last week A129818, and called twisted number sequences. Sequence A038504 presents an almost equivalent array (the difference is the 0 at the beginning of the sequence). Instead of 1 1 0 0 (equivalent to 1 1 1 0 0 0), the beginning of the diagonal is 0 1 1 0 0. It is a twist numbers sequence. Many formulas were discovered by Paul Barry.
Naturally, I studied too delate sequences.
Note that for  1  1  1  2  3  4  5 ,  the sequence and its principal diagonal :
       1   2   4   9   20   44   96   209   455   991
       1   1   1   2    1    3    1     5     0     9
It can be defined by 
u(n+1) = 2*u(n)+v(n) ,  v(n) = 0   0   1   2   4   8   17  37  81  177
     n=0...             diff =   0   1   1   2   4   9   20  44  96 = 0 1 u(n)
I wrote what I exposed here on two difference manuscripts which can be analysed (paper and ink).