cp's OEIS Frontend

a(0)=1. For k>0, a(k) = (k+1) mod 2 + A008483(k+3).

For k>=0, a(k) = A040001(k) + A165652(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.

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.

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.