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
12 Gábor Kallós Definition. Let a and b integers, with 0 < a, 6 < 9. Then we can get the k-th element in the n-th row of the a6-based triangle if we add the k — 1-th elemetn in the n — 1-th row 6-times to the k-th element in the n — 1-th row a-times. If k — 1 < 0 or k > n — 1 (id est the element in the n — 1-th row does not exist according to the traditional implementation) then we consider this element to be 0 (Figure 2.). The indices in the rows and columns of the triangle run from 0. 1 4 7 16 56 49 64 336 588 343 Figure: 2: The -based triangle Example. In the third row of the 47-based triangle 64 = 7 • 0 + 4 • 16 and 336 = 7 • 16 + 4 • 56. Proposition 1. By positional addition from the n-th row of the abbased triangle we get the n-th power of ab(10a -f 6). Proof. From the expansion of (10a -f b) n we get (10a + 6) n = ^a n10 n+^a n1610 n" 1+---+^" 1^a6 w_ 110 + This is exactly the number we get after positional addition from the n-th row. The structure of the a6-based triangle is relatively simple. We have the following. Proposition 2. The k-th element in the n-th row of the ab-based 1 triangle is a n~ kb kCwhere (the number of combinations of n things taken k at a times) is the k the element in the n-th row of Pascal's Triangle. Proof. We prove by induction. In the first row we have a = a 1 • 1 and b = b l • 1. Let us now assume, that the k — 1-th element in the n — 1-th row is an-k bk-i ck-i and the k_ th elemen t in the n - 1-th row is a n~ k~ 1b kC k_ l. Then the k-th element in the n-th row by definition is ba nkb k~ lC k nZ\ +aa nk~ 1b kC*_ 1 = a n~ kb kC k nZ\ + a n~ kb kC k_ l - a n~ kb k (C kl{ 4- C k_ x) = a n~ kb kC k.
