# Document 10443755 ```Internat. J. Math. &amp; 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)
&gt;
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
&quot;3X + i&quot; 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
--g-
Although
:
(. -) /2 &lt;
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.
.
of
0&quot;
T(k)(u)
i&quot;
Thus, 0(x)is just the number of numbers not exceeding x satisfing the conjecture.
Then Crandall’s result is 0(x)&gt; cx
r,
for appropriate constants c,
&gt;
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)
&gt;
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
&lt;
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
&lt;--&gt;
a
-=
i (rood 3).
Furthermore, put
R2n Q2nlgQ5nL, Q8n’
, Qni &lt;=&gt;
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&quot;
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&quot;
n
is much simpler.
Namely, U: R
n
Moreover, since e&amp;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)
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&quot;
(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&quot;
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 ]&quot;
Thus, by (2.2),
Let
us
HOW MANY NUMBERS SATISFY THE 3X +
By
2v-
9v
&lt; -:
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) &gt; 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) &gt;
Observe that
2)
6,
This yields if y
largest root of
5
2(y-
and
1),
2
&gt; 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
&gt;
\$2 (y)
\$1(y)
O,
2 the system (2.4) becomes for y
5
where
&gt; 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
&gt; 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
&quot;
2
m:Rn
:(Y)&quot;
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)
&gt; CnX
n
,where
rn
when n tends to infinity.
REFERENCES
1.
2.
CRANDALL, R.E., On the &quot;3X + I&quot; problem Math. Comp. 32, (1972), 1281-1292.
problem and its generalizations, The American Math.
LAGARIAS, J.C., The 3X +
Moth! 92, (1985), 3-23.
```