A simple example of a transcendental entire function that together with all its derivatives assumes algebraic values at all algebraic points

Abstract

Let { iz } = { z1, z2, Zs, } be an enumeration of all algebraic numbers [1]. Construct a sequence [21 { ,} = {I1, t2, *, * * = I z1, z1, z2, z1, z2, Zs, z *, } so that all of the algebraic numbers appear an infinite number of times in {I X}. Then, for algebraic numbers an with O 1, the series for f(z) converges absolutely and uniformly in I zl 5 R < X and If(z) I ? max { e, elst }. Since f(m) (cj) is a polynomial of Rj with algebraic coefficients an and { rj } contains all algebraic numbers infinitely many times, f(m) (r) must be an algebraic number for any algebraic number t. If we ask the general question: For what sets, S, of complex numbers do there exist transcendental entire functions which, together with all their derivatives, map S into S?, we see immediately that the above construction can be applied to any dense denumerable set, or to any denumerable ring which has 0 as a limit point, such as the ring of rationals. A similar method can be applied to discrete infinite rings such as the ring of integers. The question for nondenumerable nonclosed rings S remains open.