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

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. 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 Lucas triangle(1,2) 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 Lucas triangle A029635 along with the sum of each of the rows corresponding to the Sloane number A042950 which can be generated by the formula 3(2n), starting at n=0 for row 2, given an initial number of 2. The next triangle in the series, Pascal (1,3) triangle, where m = 3, will be discussed in Part II.

The Lucas Triangle

Table I (Triangle m=2)
2 2
12 3
132 6
1452 12
15972 24
16141692 48
1720302511 296
1827505536 132192

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 2.

We can produce the coefficients of the Lucas triangle(1,2) using a partial sums MacLaurin series which is a power series generated by a function f:

f(0) + f '(0)/1! x + f ''(0)/2! x2 + f '''(0)/3! x3 + ... + f n(0)/n! xn + ...

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

f(x) = (1 + 2x)(1 + x)n
f '(x) = 2(1 + x)n + (1 + 2x)(n)(1 + x)n-1
f ''(x) = 4n(1 + x)n-1 + (1 + 2x)(n)(n-1)(1 + x)n-2
f '''(x) = 6n(n-1)(1 + x)n-2 + (1 + 2x)(n)(n-1)(n-2)(1 + x)n-3
.
.
.
f k(x) = 2kn(n-1)...(n-k+1)(1 + x)n-k+1 + (1 + 2x)(n)(n-1)(n-2)...(n-k+1)(1 + x)n-k

Since we are given the first row as 2, 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:

at n=0
f (0) = 1(1) = 1
f '(0) = 2(1) + 0 = 2
f(0) + f '(0)/1! x = 1 + 2x

at n=1
f (0) = 1(1) = 1
f '(0) = 2(1) + 1 = 3
f ''(0) = 4(1) + 0 = 4
f(0) + f '(0)/1! x + f ''(0)/2! = 1 + 3x + 2x2

at n=2
f (0) = 1(1) = 1
f '(0) = 2(1) + 2 = 4
f ''(0) = 4(2) + 2(1) = 10
f '''(0) = 6(2) + 0 = 12
f(0) + f '(0)/1! x + f ''(0)/2! x2 + f '''(0)/3! x3 = 1 + 4x +5x2 + 2x3

Alternatively, we can compute the coefficients of the Lucas triangle using the binomial theorem and multiplying each of the terms by (1 + 2x):

A binomial theorem

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 Lucas sequence A000032 are generated from the Lucas triangle. The sequence shown in the first row of Table II and its ascending diagonals are depicted in bicolor:

Table II (Lucas/Diagonals)
SD21 347 1118 294776 123
2
12
132
145 2
159 72
1614 1692
1720 3025112
182750 5536132
1935 771059149152
11044112182196 14064172

Go to Part V. Go back to homepage.

Copyright © 2021 by Eddie N Gutierrez. E-Mail: enaguti1949@gmail.com