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In this case, ${2n-2<p-4}$, so that the *lowest* order in the key equation (\[KeyEq1\]) is on the *left hand* side, namely $\Delta^l$ with ${l=2n-2}$, and this yields the condition $$n(n+1)=0 \,.
\label{KeyEq1CaseI}$$ There are thus only two possible cases, namely ${n=0}$ and ${n=-1}$. Next, it is convenient to apply the equation (\[KeyEq3\]) whose lowest orders on its both sides are $$\big[6n(n+p-1)+p(p-1)\big]c_0\,\Delta^{n+p-2}+\cdots=-2\,\Delta^{n}+\cdots \,.
\label{KeyEq3CaseI}$$ For ${n=0}$, these powers are ${\Delta^{p-2}}$ and ${\Delta^{0}}$, respectively, but ${p-2>2n=0}$ by the definition of Case I. The lowest order ${0=-2\Delta^{0}}$ thus leads to a contradiction. Only the possibility ${n=-1}$ remains, for which reduces to $$(p-3)(p-4)c_0\,\Delta^{p-3}+\cdots =-2\,\Delta^{-1}+\cdots
\,.
\label{KeyEq3CaseIn=-1}$$ Since ${c_0\ne0}$, the only possibility is ${p=2}$, in which case ${c_0=-1}$.

**To summarize**: The only possible class of solutions in Case I is given by $$[n,p]=[-1,2]\qquad \hbox{with}\quad c_0=-1\,.
\label{CaseI_summary}$$

**Case II**
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In this case, ${2n-2>p-4}$, so that the *lowest* order in the key equation (\[KeyEq1\]) is on the *right hand* side, namely $\Delta^l$ with ${l=p-4}$, and this gives the condition $$p(p-1)(p-2)(p-3)=0 \,.
\label{KeyEq1CaseII}$$ Thus there are four possible cases, namely ${p=0}$, ${p=1}$, ${p=2}$, and ${p=3}$. Equation has the lowest orders on both sides the same as given by equation , that is $$\begin{aligned}
\hbox{for}\quad p=0:\qquad &
