the Andrews determiant Theorem: Let A(m) be the (m+1) by (m+1) matrix whose (i,j) entry is a(i,j):=, binomial(i + j, j) (0<=i,j<=m) , and let I(m) be the (m+1) by (m+1) identity matrix Let b(m) (m=1,2, ...) be the sequene annihilated by the 3 (3 m + 5) (3 m + 1) recurrence operator , - --------------------- + M 4 (2 m + 3) (2 m + 1) with the initial condition b(1)= , 1, . Then [det (I(m)+A(m))]/[det (I(m-1)+A(m-1))] = b(m) . Semi-Rigorous Proof: Let c(m,n) be the unique vector of length m+1 (c(m,0), ..., c(m,m)) satisfying c(m,m)=1, c(m,n)=0, for n>m, and annihilated by 2 (n + 2) (n + 1 + m) (m - n) the operator, -------------------------------- - 2 (n + 1) (n + 3 + m) (m - 2 - n) 2 2 2 2 2 2 3 4 (-6 + m + m - 21 n + 3 m n + 3 m n - 25 n + m n + m n - 12 n - 2 n ) / 2 2 N / ((n + 1) (n + 3 + m) (m - 2 - n)) + N , . / I claim that (1) Sum(c(k,n)*a(m,n),n=0..m)=0 for 0<=k