# Can the result of a change of one simple rule to build a Pascal's triangle be mathematically explained?

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.

• The row sums give the Pell numbers $A000129$. Commented Mar 13, 2019 at 15:49
• Thanks. I did some additions of rows but since I wasn't familiar with Pell numbers, I missed that. Commented Mar 13, 2019 at 15:59
• en.wikipedia.org/wiki/Delannoy_number Commented Mar 13, 2019 at 16:05
• @darijgrinberg, yes I did see that. A lot of references are given in the corresponding link in the OEIS database for those interested in reading more about the topic of Delannoy numbers. Commented Mar 13, 2019 at 19:03

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

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.

• thanks for this amazing answer. Commented Mar 14, 2019 at 23:49
• @user25406: I delighted myself time ago to find relations among "classical" numbers, and found that the matricial approach offers a vast scenario ! Commented Mar 15, 2019 at 0:07
• @G Cab, your matricial approach can become, if it's not already, the basis for dealing with Pascal's and Pascal's like triangles mathematics. Commented Mar 15, 2019 at 12:03
• well, although not widely known as it deserves, the matricial approach to 2D Numbers like Binomial, Stirling, Eulerian etc. is quite known. Re. for instance to 1 , 2 ... Commented Mar 15, 2019 at 15:27
• @user25406: also have a look at [this](mathoverflow.net/questions/277223 ) to better understand the usefulness of matrix notation. Commented Mar 15, 2019 at 15:34

This is the doorway to whole world of Number-Theory, Combinatorics pleasure !

With regards to your third question, a favourite of mine is the Trinomial Coefficients, where each term is the sum of the three above,

$$1$$ $$1, 1, 1$$ $$1, 2, 3, 2, 1$$ $$1, 3, 6, 7, 6, 3, 1$$ $$1, 4, 10, 16, 19, 16, 10, 4, 1$$

This expands $$(1+x+x^2)^n$$ in much the same way as Pascal's Triangle expands the Binomial's $$(1+x)^n$$

So, for example, that last row tells us that, $$(1+x+x^2)^4$$ $$=1+4x+10x^2+16x^3+19x^4+16x^5+10x^6+4x^7+x^8$$

There are other Multinomials, of course, of higher orders.

I'm looking forward to the other answers that your interesting question will generate.

• if I understand, to get $2$ on the 3rd row, we add $0+1+1$? Commented Mar 13, 2019 at 15:43
• Yes, that's right, you can think of it like that; just as you can on the edge of Pascal's Triangle, where a 1 on the left hand edge can be thought of as $0+1$ from the line above. Commented Mar 13, 2019 at 16:54