Pascal Type Triangles(1,m) (Part II)

Triangles with Ascending Diagonals

An interesting property of Pascal's triangle is that its diagonals sum to the Fibonacci sequence as shown online. (Just type in Pascal Fibonacci image). It can be stated here that an infinite number of Pascal type triangles, where the last number on a row may be any integer greater than zero, may be constructed as was shown in Part I. In addition, a method for their construction uses the expression (1 + mx)(1 + x)ⁿ where m > 0 and (1 + x)ⁿ is the polynomial used to generate the Pascal triangle. First the Pascal triangle(1,3) will be constructed and it will be shown that its coefficients can also be generated using the partial sums of a MacLaurin series as well as the binomial expansion of (1 + mx)(1 + x)ⁿ. Table I shows the known Pascal triangle(1,3) A095660 along with the sum of each of the rows corresponding to the Sloane number A198633 which can be generated by the formula 2ⁿ⁺², starting at n=0 for row 2, given an initial number of 3.

The Pascal Triangle(1,3)

As can be seen, just as in the Pascal triangle, every number in each row of the triangle is obtained by adding the two numbers nearest it (diagonally) in the row above, except the two end numbers, which are 1 and 3.

We can produce the coefficients of the triangle(1,3) using a partial sums MacLaurin series which is a power series generated by a function f:

To derive the series, we first list the function and its derivatives and using the product rule on f(x) and its derivatives:

Since we are given the first row as 3, for the three examples below we start the evaluation at the second thru fourth rows of the triangle, treating row 2 as n=0, and then sum up the partial sums to give the coefficients of each of these rows:

Alternatively, we can compute the coefficients of the Pascal triangle(1,3) using the binomial theorem and multiplying each of the terms by (1 + 3x):

where (n/k) is the number of combinations.

Table I can also be displayed in irregular triangle array format and to show that the ascending diagonals corresponding to the Fibonacci sequence A104449 are generated from the Pascal triangle(1,3). The sequence shown in the first row of Table II and its ascending diagonals are depicted in bicolor:

Go back to Part I. Go back to homepage.