
Re: Nth Order Pascal
Triangle
Posted: Aug 12, 1997 4:07 PM 

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
(N1)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(i1,j) + a(i1,j1) + a(i1,j2) + ...
+
a(i1,jN+1)
for i > 2 or i = 2 and j < N we have by
definition
a(i,j) = a(i1,j) + [ sum of the remaining entries (
on
the left side of a(i1,j) in the (i1)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(N1).
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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