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A168510 Products across consecutive rows of the denominators of the Leibniz harmonic triangle (A003506).

%I A168510 #24 Mar 11 2018 03:22:43
%S A168510 1,4,54,2304,300000,116640000,133413966000,444110104166400,
%T A168510 4267295479315169280,117595223746560000000000,
%U A168510 9245836018244425723200000000,2065215715357207851951980544000000
%N A168510 Products across consecutive rows of the denominators of the Leibniz harmonic triangle (A003506).
%C A168510 As in A001142, lim_{n->inf} (a(n)a(n+2))/a(n+1)^2 = e, demonstrating an underlying relation between A003506 and Pascal's triangle A007318. Unlike A001142, in this case the function is asymptotic from above.
%H A168510 A. Bogomolny, <a href="http://www.cut-the-knot.org/Curriculum/Combinatorics/LeibnitzTriangle.shtml">Cut The Knot: Leibniz and Pascal Triangles</a>
%H A168510 H. J. Brothers, <a href="https://doi.org/10.1017/S0025557200004447">Pascal's prism</a>, The Mathematical Gazette, 96 (July 2012), 213-220.
%F A168510 a(n) = n!*Product_{k=1..n} k^(2k-n-1).
%F A168510 a(n) = Product_{j=1..n} Product_{k=2..j} ((1-1/k)^-k).
%F A168510 a(1) = 1; a(n) = a(n-1)*Product_{k=2..n} ((1-1/k)^-k).
%F A168510 a(n) ~ A^2 * exp(n^2/2 - 1/12) * n^(n/2 + 1/6) / (2*Pi)^(n/2), where A is the Glaisher-Kinkelin constant A074962. - _Vaclav Kotesovec_, Oct 22 2017
%F A168510 a(n) = Product_{k=0..n-1} (n-k)^(n-2k). - _Peter Munn_, Mar 07 2018
%e A168510 For n=3, row 3 of A003506 = {3, 6, 3} and a(3)=54.
%e A168510 a(5) = 5^5 * 4^3 * 3^1 * 2^-1 * 1^-3 = 5^5 * 3 * 2^5 = 300000. - _Peter Munn_, Mar 07 2018
%t A168510 Table[n! Product[k^(2 k - n - 1), {k, 1, n}], {n, 1, 12}]
%t A168510 Table[Product[Product[(1 - 1/k)^-k, {k, 2, j}], {j, 1, n}], {n, 1, 12}]
%t A168510 (* or *)
%t A168510 a[1] = 1; a[n_] := a[n - 1] Product[(1 - 1/k)^-k, {k, 2, n}]
%Y A168510 Cf. A003506, A001142, A007318. For n >= 1, a(n) = n!*A001142(n).
%K A168510 easy,nonn
%O A168510 1,2
%A A168510 _Harlan J. Brothers_, Nov 27 2009