%I #40 Jul 18 2022 22:49:11
%S 1,1,1,1,2,3,6,11,24,51,120,283,716,1833,4948,13561,38788,112745,
%T 339676,1039929,3283876,10532747,34717276,116158851,398257012,
%U 1385117947,4925094508,17752742867,65297807204,243319812785,923739847132,3550638576721,13885783706324
%N Number of up-down involutions of length n.
%C This resulted from a question from _Richard Ehrenborg_ and Margie Readdy.
%H Alois P. Heinz, <a href="/A072187/b072187.txt">Table of n, a(n) for n = 0..500</a> (terms n = 1..50 from Vladeta Jovovic)
%H D. Zeilberger, <a href="https://sites.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/ehrenborg.html">I Am Sorry, Richard Ehrenborg and Margie Readdy, About Your Two Conjectures, But One Is FAMOUS, While The Other Is FALSE</a>; <a href="/A072187/a072187.pdf">Local copy</a> [PDF file only, no active links]
%F G.f.: Sum_{n>=0} a(2n+1)x^(2n+1) = Sum_{i,j >= 0) arctan(x)^(2i+1) (log((1+x^2)/(1-x^2)))^j E(2i+2j+1)/((2i+1)!j!4^j), where E(2i+2j+1) is an Euler number (A000111). There is a similar but more complicated generating function for a(2n). - _Richard Stanley_, Jan 02 2006
%e a(3)=1 since among the four involutions of length 3 (123, 213, 321, 132), only one is up-down (132).
%Y Cf. A000085, A000111.
%K nonn
%O 0,5
%A _Doron Zeilberger_, Jul 01 2002; more terms, Dec 09 2003
%E More terms from _Vladeta Jovovic_, May 16 2007
%E a(0)=1 prepended by _Alois P. Heinz_, Aug 07 2018