OFFSET
0,2
FORMULA
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).
KEYWORD
nonn
AUTHOR
Jason Kimberley, Jan 11 2011
STATUS
approved