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27 April 2024
 
  » arxiv » 2201.13253

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On Two Families of Generalizations of Pascal's Triangle
Michael A. Allen ; Kenneth Edwards ;
Date 31 Jan 2022
AbstractWe 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 $(n+k) imes1$ 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 ${1,2,ldots,n-m}$ 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 $(n imes1)$-boards to be a Riordan array.
Source arXiv, 2201.13253
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