Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1997. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 24)

KALLOS, G., The generalization of Pascal's triangle from algebraic point of view

14 Gábor Kallós In the combinatorical literature the 11... 1-based triangles (

k pieces of 1 digits) with name order k (Vilenkin [2]) or k-th Pascal's triangle (Gerőcs [3]) can be found. The authors gave this different definition because of different approach. Theorem 1. From the n-th row of the aQa\a

2 ... a

m2

am-i-based triangle after positional addition we get the n-th power of the base-number a^a\a

2 .. . a

m_2fl

m-i is obviously in the first row of the triangle. Let us now assume, that in the n—1-th row (n > 1) we have the elements &o

, b\

, b

2,..., 1, b

p (where p equals to m| (m — 1 )(n — 2) — 1 = mn — m — 71+1, because in the first row there are m pieces of elements and in every new row there are m — 1 pieces more), and from these elements with positional addition we get the n — 1-th power of the number aQü\a

2 •••am-2

am-iThen we can write out (a

o10

m_

1 + ailO

m"

2 + • • • + a

m_

210 + a

m_i)

n as (M0

P + M 0

P + 6110

p_

1 + ••• + 6p_i 10 + frpJaolO™"

1 + (6

o10

p + bi lO?"

1 + • • • + b

p_ jlO + b

p)a

! 10

m"

2 + (6

o10

p + brio?'

1 + • • • + 6

p_il0 + &

p)a

m_

210 + (6

o10

p + 10

p—

1 + • • • + 6p_

x10 + &p)a

m_i. By adding these expressions (using that p = mn — m — n + 1) we get a

o6

o10

m

n"

n + (fl

06i + ai6o)10

mn-n1 + (a

0b

2 + fl

l6i + a

2&

o)10

m

nn2 + • • • + (a

m_i&

0 + a

m2bi + a

m_

36

2 • • • + a

lb

m_

2 + aoö

m-i)10

mn_

m"

n"

1 + f- K-2b

p + a

m_i&

p_i)10 + a

m_iöp. And this is exactly the number we get after positional addition from the n-th row of the triangle. Consideration of effectivity. This method is easy to algorithmize, it is enough to store the proceeding row and the base-number to determine one row of the triangle. In the row we work with relatively small numbers (compare with the final result), and we have to multiply only with digits. However, to reach the n-th row we need to determine (2 + (n — l)(m — 1)) j = 0(n

2m) elements. So obviously, if we need only the nthe power of the basenumber some other methods are more effective (Knuth [4]). However, if we need all the (non-negative integer) powers up to n of the base-number this method is competitive. It is especially interesting that with this method the first some powers of a base number of a few digits can even be determined