Yi-Wen Liu, a former student of Julius Smith (Stanford Dept of Music) noticed, in effect, binom(4*(n+1)^2,(n+1)*(2*n+1))/binom(2*(2*n^2+4*n+1),n*(2*n+3)) = 4 4(n+1)^2 ( ) (n+1) (2n+1) ---------------- = 4 . 2 (2n^2+4n+1) ( ) n (2n+3) I just found binom((2*n+1)^2,(n+1)*(2*n+1))/binom(4*n^2+4*n-1,n*(2*n+3)) = 4 (2n+1)^2 ( ) (n+1) (2n+1) --------------- = 4 . 4n^2+4n-1 ( ) n (2n+3) Both of these are trivial to prove, but perhaps not to find, and I think at least one of them was discussed here a few years ago. There are more. Have they been characterized? Less obvious: binom(fib(2*n+4)-fib(2*n)-2,fib(2*n+1)-1)/binom(fib(2*n+4)-fib(2*n)-4,fib(2*n+1)-2) fib(2n+4)-fib(2n)-2 ( ) fib(2n+1)-1 --------------------- = 5 , fib(2n+4)-fib(2n)-4 ( ) fib(2n+1)-2 which is equivalent to the identity (fib(2*(n+2))-fib(2*n)-3)*(fib(2*(n+2))-fib(2*n)-2)/((fib(2*n+1)-1)*(fib(2*n+3)-1))=5 (fib(2 (n + 2)) - fib(2 n) - 3) (fib(2 (n + 2)) - fib(2 n) - 2) --------------------------------------------------------------- = 5 . (fib(2 n + 1) - 1) (fib(2 n + 3) - 1) Known? --rwg ISOMETRIC EROTICISM ISOTONIC COITIONS