A279055 - OEIS

%I #34 Feb 21 2023 11:28:43

%S 1,2,9,80,1240,30240,1071504,51996672,3307723776,266872320000,

%T 26615381760000,3214252921651200,462189467175321600,

%U 78024380924038348800,15279632043682406400000,3435553774431004262400000,879010223384483132866560000,253916900613208108255150080000

%N Self-convolution of squares of factorial numbers (A001044).

%C a(n) = (n!)^2 * Sum_{i=0..n} (binomial(n,i)^(-2)).

%C Consider a triangle ABC with area p. Let points X, Y, Z be randomly and uniformly chosen on sides BC, CA, BA. Let r = area of XYZ. Then the average or expected value of (r/p)^n = a(n)/(n!^2 * (n+1)^3).

%C a(n) = (3*(n+1)^4 *(n!)^4 /(2n+3)!) * Sum_{i=1..n+1} ((1/i)* binomial(2i, i)), see Sprugnoli Formula 5.2 as noted by Markus Scheuer.

%H Seiichi Manyama, <a href="/A279055/b279055.txt">Table of n, a(n) for n = 0..253</a>

%H Arman Maesumi, <a href="https://arxiv.org/abs/1804.11007">Triangle Inscribed-Triangle Picking</a>, arXiv:1804.11007 [math.GM], 2018.

%H R. Sprugnoli, <a href="http://www.dsi.dsi.unifi.it/~resp/GouldBK.pdf">Riordan Array Proofs of Identities in Gould's Book</a>.

%F a(n) = Sum_{i=0..n} (i! * (n-i)!)^2.

%F a(n) ~ 2*(n!)^2. - _Vaclav Kotesovec_, Dec 05 2016

%F a(n) = A001044(n)*A100516(n)/A100517(n). - _Alois P. Heinz_, Feb 21 2023

%t Table[Sum[(k!*(n-k)!)^2, {k, 0, n}], {n, 0, 20}] (* _Vaclav Kotesovec_, Dec 05 2016 *)

%Y Cf. A000142, A001044, A003149, A100516, A100517.

%K nonn

%O 0,2

%A _Arman Maesumi_, Dec 04 2016

%E Definition clarified by _Georg Fischer_, Feb 21 2023