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a(n) = A000085(4*n+2)/2^(n+1).
4

%I #21 Jul 11 2018 10:35:55

%S 1,19,1187,149405,31166057,9670072483,4163946939067,2370770585582221,

%T 1722046856020416785,1552401874990891104371,1699257737580930574489619,

%U 2218555640616875773883091901,3404174268230266459851637679353,6062646848508401565245592651382915,12398960005973049406349011215379703723

%N a(n) = A000085(4*n+2)/2^(n+1).

%H Seiichi Manyama, <a href="/A316332/b316332.txt">Table of n, a(n) for n = 0..210</a>

%H S. Chowla, I. N. Herstein and W. K. Moore, <a href="http://dx.doi.org/10.4153/CJM-1951-038-3">On recursions connected with symmetric groups I</a>, Canad. J. Math. 3 (1951) 328-334.

%H Dongsu Kim and Jang Soo Kim, <a href="http://dx.doi.org/10.1016/j.jcta.2009.08.002">A combinatorial approach to the power of 2 in the number of involutions</a>, J. Comb. Theory Ser. A 117 (8) (2010): 1082-1094.

%Y Cf. A000085, A316330, A316331, A316333.

%K nonn

%O 0,2

%A _N. J. A. Sloane_, Jul 09 2018, following a suggestion from _Doron Zeilberger_