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Internat. J. Math. & Math. Sci. VOL. 12 NO. 4 (1989) 791-796 791 HOW MANY NUMBERS SATISFY THE 3X + 1 CONJECTURE? I. KRASIKOV School of Mathematlcal Sciences Tel-Avlv University Is rae I (Received March 25, 1988 and in revised form December 12, 1988) ABSTRACT. Let 0(x) be the number of numbers not exceeding x satisfy the 3X + We obtain a system of difference inequalities on functions closely conjecture. related to 0. Solving this system in the simplest case, we establish 0(x) > cx 3 7 This improves a result of Crandall KEY WORDS AND PHRASES. [I]. 3X + conjecture, residue class, difference inequality. 1980 AMS SUBJECT CLASSIFICATION CODES. IA, IB. I. INTRODUCTION. The famous conjecture of Collatz-Kakutanl, also known as the Syracuse or the "3X + i" problem, claims that the sequence + 3a a n 2 an+l T(Un) (1.1) --,n validity. following well-known -=- a n converges to the cycle (1,2) for any s The (mod 2) 0 (mod 2) 0 heuristic Z + argument serves Consider T as though it were a random walk. as an evidence for its It is natural to suppose that odd and even numbers appear independently, with probability I/2 at each Jump. 792 I. KRASIKOV Then T of (n) T() (0) should co,verge since the mathematical expectation is about --g- Although : (. -) /2 < this conjecture seems be intractable at present, to some supporting An interesting review on this problem can be found in In particular, Crandall [I] proved that the conjecture is true for many values {u x} 0 and u for some k Namely, set 9(x) results have been obtained. [2]. of 0" T(k)(u) i" Thus, 0(x)is just the number of numbers not exceeding x satisfing the conjecture. Then Crandall’s result is 0(x)> cx r, for appropriate constants c, > r However, 0. his proof gives a very poor value for r, about 0.05. Here we derive a system of difference inequalities on functions closely related (Lemma 4). to 0 Solving this system in the simplest case, we 3 establish 0(x) > Actually our proof gives a little more, namely: cx v E given any or 2 (mod 3) that is not in a cycle, for all x ) 3 {n where c o vx: T Ik(n) c0 x7, I} v for some k is a positive constant independent of v. In some sense the proof may be regarded as an attempt to formalize the above mentioned heuristic argument. 2. RESULTS. Consider the [(T(v), v)}, set E induced some infinite directed graph G on Tk(n) who.e edges are oriented from T(v) G of subgraph v and the vertex set V Ti(n) whose set vertex x for 0 k. to consists of That it is, v. all < v. G(v,=). We also put G(v) + the edge and Denote by G(v,x) an integers n consists of all whose trajectory hits v and remains below x the entire time. the empty set if x Z such that integers n In particular G(v,x) is Observe that G(v) has at most one cycle since the in degree of each vertex, but may be v, is one. Moreover, if v does not lie in a cycle of G then G(v) is a tree. Here we prefer to deal with U, the mapping inverse to T, namely: 2a, 0,1 (rood 3) u(=) (2.1) 2a U 2a- -------, a Since only numbers a--2 (rood 3) have iterates under U T -I restricted consider values of a (rood 9). to 2 (rood 3) two integers inverses under 2 (rood 3). T, we wish To do to analyze this we must HOW MANY NUMBERS SATISFY THE 3X Let S n be a complete of system resldue + CONJECTURE? classes 793 modulo 3 n. We split S as n follows: S 2 U n Ri a R n’ where i=O i n <--> a -= i (rood 3). Furthermore, put R2n Q2nlgQ5nL, Q8n’ , Qni <=> where R and U: R Obviously, U: R n n n the four following operators: U R 2 R 2 Ul(a) n’ n 5 o R U3: R2n-I’ U3 8 R n-I 2 The action of U on R 2 can be split into n 23 23 4 (a) 2a U4(e) n-l’ 2 n 9). 4 U () 2 U2: Qn U4: Qn R c :_ i (rood 3 The following lemma is an easy exercise in elementary number theory: LEMMA I. 2 U (i) 2 R ia a bljectlon R n n smallest positive integer such that U (ii) U (lii) U 3 4 is a bljectlon Qn2R2n-l" is a bljectlon Qn The action of U on R R U: R 2 n n (I) If v R n n-1 then is the Rn-l" n is much simpler. Namely, U: R n Moreover, since e&R n implies U(a) are bijections. LEMMA 2. . 2 n 2 8 and R n e R Moreover, if (e) R 2a n and R n we get then G(v) is a chain. Now we define the functions we deal with in this paper. Let v E m (mod 3n). f(v,x) We set f(v,x) redundant notation fm(v,x) instead of n difference inequalities that follow.) Observe that for v f (v,x) fm(v,x) n ( G(v,x) f(v,x) is to l" (The reason for using the simplify the statement of the x + [log2 + f2m(2v,x), n (2.2) n v m I. e Rn (2.3) I. KRASIKOV 794 {w} be the set of those vertices of G which do not belong to W for all K 0. Then G(w) is a tree and we set Furthermore, let W a cycle. m(y) n Note that for any m (rood uk(4) For instance, inf vW =- _m(y) LEMMA 3. PROOF inf{f(v,2Yv): 2 (rood 3) and n, the set [v: v because 2kv 3n)} 2Yv) fro(v, n ve W and v G(u), v E m is nondecreasing function of y. Obviously, fm(v,x) is a nondecreaslng function of x. m LEMMA 4. For y m n where a 4m (y- 2) + n (y) if v Hence, qn,m( y) 4mn (y- 2) + [y + a], me n n4m (y _m(y). 8 Qn (y + a- I), m 2 Qn e Qn8 (24) 5 Qn 1.585 and mln follows If v nm (y), m+3 n n-1 (Y)’ nm+2" 3 n-1 (y)). :(y). from the definition of 5 mE Qn then, by (2.1), if v x, immediately m (rood 3n), 2v 3- ,x) f(v,x) f4m(4v,X)n + + x, 2), m a n 0 (rood 3) then G If m e 2v-I 3 (Y + 2) + demonstrate (2 4) 2v3 n-I n.4m( y- (25) PROOF 4m-2 3 m( y) Iog23 (y). 0, mn-I (y) If 3n)}. is in this set and 2 is a primitive root (rood 3 n) for all n. inf fm(v 2Yv) is nondecreaslng function of y. (y) n n The following lemma gives important recurrent inequalities on Hence, m (rood is a chain by lemma 2. [log 2 2v 2) + [y + a]. then G(v,x) is a forest. Hence, 3x -i ]" Thus, by (2.2), Let us HOW MANY NUMBERS SATISFY THE 3X + By 2v- 9v < -: CONJECTURE? 2Yv, 0 and x and by lemna 3 we get, if y 795 then 2v-1 n’n(Y) fm(vn inf ,x) (f4m(4vn inf fn_13 (2v_,x)) ,x) + 2v-I 2v-I f4m(4v’n inf The 2 Qn case m fn_13 (2v3-I x) + inf n_13 (y + 2) + QS.n slmllarly to the case m considered may be O4mn (y ,x) a- I). We omit the details THEOREM I. 8 2 2(y) #2(y- 2(y) , 2 mln 5 8 2 (y) 2 (y since I (y + 2 2 I (y + 2 rain ($ 2 2 5 2. 2) > 0 if y min 2 (y) 2(y- 4) + 2 (y 2 6) + I (y + 22(y- 6) + 22(y + The initial conditions % -6 $ 21 (y) for y 2 by Hence, 2 2 2 2(y 2)) (2(Y)’ 2 (y y X-5 Finally, we obtain 0() a + - I) + 21(y + I) + 2 !(y + - 5) + 22(y + 2 imply 2(y) n + I, one has 2(0) + l(y 2 2 induction on n, that for n REMARK. 5 2 (y)) (y), 8 $2(y) > Observe that 2) 6, This yields if y largest root of 5 2(y- and 1), 2 > I(Y)’ I) I (y) I) $2 (y) 5 ($2(y), 2(y), 2(y)). 2) + a - 2l(y + 8 2 + a- 2), Ot(y 2), y 2 2 2) + (y-2) + 2 > $2 (y) $1(y) O, 2 the system (2.4) becomes for y 5 where > c2x 0(x) For n PROOF. 3 7 for y 2) 2) 4). 6, whence one proves by ,22(y) c!%Y, where k 1.3534 is the + log2k c2x 3 > c2x wheme log2k 0.436. Although system (2.4) seems to be very complicated and we were unable to I. KRASIKOV 796 3, averaging it over all residue classes modulo 3 solve it for n n-I looks much more attractive. Namely, define F (y) n 3 -n+l " 2 m:Rn :(Y)" and 4 we get Using lemmas 3n-I Fn(Y)=m( R m(n y) 2 n 3n-lFn (y m m- R2n 2) + m (y-2) + ). 2 n_l(y + a-2) + 2 n_l(y+a-l) mgR m-R n-I n-I 3n-2Fn_l(y + 2) + a- 3n-2Fn_l (y + a 1). Thus, F (y) n Fn(Y- 2) + Fn_l(y + c- 2) Observe that the associated limit equation the smallest positive root. + Fn_l(y + 4 -2 + ct- l). a-2+ -I has k 2 as Therefore, one might expect that the solution of the difference ineqalities gives 0(x) > CnX n ,where rn when n tends to infinity. REFERENCES 1. 2. CRANDALL, R.E., On the "3X + I" problem Math. Comp. 32, (1972), 1281-1292. problem and its generalizations, The American Math. LAGARIAS, J.C., The 3X + Moth! 92, (1985), 3-23.