A184325 The number of disconnected 2k-regular simple graphs on 4k+5 vertices.

1, 3, 8, 25, 100, 550, 4224, 42135, 516383, 7373984, 118573680, 2103205868, 40634185593, 847871397697, 18987149095396, 454032821689310, 11544329612486760, 310964453836199398, 8845303172513782781

Offset

0,2

Links

Table of n, a(n) for n=0..18.

Formula

a(0)=1. For n > 0, a(n) = A051031(2k+4,3) + A051031(2k+3,2) = A005638(k+2) + A008483(2k+3).

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.

Example

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.

Crossrefs

This sequence is the (even indices of the) fourth highest diagonal of D=A068933: that is a(n) = D(4k+5, 2k).

Cf. A184324(k) = D(2k+4, k) and A184326(k) = D(2k+6, k).

Sequence in context: A357591 A182927 A012408 * A051403 A004205 A268114

Adjacent sequences: A184322 A184323 A184324 * A184326 A184327 A184328

Keyword

nonn

Author

Jason Kimberley, Jan 11 2011

Status

approved