A122067 - OEIS

%I #13 Oct 09 2024 10:59:53

%S 1,8,1024,2097152,68719476736,36028797018963968,

%T 302231454903657293676544,40564819207303340847894502572032,

%U 87112285931760246646623899502532662132736,2993155353253689176481146537402947624255349848014848

%N a(n) = 2^A014105(n).

%C a(n) is the number of simple labeled graphs on 2(n+1) nodes such that every vertex has odd degree.  The complements of these graphs are precisely the Eulerian graphs on 2(n+1) nodes. a(1) = 8 because we have: K_4; K_1,3; and K_2 + K_2 with 1,4, and 3 labelings respectively: 1 + 4 + 3 = 8. Cf. A006125. - _Geoffrey Critzer_, Feb 16 2020

%H Alois P. Heinz, <a href="/A122067/b122067.txt">Table of n, a(n) for n = 0..40</a>

%F a(n) = (-1)^floor(n/2)/Product_{i=1..2*n} cos(i*Pi/(2*n+1))^i.

%p a:= n-> 2^(n*(2*n+1)):

%p seq(a(n), n=0..10);  # _Alois P. Heinz_, Feb 16 2020

%t Table[2^(Binomial[n, 2] - (n - 1)), {n, 2, 20, 2}] (* _Geoffrey Critzer_, Feb 16 2020 *)

%o (PARI) a(n)=2^(n*(2*n+1))

%Y Cf. A006125, A014105.

%K nonn

%O 0,2

%A _Benoit Cloitre_, Oct 15 2006