Grrwdlngen d. Jlalh. CLASSES OF ONE-ARGUMENT RECURSIVE FTJNCTIONS by NADEJDA V. GEOR~IEVA in Sofia (Bulgaria) GRZEGORCZYK'S [l] sequence 8" (n 2 0) of increasing classes of functions has the following properties: 8 n + l contains gn, gn+l + &; each gn+l(x, x), where gn(x, y) is the function by which Bn is defined, majorizes the classe 8; of all the one argument functions in 8"; US" is the class of all primitive recursive functions.RITCHIE has proved in 121 that 82, n 2 2, is the smallest class of functions containing s(~), E(x), [x1j2] and fn(x, x) (ACKERMANN'S function), which is closed under addition, multiplication, composition and limited iteration.Here, we define and study a sequence Sn (n 2 0) of classes of one-argument functions the definition of which seems more simple than this of 8; and we prove that 3'n+1 = >l for n 2 2.In the first section below, we define the classes S,,, n 2 0, and develop some of their properties. In the second section, we consider $Po, S1 and di", more precisely. In the last section we establish the equality of 9'n+l and 8;+* for n 2 2.
1. The classes gnLet us setting f&) = S(x) = x + 1 and f r + l ( O ) = f r ( 1 ) , f r + l ( x + 1 ) = f r ( f r t l ( X ) ) > for r 5 0 .Here are including for reference two of monotonicity properties of the fr(x), which were established in [5]. We shall refer to them simply as [5].f r ( x ) < f r + l ( X ) , for T 2 0 and fr(z) < fr(x + l ) , for r 5 0 .Although we have got these properties of f&), now we shall prove three lemmas more. Lemma 1.1. For all x and r 2 2, 2fr(z) < fr(fr(x)).Proof. We shall proceed by induction on r, beginning at T = 2. 2 * f 2 ( z ) = 2(2Z + 3) < 2(2z + 3) + 3 = f z ( f 2 ( x ) ) .
* f r + l ( O ) = * f r ( 1 ) < f r ( f r ( 1 ) ) = f r ( f r + l ( O ) ) < fr+x(fr+l(O))and 2 * f r t l ( x + 1) = 2 * f r ( f r + 1 ( 2
) ) < f r ( f r ( f r t l ( X ) ) ) = f A f r t l ( x + 1)) < f r t l ( f r + l ( x + 1)).Assume the result holds for r. Using [5] we see that:Lemma 1.2. For each k 2 0 and r 2 1 there is an integer xo, 5 w h that fr(s + k ) < Proof. We have x + k < f2(xk -I) 5 frtl(xkl), for x > 3k -1 andThe function f(x) is defined from g(z) and j(s) by limited iteration, if f(0) = 0, j(x) for all x. This wiU be written sometimes for short < fr+l(zk), for all z 2 20. therefore fi(z + k ) < f r ( f r + l ( xk -1)) = f r t i ( xk).f(x + 1) = g(f(z)) and f ( x ) as m) = Sz(O), f(4 5 i(4. and f:' 2(1) = f r ( f r + l ( X ) ) = f r + l (~ + 1)-Lemma 1.3. For r 2 0 and x 2 0, g+'(l) = f r + l ( x ) .Proof. The definition of the function fr(x) and the induction on x give f r ( l ) = f , + l ( 0 )