Notation: R(a,k) denotes the rising factorial a(a+1)...(a+k-1) Theorem: Let A(m) be the (m+1) by (m+1) matrix whose (i,j) entry is a(i,j):=, binomial(i + j + p, 2 i - j + 1) (0<=i,j<=m) . Let b(m):= /729\m 2 p 2 p 2 p |---| R(2/3 + ---, m) R(1/3 + ---, m) R(--- + 1, m) R(p/3 + 1/2, m) \256/ 3 3 3 / 2 R(p/3 + 5/6, m) R(p/3 + 7/6, m) p / (R(p/2 + 3/4, m) R(p/2 + 5/4, m) / R(p/2 + 1/4, m) R(p, m) R(3/2, m)), . Then [det A(m)]/[det A(m-1)] = b(m) and hence det [A(m)]=b(1)b(2)...b(m) . Rigorous Proof: Let c(m,n) be n (-4) R(p, n) R(-m, n) R(p + 1/2 + m, n) R(1, m) R(-2 m, m) R(2 p + 1 + 2 m, m) / m / (R(1, n) R(-2 m, n) R(2 p + 1 + 2 m, n) (-4) R(p, m) R(-m, m) / R(p + 1/2 + m, m)), . Note that c(m,m)=1, c(m,n)=0, for n>m . I claim that (1) Sum(c(k,n)*a(m,n),n=0..m)=0 for 0<=k=0 . These two statements imply the theorem since (1) is the defining property for the cofactors of the m-th row up to normalization, and with the normalization such that c(m,m)=1, which is tantamount to dividing the cofactors of the m-th row by the cofact\ or of the (m,m) entry, which is det(A(m-1)), the left of (2) is det[A(m)]/det[A(m-1)]. According to EKHAD, the left side of (1) is annihilated by the operator 3 2 2 -(m + p + 1) (m - p - 2 k + 1) (m + p + 2 k + 2) + (-8 m - 8 m p + 4 m p 2 3 2 2 2 + 20 m p k + 20 m k + 4 p + 16 p k + 16 p k - 44 m - 26 m p + 10 m k 2 2 + 10 p + 40 p k + 32 k - 82 m - 22 p + 16 k - 52) M 2 - 4 (2 m + 5) (m - k + 2) (2 m + 2 p + 2 k + 5) M 2 the certificate being, (6 m p - 5 m - 11 + 2 p + 11 p + 19 n + 9 n m + 3 p n) (-m - 2 + 2 n + p) (-m + 2 n + p - 1) n (2 k + n + 2 p) (-2 k + n - 1)/( (-2 m + n - 2) (-2 m - 3 + n) (-2 m - 4 + n) (-2 m - 5 + n)), . This implies that indeed the sum is zero for k