 Solution

Because $\Delta K=0$, the $(\tilde{C}-\tilde{B})K^2$ terms cancel between the initial and final states. The two positive-shift branches are consequentlyThe $+2$ branch gives shifts $6\tilde{B},10\tilde{B},14\tilde{B},\dots$, as for the diatomic. The $+1$ branch fills in a series spaced by $2\tilde{B}$, so the combined nonzero series is $4\tilde{B},6\tilde{B},8\tilde{B},10\tilde{B},\dots$.

There is a boundary qualification to the abbreviated rules printed in the question: for an ordinary symmetric polarizability tensor, $\Delta J=\pm 1$ has zero strength when $K=0$. Thus $J=0,K=0\to J=1,K=0$ does not produce a $2\tilde{B}$ line; the first $+1$ transition can start at $J=1,|K|=1$ and has shift $4\tilde{B}$. This follows from the rank-two angular-momentum coupling: the all-zero-projection coupling for $J$ and $J+1$ vanishes because $J+2+(J+1)$ is odd. If one applies only the printed inequalities without this matrix-element restriction, $2\tilde{B}$ is a formal position, but its intensity is zero.

The lower panel of the figure in part (a)(ii) marks the two branches and their overlaps. The useful spacing is $2\tilde{B}$, not the diatomic value $4\tilde{B}$.