Eric Weisstein's World of Mathematics, <a href="httphttps://mathworld.wolfram.com/AdjacencyMatrix.html">Adjacency Matrix</a>

Eric Weisstein's World of Mathematics, <a href="httphttps://mathworld.wolfram.com/DoubleFactorial.html">Double Factorial</a>

Eric Weisstein's World of Mathematics, <a href="httphttps://mathworld.wolfram.com/Erf.html">Erf</a>

Eric Weisstein's World of Mathematics, <a href="httphttps://mathworld.wolfram.com/LadderRungGraph.html">Ladder Rung Graph</a>

Eric Weisstein's World of Mathematics, <a href="httphttps://mathworld.wolfram.com/NormalDistributionFunction.html">Normal Distribution Function</a>

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S. Goodenough and C. Lavault, <a href="httphttps://www.combinatoricsdoi.org/ojs/index10.php/eljc/article/view37236/v22i4p165264">Overview on Heisenberg—Weyl Algebra and Subsets of Riordan Subgroups</a>, The Electronic Journal of Combinatorics, 22(4) (2015), #P4.16.

Helmut Prodinger, <a href="https://www.combinatoricsdoi.org/ojs/index10.php/eljc/article/view37236/v3i1r291253">Descendants in heap ordered trees or a triumph of computer algebra</a>, The Electronic Journal of Combinatorics, Volume 3, Issue 1 (1996), R29.

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b(n) = a(n) / (n! 2^n) = Sum_{k = 0..n} (-1)^n binomial(n,k) (-1)^k a(k) / (k! 2^k) = (1-b.)^n, umbrally; i.e., the normalized double factorial a(n) is self-inverse under the binomial transform. This can be proved by applying the Euler binomial transformation for o.g.f.s Sum_{n >= 0} (1-b.)^n x^n = (1/(1-x)) Sum_{n >= 0} b_n (x / (x-1))^n to the o.g.f. (1-x)^{-1/2} = Sum_{n >= 0} b_n x^n. Other proofs are suggested by the discussion in Watson on pages 104-5 of transformations of the Bessel functions of the first kind with b(n) = (-1)^n binomial(-1/2,n) = binomial(n-1/2,n) = (2n)! / (n! 2^n)^2. - Tom Copeland, Dec 10 2022 (Subscript corrected Dec. 16, 2024)

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b(n) = a(n) / (n! 2^n) = Sum_{k = 0..n} (-1)^n binomial(n,k) (-1)^k a(k) / (k! 2^k) = (1-b.)^n, umbrally; i.e., the normalized double factorial a(n) is self-inverse under the binomial transform. This can be proved by applying the Euler binomial transformation for o.g.f.s Sum_{n >= 0} (1-b_n.)^n x^n = (1/(1-x)) Sum_{n >= 0} b_n (x / (x-1))^n to the o.g.f. (1-x)^{-1/2} = Sum_{n >= 0} b_n x^n. Other proofs are suggested by the discussion in Watson on pages 104-5 of transformations of the Bessel functions of the first kind with b(n) = (-1)^n binomial(-1/2,n) = binomial(n-1/2,n) = (2n)! / (n! 2^n)^2. - Tom Copeland, Dec 10 2022 (Subscript corrected Dec. 16, 2024)

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