Collatz-Intro - The primitive loop
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Workshop |
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(depending
on a empirically checked assumption)
3) a modular view
It is very
natural, to approach the critical diophantine equation with modular
arguments. If this equation cannot be satisfied in integers, no primitive
loop of any length is possible. Unfortunately
relies the modular approach (in my current path) on an assumption, which is
not yet proven, and is equivalent to a critical inequality steming from the
waring-problem, which is very likely true, but which truth could not yet been
proven explicitely. But for the
very unlikely case, that a counter-example exists, another condition must be
satisfied - and this is one on the value of A in relation to N. Since I was
not experienced in approximation-theory, but as well as I was not aware, that
a first breakthrough was already made in disproving that type of loop in 1978
by Ray Steiner diproving the 1-cycle, 2000 by John Simons disproving the
2-cycle and 2003 de Weger/Simons disproving all m-cycles up to m=68 with such
means, I approached that problem from the modular view. I came pretty
far and found a table which beautifully exhibits the impossibility of the
primitive loop on the basis of being no integer solutions in that table. However,
*that* in that table were no integers was depending on an assumption, which
in turn I could *not*prove. It came out to be the same condition, which is
known in the waring-problem, and again seems to deserve approximation-theory
to be proven. But about that theme read the next chpater about
approximation-arguments. |
3.1) the
two-way-modulus-table
Starting
point is the divisibility table with the residue classes modulo powers of 2. I'm
presenting arguments, why the "primitive loop" cannot exist for
modular reasons, in short: because [Eq 1]: i*3^N - 1 cannot hold
for any N,i > 1 , since the rhs is never an integer , and even less a
power of 2. If I rewrite
this formula as a sum of fractions, then -for the rhs- we get this: 3^N*i-1 3^N*i-1 3^N*i-1 --------- + ------- + -------- 3^N 3^N 3^N 1 1 1 ---- + ---- + --- + ... - -----
- ----- - ------- For a first
step-in we assume the case i=1 (we will later include a variable i again),
getting 3^N 3^N 3^N 1 1 1 ---- + ---- + ---
+ ... - ----- -
--- - --- .... This looks
like a periodic digit-representation in the numbersystem to the base 2^N: let z = 3^N q = 2^N // the base of the numbersystem then d = floor(z/q) and r = z-d*q then we can
rewrite this in digits z - 1 1 1
1 1 1
1 or as
string of digits 0.z z z z z z z z.... - 0.1 1 1 1 1 1... Since z is
d*q + r we rewrite that as d.d d d d d d d d ... and 2^S in
that system is 2^(S-N) = 2^A , give that the letter e: e.0 0 0 0... = d.d d d d ... + 0.r r r r ... - 0.1 1 1 1... This can only
be an equation, if d + r = q, thus making the rhs a periodic number of the
highest residue-class of q, which assures a carry on each digit to make the
number integer in base q. So from this
consideration we have (still without respect to the free parameter i) q = d + r Empirically The arguments
that I'm presenting here look really good and suggestive. But I have to
mention, that they rely on one basic assumption, which was *not* proven by
this arguments, but seemingly needs external backing from other resources. This
is the following assumption, which uses the expression of 3^N/2^N as an irregular
fraction: let 3^N p then M + p < q That means -
in other words - the remarkable fact, that the two digits in a
base-2^N-representation not only never sum up equal to 2^N-1, but
also that their sum is always smaller - an empirical observation,
which I find very astonishing. I'm currently trying to relate this property
to that condition, which is mentioned in the Log3/Log2-observations-section. So the
generalization of the observed facts in the modular two-way-table, (which
strongly suggest, that this assumption is true) gives an argument, that there
is no primitive loop at all on modular reqirements. Another
reason, why this approach may be interesting is, that it allows to be
extended even to conacatenated PT-M-loops, (or "m-cycles")
schematically while the approximation-approach does not suffice for M>68,
means the 68-fold-combination of 1-cycles to one m-cycle, according to
results of J. Simons and B.de Weger. Since modular restrictions do not depend
on the relative size of numbers it could be, that following this way the
general m-cycle-problem can be solved. In short, the
extensions follows the idea: ·
an
1-cycle is impossible, if no entry in that table is integer (more precisiely:
a power of 2) ·
an
m-cycle is impossible, if no product of m entries of this table can result in
an integer (again more precisely: in a power of 2) (note that
the exponents of the powers-of-2 have the known further restrictions to match
about 0.59*N) I'm currently
investigating the conditions of such products. Because of
the possible value of that table, it is presented here in some detail. |
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Let x =
3^N/2^N be displayed as a irregular fraction base a power of 2: let q =
2^N and d be the integer d=floor(x/q) then let x
be represented as x = d * q + r
which
means, that in the numbersystem base(q) the both digits d and r make x x = "d r" (digits d,r
base(q)) The critical
condition for a 1-cycle is then d + r < q because it is
needed, that by division by q-1 no carry occurs in x
"d r" "d
r" "d r" = " d d.d d d d ..." + "
r.r r r r r..." = " d d.d d d d ..." and x - 1
" d d.d d d d ..." Now the
critical condition for a primitive loop says, that the result must a) be integer b) a power of 2 to make a
primitive loop possible since 3^N - 1 x - 1 In the base
q-system this means: e 0.0 = " d.d d d d d
..." Since d is
always smaller than e, one can immediately see, that it must be e-1; also
each digit of the rhs-sum must produce a carry , and the new resulting digit
must precisely 0. So this
means, to have a promitive loop, it needs, that d+r = q or a multiple d+r = i*q. Therefor it
is a most interesting question, whether d + r < q holds for the
given rhs-term for all N. If this
statement could be proven for the interesting values of x and q, then the
primitive loop is disproven as well as the "waring-problem is completely
solved" (see [mathworld/waringproblem] and mathworld/powerfrac ) . The same
problem was attacked by Kurt Mahler, who analyzed that question
systematically and formulated a criterion called Z-numbers. If there were no
Z-numbers, then this would be equivalent
to the proof of the critical condition. Mahler didn't succeed in
proving the nonexistence of such Z-numbers, but could at least state, that at
most finitely many such numbers exist. (see [mathworld/Z-Numbers]) |
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An updated
approach to display things seems the following table, where the lhs-fraction
of Eq 1 is displayed in the number-system of 2^N as an irregular fraction. The entries
are always of the form M + pi/qi where qi is the
appropriate term of i*2^N-1 and M + p /q in the term for the limit i->oo Especially
interesting was the form for i=0, which always come out to be 1 since (i*3^N - 1)/ (i*2^N-1) = -1/-1 = 1 for i=0 That entry
could be regularly added by the trick to write it as an irregular fraction as N - 1 which is true
for all N. Here follows
the table: For instance some rows for N>=2 ,
i=0..oo give i=0
i=1 i=2 i->oo M0 p0 q0
M1 p1 q1 M2 p2 q2 M p q -------------------!---------!---------!---------!-----------!----------------!------- f(N=2,i=0..oo): 2 1/-1 2
2/3 2 3/7
2 4/11 2
5/15 .. -> 2 1/4 f(N=3,i=0..oo): 3 2/-1 3
5/7 3 8/15
3 11/23 3 14/31
.. -> 3 3/8 f(N=4,i=0..oo): 5 4/-1 5
5/15 5 6/31
5 7/47 5
8/63 .. -> 5 1/16 f(N=5,i=0..oo): 7 6/-1 7 25/31 7 44/63 7
63/95 7 82/127 ..-> 7 19/32 f(N=6,i=0..oo): 11 10/-1 11 35/63 11 60/127 11 85/191 11 110/255 . ->11 25/64 f(N=7,i=0..oo): 17 16/-1 17 27/127
17 38/255 17 49/383 17
60/511 -->17 11/128 ... On a first
glance we see some very useful properties: a) The integer-part M does never change by
changings of i b) the nominators p1,p2,p3,... grow in
positive arithmetic progression c) the progression p2-p1 of the nominator
is smaller than that of the denominator q2-q1 d) the second nominator (p1 at i=1) is always equal or greater
than the integral part M e) the progressions are p2-p1 = p =
(p1-M) + 1 q2-q1 = q (which follows easily from the
difference-equation f(N,i+1)-f(N,i)) f) the first nominator p1 is always
smaller than q1-1 for (N>2) or (N=2 and i>1). These observations, if they can be
generalized, have remarkable consequences: Any
changing of i does not affect the integral part of the fraction in [Eq 1], which is a very useful property in analyzing formulae derived
from [eq 1] and Since the progression of the
nominator is always positive, but smaller than that of the denominator (which is obvious from the fact, that p/q is
the remainder), Nominator
and denominator can never have a Null or an integral ratio, iff the second nominator p1 is already
smaller than q1. The
consequence is, that we cannot find an integer in the whole two-way table,
which covers all possible combinations of results for [eq 1], which in turn
disproves the primitive loop, since that would mean, that either pi=0 or
pi=qi to have a result for rhs in [eq1], which is at least integer (it even
must be a power of 2, but that need not be discussed, if the result is no
integer at all) Now the
prerequisite for that effect is, that required,
but not proven: p0 + p < q0 + q M-1 + p < -1 + q p + M < q or
,required (but not proven) [Cond 1] p
< q - M I checked
empirical evidence for that up to N~ 15000 and had no counterexample. I didn't find
an elementary proof for that, but after some re-arranging of formulas it
seems, that this condition is nothing else than the critical inequality in
disguise. I'll have to verify this precisely. So we have, - given [cond 1] holds, then - since each combination of i and N in
equation [eq 1] [Eq 1]: i*3^N - 1 results in a
non-integer value, the equation [eq 1] describing a condition for a primitive
loop cannot be satisfied. Thus a
primitive loop of any length cannot exist - given [cond 1] holds for all N. |
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[added
20.8.04] The condition
[Cond 1] is equivalent to the critical inequality for primitive loops, so it
is proven with the Steiner-approximation-proof of nonexistence of 1-cycles. So
it *is* indeed proven, but not by elementary means. It is
possible then to proceed with that table even if the elementary proof needs
awaiting... Given: s = M*q + p so M = floor(s/q) and M+1 = ceil(s/q) cond1: p < q - M Then : M*q + p < M*q - M + q s
< (M+1)*q - M s
< M(q-1) + q s - q
< M(q-1) s - q -----
< M q - 1 s - q 1 + -----
< M +1 = ceil(s/q) q - 1 s - 1 -----
< ceil(s/q) q - 1
q
s - 1 q - * -----
< ceil(s/q)* --- s
q - 1 s Now with s =
3^N, q = 2^N , rhs = powerceil2(3^N)/3^N
this exactly the critical inequality for the primitive loop with i=1: 2^N
3^N - 1 powerceil2(3^N) -
* ------- < ---------------- 3^N
2^N - 1 3^N This
formula is proven via the proof done by Steiner, so cond1 is proven, * so the
p-values for the column i->oo holds, * so the
column for i=1 hold, * and so
all columns hold. Since all
columns and rows hold, * there is
no integer in this table, and the
table can be used in further considerations. |
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last update: 20. Aug. 2004 |