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a(n) = Sum_{k=1..n} floor(k/2)! * floor((n - k)/2)! binomial((n-floor(k/2)-1), n-k).
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%I #7 Nov 26 2020 03:20:09

%S 1,2,3,6,12,26,62,148,396,1044,3024,8784,26928,85824,274320,954720,

%T 3149280,11910240,40253760,164643840,567181440,2497703040,8736698880,

%U 41250263040,146090649600,736680268800,2635858713600,14145091430400,51047113420800,290574650419200

%N a(n) = Sum_{k=1..n} floor(k/2)! * floor((n - k)/2)! binomial((n-floor(k/2)-1), n-k).

%H Jonathan Fang, Zachary Hamaker, and Justin Troyka, <a href="https://arxiv.org/abs/2009.00079">On pattern avoidance in matchings and involutions</a>, arXiv:2009.00079 [math.CO], 2020. See Theorem 1.6 (c).

%t Array[Sum[Floor[k/2]! Floor[(# - k)/2]! Binomial[(# - Floor[k/2] - 1), # - k], {k, #}] &, 30]

%o (PARI) a(n) = sum(k=1, n, (k\2)! * ((n-k)\2)! * binomial(n-k\2-1, n-k)); \\ _Michel Marcus_, Nov 25 2020

%K nonn

%O 1,2

%A _Michael De Vlieger_, Nov 25 2020