OFFSET
0,3
FORMULA
a(0)=1. For k>0, a(k) = (k+1) mod 2 + A008483(k+3).
Proof: Let C=A068934, D=A068933, and E=A051031. Now a(n) = D(2k+4, k) = C(k+1, k) C(k+3, k) + A000217(C(k+2,k)), from the disconnected Euler transform. C(k+1, k)=1 because K_{k+1} is connected and the unique k-regular graph on k+1 vertices. For k > 1, since D(k+3,k)=0, then C(k+3,k) = E(k+3,k) = E(k+3,2) = A008483(k + 3). Also, for k >0, since D(k+2,k)=0, then C(k+2,k) = E(k+2,k) = E(k+2,1) = (k+1) mod 2. With the examples below and A165652(n)=0 for n < 6 = offset, QED.
EXAMPLE
The a(0)=1 graph is 4K_1. The a(1)=1 graph is 3K_2. The a(2)=2 graphs are C_3+C_5 and C_4+C_4.
PROG
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jason Kimberley, Jan 11 2011
STATUS
approved