Ronald Brady

Posts: 43
Registered: 12/8/04
Re: Nth Order Pascal Triangle
Posted: Aug 12, 1997 4:07 PM
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Nth Order Pascal Triangle: General Formulation

An Nth Order Pascal Triangle will be an array of numbers
characterized by the following:

1) There will be N entries in the 1st row- all of them
equal to 1. In general there will be (N-1)k +1 entries
in the kth row.

2) The first N elements of the 2nd row will be simply
1,2,3,...,N. Let any other entry, appearing in the ith
row and the jth column, be denoted by a(i,j). Then by
definition:

for i > 2 or i = 2 and j > N or j = N we have

a(i,j) = a(i-1,j) + a(i-1,j-1) + a(i-1,j-2) + ... +

a(i-1,j-N+1)

for i > 2 or i = 2 and j < N we have by definition

a(i,j) = a(i-1,j) + [ sum of the remaining entries ( on

the left side of a(i-1,j) in the (i-1)th row: if

any].

If the intersection of the ith row and the jth column is a
blank space then a(i,j) is considered to be zero.

Conjecture: given the definitions above it is now conjectured
that the sum of the entries in the kth row of the Nth Order
Pascal Triangle is given by N^k. It is further conjectured that
( motivated by a note from Kevin Brown for the case of N = 3)
the entries in the kth row of the Nth Order Pascal Triangle
corresponds to the coefficients of x^r in the expansion of
(1 + x +x^2 + x^3 + ... + x^(N- 1))^k, where r takes on integer
values between 0 and k(N-1).

The triangles for the cases where N=2 and N=3 have been present-
ed in a previous post. The triangle for N=4 is presented below.

1 1 1 1
1 2 3 4 3 2 1
1 3 6 10 12 12 10 6 3 1
1 4 10 20 31 40 44 40 31 20 10 4 1
-
It will be seen that the sum of the entries in the respective
rows are as follows:

1st row sum = 4

2nd row sum = 16 = 4^2

3rd row sum = 64 = 4^3

4th row sum = 256 = 4^4




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