A184324 -id:A184324 - cp's OEIS frontend

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

Jason Kimberley, Jan 11 2011

nonn

			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.

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).

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.

A184326 The number of disconnected k-regular simple graphs on 2k+6 vertices.

Original entry on oeis.org

1, 1, 4, 9, 25, 66, 297, 1562, 10901, 88238, 806174, 8037887, 86228020, 985884104, 11946634677, 152808994328, 2056701656260

Offset: 0

Jason Kimberley, Jan 15 2011

			The a(0)=1 graph is 6K_1. The a(1)=1 graph is 4K_2. The a(2)=4 graphs are 2C_3+C_4, 2C_5, C_4+C_6, and C_3+C_7.

This sequence is the fifth highest diagonal of D=A068933: that is a(n)=D(2k+6, k).

Cf. A184324(k) = D(2k+4, k) and A184325(k) = D(4k+5, 2k).

a(0)=1, a(1)=1, a(2)=4, a(3)=9. For n>3, a(n) = A033301(k+5) + ((k+1)mod 2)*A005638(k div 2 + 2) + A000217(A008483(k+3)).

Proof: Let C=A068934, D=A068933, and E=A051031. Now a(n) = D(2k+6,k) = C(k+1,k)C(k+5,k) + C(k+2,k)C(k+4,k) + A000217(C(k+3,k)), from the disconnected Euler transform. Notice that D(k+i,k)=0 provided k+i < 2k+2; that is k > i-2. So if i <= 5 and k > 3, then D(k+i,k)=0. Hence for k > 3, a(n) = E(k+1,k)E(k+5,k) + E(k+2,k)E(k+4,k) + A000217(E(k+3,k)) = E(k+1,0)E(k+5,4) + E(k+2,1)E(k+4,3) + A000217(E(k+3,2)). We have E(k+1,0)=1, and E(k+2,1)=(k+1)mod 2. For even k, E(k+4,3)=A005638(k div 2 + 2); for odd k, E(k+2,1)=0. QED.

cp's OEIS Frontend

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

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