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 A184326 The number of disconnected k-regular simple graphs on 2k+6 vertices. 2
 1, 1, 4, 9, 25, 66, 297, 1562, 10901, 88238, 806174, 8037887, 86228020, 985884104, 11946634677, 152808994328, 2056701656260 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 LINKS FORMULA 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. EXAMPLE 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. CROSSREFS 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). Sequence in context: A007598 A121648 A133022 * A217590 A279397 A176497 Adjacent sequences:  A184323 A184324 A184325 * A184327 A184328 A184329 KEYWORD nonn,hard,more AUTHOR Jason Kimberley, Jan 15 2011 STATUS approved

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Last modified July 15 16:29 EDT 2019. Contains 325049 sequences. (Running on oeis4.)