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On Two Families of Generalizations of Pascal's Triangle
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Michael A. Allen and Kenneth Edwards

Physics Department

Faculty of Science

Mahidol University

Rama 6 Road

Bangkok 10400

Thailand

**Abstract:**

We consider two families of Pascal-like triangles that have all ones on
the left side and ones separated by *m* – 1 zeros on the right side. The
*m* = 1 cases are Pascal's triangle and the two families also coincide when
*m* = 2. Members of the first family obey Pascal's recurrence everywhere
inside the triangle. We show that the *m*-th triangle can also be obtained
by reversing the elements up to and including the main diagonal in each
row of the (1/(1 – *x*^{m}), *x*/(1 – *x*))
Riordan array. Properties of
this family of triangles can be obtained quickly as a result. The (*n*,
*k*)-th entry in the *m*-th member of the second family of triangles is the
number of tilings of an (*n* + *k*) × 1 board that use *k* (1, *m* – 1)-fences
and *n* – *k* unit squares. A (1, *g*)-fence is composed of two unit square
sub-tiles separated by a gap of width *g*. We show that the entries in the
antidiagonals of these triangles are coefficients of products of powers
of two consecutive Fibonacci polynomials and give a bijective proof that
these coefficients give the number of *k*-subsets of
{1, 2, ... , *n* − *m*}
such that no two elements of a subset differ by *m*. Other properties
of the second family of triangles are also obtained via a combinatorial
approach. Finally, we give necessary and sufficient conditions for any
Pascal-like triangle (or its row-reversed version) derived from tiling
(*n* × 1)-boards to be a Riordan array.

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(Concerned with sequences
A000045
A000079
A000217
A001045
A006324
A006498
A007318
A011973
A059259
A077947
A079962
A115451
A118923
A123521
A157897
A335964
A349839
A349840
A349841
A349842
A349843
A350110
A350111
A350112.)

Received January 18 2022;
revised versions received May 26 2022; June 21 2022.
Published in *Journal of Integer Sequences*,
July 29 2022.

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