number of boxes in the short column is always an even number.

Two Impurities
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For a given value of $a_{\rm max}$ there are $2+2a_{\rm max}$ states in total. There are ${3\over 2}a_{\rm max}+1$ zero eigenvalues (corresponding to supersymmetric states). The remaining eigenvalues are $$\lambda_i =8g_{YM}^2 i\qquad i=1,2,\cdots,{a_{\rm max}\over 2}+1\, .$$

Three Impurities
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For a given value of $a_{\rm max}$ there are $1+3a_{\rm max}$ states in total. There are $2 a_{\rm max}$ zero eigenvalues (corresponding to supersymmetric states). The remaining eigenvalues are $$\lambda_i =8g_{YM}^2 i\qquad i=1,2,\cdots,{a_{\rm max}\over 2}\, ,$$ each with a degeneracy of two and a single maximum eigenvalue $\lambda=4a_{\rm max}g_{YM}^2+8g_{YM}^2$. This degeneracy almost certainly indicates a symmetry enhancement in the large $N$ limit.

Four Impurities
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For a given value of $a_{\rm max}$ there are $1 +{9\over 2}a_{\rm max}$ states in total. There are ${5\over 2}a_{\rm max}-1$ zero eigenvalues. The eigenvalues are again evenly spaced with a level spacing of $8g_{YM}^2$ and they are again degenerate. The low lying eigenvalues $$\lambda_i = 8 g_{YM}^2 i\qquad i=1,2,\cdots,{a_{\rm max}\over 2}\, ,$$ have a degeneracy which alternates between 3 and 4. Thus, we find three eigenvalues $\lambda =8g_{YM}^2$, followed by four eigenvalues $\lambda=16g_{YM}^2$, followed by three eigenvalues $\lambda=24g_{YM}^2$, followed by 