THE ZEROS OF THE ROW POYLNOMIALS OF A185417 Define a polynomial sequence S(n,x) recursively by (1)... S(n+1,x) = x*S(n,x-1)+(x+1)*S(n,x+1) with S(0,x) = 1. The first few polynomials are S(0,x) = 1 S(1,x) = 2*x+1 S(2,x) = 4*x^2+4*x+3 S(3,x) = 8*x^3+12*x^2+26*x+11. We shall show that the polynomials S(n,-x) satisfy a Riemann hypothesis: that is, for each n the zeros of the polynomial S(n,-x) lie on the vertical line Re x = 1/2 in the complex plane. The following proof is adapted from similar proofs in the literature. PROOF: Let Q(n,x) = S(n,x-1/2). It is sufficient to show that for each n the polynomial Q(n,x) has all its zeros on the imaginary axis. The proof is by induction. Certainly Q(1,x) = 2*x has its zeros on the imaginary axis. Let n >= 1, and let r_1,...,r_n be the zeros of Q(n,x), which we assume to lie on the imaginary axis. Then Q(n,x) factorizes as (2).. .Q(n,x) = c*(x-r_1)...(x-r_n) for some nonzero constant c. We use an argument by contradiction to show that the zeros of Q(n+1,x) also lie on the imaginary axis. From (1) the recurrence satisfied by the Q polynomials is (3)... Q(n+1,x) = (x-1/2)*Q(n,x-1)+(x+1/2)*Q(n,x+1). Let s be a zero of Q(n+1,x). Then by (2) and (3) (4)... |(s-1/2)*(s-1-r_1)*...*(s-1-r_n)| = |(s+1/2)*(s+1-r_1)*...*(s+1-r_n)|. Suppose now s is not purely imaginary, say Re(s) > 0 (the case Re(s) < 0 is handled similarly). Then clearly (5)... |s+1/2| > |s-1/2|. Since Re(s) > 0 and by the induction hypothesis Re(r_i) = 0 we have (6)...|s+1-r_i| > |s-1-r_i| for each i. Multiplying the inequalities (5) and (6) together gives a contradiction to (4). Hence the zeros of Q(n+1,x) are pure imaginaries and the induction goes through. Peter Bala