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%I #6 Oct 13 2012 00:23:22
%S 1,3,8,25,100,550,4224,42135,516383,7373984,118573680,2103205868,
%T 40634185593,847871397697,18987149095396,454032821689310,
%U 11544329612486760,310964453836199398,8845303172513782781
%N The number of disconnected 2k-regular simple graphs on 4k+5 vertices.
%F a(0)=1. For n > 0, a(n) = A051031(2k+4,3) + A051031(2k+3,2) = A005638(k+2) + A008483(2k+3).
%F Proof: Let C=A068934, D=A068933, and E=A051031. Now a(n) = D(4k+5,2k) = C(2k+1, 2k) C(2k+4,2k) + C(2k+2,2k) C(2k+3,2k), from the disconnected Euler transform. For n > 1, D(2k+1,2k) = D(2k+2,2k) = D(2k+3,2k) = D(2k+4,2k) = 0. Therefore, a(n) = E(2k+1, 2k) E(2k+4,2k) + E(2k+2,2k) E(2k+3,2k) = E(2k+1,0) E(2k+4,3) + E(2k+2,1) E(2k+3,2). Note that E(2k+1,0) = E(2k+2,1) = 1. Checking a(1) = E(6,3) + E(5,2), QED.
%e The a(0)=1 graph is 5K_1. The a(1)=3 graphs are 3C_3, C_3+C_6, and C_4+C_5.
%Y This sequence is the (even indices of the) fourth highest diagonal of D=A068933: that is a(n) = D(4k+5, 2k).
%Y Cf. A184324(k) = D(2k+4, k) and A184326(k) = D(2k+6, k).
%K nonn
%O 0,2
%A _Jason Kimberley_, Jan 11 2011
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