TITLE: Can the result of a change of one simple rule to build a Pascal's triangle be mathematically explained?
QUESTION [5 upvotes]: The only rule that is changed is this: when adding two numbers from row $(n-1)$ to get the number $N_n$ of row $n$ we also add the number from row $(n-2)$ that is aligned (vertically) with the position of the number $N_n$.
We start with the same triangle $$1$$
$$1---1$$
The next line will be $$1---3---1$$ because we added the first $1$ aligned with the result of adding $1+1$ from row $2$. If we keep using this rule, we will get a Pascal's triangle whose diagonals are very interesting. Since I don't know how to format a Pascal's triangle with mathjax, I will simply list few diagonals.
The first diagonal is:
$$1--3--5--7--9--11--13--15--15--17--19--21...$$
The second diagonal is given by $A001844$ which gives centered square numbers.
$$1--5--13--25--41--61--85--113--145--181--221...$$
The third diagonal is given by $A001845$ which gives centered octahedral numbers (also called crystal ball sequence for cubic lattice).
$$1--7--25--63--129--231--377--575--833--1159...$$
The fourth diagonal is given by $A001846$ which gives centered 4-dimensional orthoplex numbers (also called crystal ball sequence for 4-dimensional cubic lattice).
$$1--9--41--129--321--681--1289--2241--3649...$$
The fifth diagonal is given by $A001847$ which gives crystal ball sequence for 5-dimensional cubic lattice numbers. The sixth diagonal is given by $A001848$ which gives crystal ball sequence for 6-dimensional cubic lattice numbers. I suppose the next sequence will give the 7-dimensional cubic lattice numbers ( I checked it ) and higher.
Some of the formulas to generate the numbers are given in the OEIS sequences given above.  
The first $7$ rows of the this triangle look like this:
$$1$$
$$1---1$$
$$1---3---1$$
$$1---5---5---1$$
$$1---7---13---7---1$$
$$1---9---25---25---9---1$$
$$1---11---41---63---41---11---1$$
Note the pattern that repeats indefinitely formed by multiplying numbers under the starting $1$ at the top in the following way:
$$1*3+1*1=4=2^2=(1+1)^2$$
$$3*13+5*5=64=8^2=(3+5)^2$$
$$13*63+25*25=1444=38^2=(13+25)^2$$.
The triangle has other properties that deserve to be mentioned. If we add term by term the first diagonal and the second to get:
$$(1+1),(3+5), (5+13), (7+25), (9+41), (11+61), (13+85)...$$ we get the sequence $2n^2$.
If we add the second diagonal and the third term by term, we get the sequence A035597 which gives the number of points of L1 norm 3 in cubic lattice Z^n. Its formula is ($4n^3+2n)/3$:
$$(1+1), (5+7), (13+25), (25+63), (41+129), (61+231)...$$
But we can also get new numbers by multiplying term by term the first and second diagonals. We get OEIS A005917 Rhombic dodecahedral numbers: $a(n) = n^4 - (n - 1)^4$
$$1, 15, 65, 175, 369, 671, 1105, 1695, 2465, 3439...$$
There are probably more hidden patterns waiting to be found in this triangle. Only a systematic search can find them.  
There are many questions that come to mind.  
1-How come one simple modification of a rule provides such a change.
2-What mathematics (formulas, theorems...) is common to both triangles (assuming some features are common to both triangles which is not obvious at all at this point).
3-Have the effect of other modifications of the rule to build a Pascal's triangle been systematically studied before? ( For example, one can think of a classical Pascal's triangle where the result is squared...).  
If someone can think of more meaningful tags, please add them or change them.

REPLY [2 votes]: a) In dealing with such triangular arrays, it is convenient to arrange them
as a Lower Triangular array (matrix), indexed from $0$. That greatly simplifies notation, and allows
matrix "tools" to be applied.
b) In this scheme, the LT Pascal matrix ($\bf P$) is defined by the recurrence
$$
\left\{ \matrix{
  p_{\,0,\,m}  = \left[ {0 = m} \right] \hfill \cr 
  p_{\,n,\,m}  = p_{\,n - 1,\,m}  + p_{\,n - 1,\,m - 1}  \hfill \cr}  \right.
$$
where $[P]$ denotes the Iverson bracket,
or more compactly as
$$
p_{\,n,\,m}  = \left[ {0 = n} \right]\left[ {0 = m} \right] + \left[ {1 \le n} \right]\left( {p_{\,n - 1,\,m}  + p_{\,n - 1,\,m - 1} } \right)
$$
Note that the Initial Conditions are as much qualifying as the recurrence itself.
c) The LT "modified" Pascal matrix ($\bf Q$) you propose will read
$$
q_{\,n,\,m}  = \left[ {0 = n} \right]\left[ {0 = m} \right] + \left[ {1 \le n} \right]\left( {q_{\,n - 1,\,m}  + q_{\,n - 1,\,m - 1} } \right)
 + \left[ {2 \le n} \right]q_{\,n - 2,\,m} 
$$
Note that it is a second degree recurrence, instead of a first degree, and that it involves three 
precursors instead of two.
No doubt that it will produce quite a different result, and indeed much interesting, and here I will limit to
summarily describe some, in addition to those already indicated.
The matrix ($0 \ldots 10 \times 0 \ldots 10$) is

d) The double o.g.f. will be
$$
G(x,y) = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,m} {q_{\,n,\,m} x^{\,n} y^{\,m} } }  = {1 \over {1 - x\left( {1 + y} \right) - x^{\,2} }}
$$
which for $m=0 \to y=0$ is that of the Fibonacci Numbers, which are in fact in the first column.
The following columns are convolutions of the Fibonacci N.    
Putting $y=1$, we get the o.g.f. of the row sums, which corresponds to that of the Pell Numbers (shifted by one).
e) In relation to the Pascal matrix $\bf P$, it turns out that both are
similar to the bidiagonal matrix $\bf I + \bf E$, and thus are similar to each other.
$$
{\bf I + E} = \left( {\matrix{
   1 & 0 & 0 &  \cdots   \cr 
   1 & 1 & 0 &  \cdots   \cr 
   0 & 1 & 1 &  \cdots   \cr 
    \vdots  &  \ddots  &  \ddots  &  \ddots   \cr 
 } } \right)
$$
Moreover
$$
{\bf Q}\,{\bf P}^{\, - \,{\bf 1}}  = {\bf A}\quad \left| {\;a_{\,n,\,m} :\left[ {0 = \left( {n + m} \right)\bmod 2} \right]\left( \matrix{
  {{n + m} \over 2} \cr 
  m \cr}  \right)} \right.
$$
.. and much else.