In this note we prove that j! 2N (N+j−1j) 2F1(−j,−2j−N−j+1;−1)=∑l=0N(Nl)∏i=0j−12(2i+1+l),j!2N(jN+j−1)2F1(−j,−2j−N−j+1;−1)=l=0∑N(lN)i=0∏j−12(2i+1+l), where NN and jj are positive integers, which resolves a question posed by Ercolani, Lega, and Tippings.
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