# 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 $A000129$. 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. Mar 13, 2019 at 15:59
• en.wikipedia.org/wiki/Delannoy_number 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. 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. 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 ! 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. 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 ... 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. 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 $0+1+1$? 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. Mar 13, 2019 at 16:54