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

%I #22 Jul 11 2018 10:36:31

%S 1,13,655,71063,13237457,3748521653,1495006933759,796798642614895,

%T 546144645571635169,467512355698028529821,488384275088035513080239,

%U 611057820865315450415912327,901643505614430586911510015025,1548711768835068239482321088560837

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

%H Seiichi Manyama, <a href="/A316331/b316331.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, A316332, A316333.

%K nonn

%O 0,2

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