rwg> The integerhood of a 3F2 summand I mentioned follows from the conjecture that n /===\ k - j + 1 k + 2 j + 2 k + 2 j + 3 | | (1 - q ) (1 - q ) (1 - q ) | | ------------------------------------------------------ | | j j + 1 j + 2 j = 1 (1 - q ) (1 - q ) (1 - q ) is a polynomial in q for all n,k >0. This seems to be hard!? I'll bet there's a theorem that for an expression like this, you only need to check a dozen or so consecutive n and k --rwg Chorasmian maraschino harmonicas inspissate antisepsis. Should the exponent on the first q in the numerator be k+j+1? As it stands, it looks like k=1 n=3 (and hence the j=3 term) will create a q^-1 that won't disappear. --- Wrt the pyramid puzzle, I recall seeing an article about it Science News, maybe 5-10 years ago. It may have been the Math SAT? Rich rcs@cs.arizona.edu