Maybe this is all that has to be said about the infamous Conjecture: that it is

*) Sample of Pure Applicable Mathematics

Division byFor any positive integerNa sequenceScan be defined by putting_{i}Sand for all_{0}= Ni > 0:Sif_{i}= S_{i-1}/ 2Sis even_{i-1}Sif_{i}= (S_{i-1}* 3 + 1) / 2Sis odd_{i-1}

And suppose the number1 2 3 4 5 6 7 ... 132 133 134 ... 123421412

Now suppose that the number of elements inIf T = even then the number of even integers inSisT/2and the number of odd integers inSisT/2If T = odd then the number of even integers inSis(T-1)/2and the number of odd integers inSis(T+1)/2

In the limiting case, of an infinitely large set S, we find, quite as expected:If T = even then P(even) = (T/2) / T = 1/2 and P(odd) = (T/2) / T = 1/2 If T = odd then P(even) = ((T-1)/2) / T -> 1/2 and P(odd) = ((T+1)/2) / T -> 1/2

And this property does not change, no matter how large the set S will become. Such an limiting case - an almost infinitely large set S of contiguous natural numbers - could easily be renamed to: THE NATURAL NUMBERS. For who on Earth can possibly explain me what the difference would be with a "truly" infinite set of natural numbers ?P(even) = P(odd) = 1/2

But "NO", says the Pure Mathematician. The following reasoning has been coined up by Cantor, and has been swallowed quite uncritically ever since, by the entire mathematical community. A one-to-one relationship (bijection) is assumed to exist between all Natural numbers and all Even numbers:

I would say now: Give me more, give me more ! But no, these theorists have already drawn their "inevitable" conclusion: the cardinality of the set of all natural numbers must be equal to the cardinality of the set of all even numbers. In layman's terms: there are1 2 3 4 5 6 7 8 9 10 11 12 ...2 4 6 8 10 12 14 16 18 20 22 24 ...

Let's re-create

My answer is simple: this only proves that sound scientific reasoning gradually has been replaced by

This results in:S_{i}= S_{i-1}/ 2

Hence the probability that the result of the division will become even is equal to the probability that it will become odd. Both probabilities are equal toS_{i-1}= 2 4 6 8 10 ... 132 134 136 138 ...S_{i}= 1 2 3 4 5 ... 66 67 68 69 ...

Rewrite the 'odd' step in the Collatz algorithm as follows:

This results in:S_{i}= S_{i-1}+ (S_{i-1}+1)/2

Hence the probability that the result of the addition will become even is equal to the probability that it will become odd. Both probabilities are equal again toS_{i-1}+1 = 2 4 6 8 10 ... 132 134 136 138 ...S_{i}= odd + 1 2 3 4 5 ... 66 67 68 69 ...

Summarizing. Given an arbitrary (that is: unknown) positive integer, which is either even or odd with a probability of 50 % , then any step in the Collatz process will generate another positive integer, which is even or odd with the same probability, namely 50 % . Numbers are enlarged by the odd steps, while they are diminished by the even steps.

We find, for example:Sand for all_{0}= Ni > 0:Sif_{i}= S_{i-1}/ 2Sis even_{i-1}Sif_{i}= (S_{i-1}* 3 + 5) / 2Sis odd_{i-1}

Which from now on will repeat itself.19 31 49 76 38 19

Not every modification of the (3x+1) problem necessarily has the property that it will loop somewhere. Let us consider, for example, the sequence:

It can easily be shown that this "(3x+3)" sequence is, in fact, completely equivalent to the original Collatz problem, where every number in the Collatz sequence corresponds withSand for all_{0}= Ni > 0:Sif_{i}= S_{i-1}/ 2Sis even_{i-1}Sif_{i}= (S_{i-1}* 3 + 3) / 2Sis odd_{i-1}

An arbitrary (part of a) Collatz sequence can be written in the form:

The proof of this Lemma is by complete induction toSwith_{D}= P_{D}. S_{0}+ Q_{D}PWhile_{D}= 3^{O}/ 2^{D}Qis defined by a (3x+1)-like algorithm:_{D}Qand for all_{0}= 01 < i ≤ D:Qif_{i}= Q_{i-1}/ 2Sis even_{i-1}Qif_{i}= (Q_{i-1}* 3 + 1) / 2Sis odd_{i-1}

Suppose that the Lemma is true for a sequence of lengthS_{0}= P.S_{0}+ Q where P = 3^{0}/ 2^{0}and Q = 0

And for even numbers:S) + 1 ) / 2 = 3_{D+1}= ( 3 ( P_{D}.S_{0}+ Q_{D}^{O+1}/ 2^{D+1}. S_{0}+ (Q_{D}.3+1)/2

S) / 2 = 3_{D+1}= ( P_{D}.S_{0}+ Q_{D}^{O}/ 2^{D+1}. S_{0}+ Q_{D}/2

Let *O = * the number of 'odd' iterations, *D = * the total length of
the sequence and *N = S _{0}* the number the sequence starts with.

Then, for an arbitrary (part of a) Collatz sequence, the following holds:

It is easy to see that Q in the previous Lemma has a maximum value, which occurs in sequences of the formS_{D}< 3^{O}/2^{D}( N + 2^{D-O})

It's easy to verify this by complete induction. The Lemma is obviously true forQ_{max}= 3^{O}/2^{O}- 1

Suppose that the Lemma is true forQ_{max}:= (3*0+1)/2 = 1/2 = 3^{1}/2^{1}- 1

Rearranging terms gives for the maximum value of S:Q_{max}= (3*(3^{O}/2^{O}-1)+1)/2 = 3^{O+1}/2^{O+1}-1

S_{D}= 3^{O}/2^{D}.N + 3^{O}/2^{O}- 1 < 3^{O}/2^{D}( N + 2^{D-O})

The the outcome afterRand for all_{0}= N + 2^{D-O}i > 0:Rif_{i}= R_{i-1}/ 2Sis even_{i-1}Rif_{i}= 3 * R_{i-1}/ 2Sis odd_{i-1}

It's possible to construct a much tighter Upper Bound by traversing the Collatz sequence in reverse order:R_{D}= 3^{O}/2^{D}( N + 2^{D-O}) > S_{D}

Resulting in the following starting value for the bounding Real Valued sequence:y = (3.x + 1)/2 ==> x = (2.y - 1)/3 < 2.y/3 y = x/2 ==> x ≤ 2.y

Any sequence of integer numbers may thus be conveniently bounded by a similar one of real-valued numbers. We have already seen, in addition, that the choices 'odd' and 'even' are in fact random choices, with equal probabilities.R_{0}= 2^{D}/3^{O}

The convergence behaviour of Collatz sequences is thus mimicked by a procedure where random decisions are made, for example by generating random numbers in the interval (0,1) and setting up the following algorithm, for some real and sufficiently large number S:

This can even be generalized to arbitraryR := Random; if R > 0.5 then S := S + S/2; if R < 0.5 then S := S - S/2;

A sloppy argument already reveals that, in the long run, there will be as many even steps as there are odd steps, on the average. Thus the behaviour ofp := Random; while true do begin R := Random; if R > 0.5 then S := S*(1+p); if R < 0.5 then S := S*(1-p); end;

Which converges to zero, given the fact thatS := (1+p)^{D/2}(1-p)^{D/2}.S = (1 - p^{2})^{D/2}.S

And we know that the accompanying Collatz sequence is entirely below it. Hence if the Real number sequence converges, then the Collatz sequence must also converge.R_{D}= 3^{O}/2^{D}.R_{0}

Now the convergence of the Real number sequence is only dependent upon the factor

Mean value and variance of this distribution are given by:P(O) = D! / ((D-O)!.O!).(1/2)^{O}.(1-1/2)^{D-O}= D! / ((D-O)!.O!) / 2^{D}

The Binomial Distribution may be approximated well by a Normal Distribution for moderate and large values of<O> = D.1/2 = D/2 ; <O^{2}> - (<O>)^{2}= D.1/2.(1-1/2) = D/4

This is the same as demanding that the starting value of the real valued upper bound sequence should be a number which is greater than 1 :3^{O}/2^{D}.x < x ==> 3^{O}/2^{D}< 1 ==> O.ln(3) - D.ln(2) < 0 ==>

O/D < ln(2)/ln(3)= 0.630929753571457

When combined with the above, this means that it will be increasingly difficult to find sequences where the ratio2^{D}/3^{O}> 1 <==> 3^{O}/2^{D}< 1

O/D < ln(2)/ln(3) = 0.630929753571457 >> 0.5

can be used for extending an arbitrary Collatz sequence to one of predetermined length1 2 1 2 1 2 1 2 1 2 ...

Meaning that3^{O}/2^{D}==> 3^{O+T/2}/2^{D+T}= 3^{O}/2^{D}. (3/4)^{T/2}if T is even ==> 3^{O+(T+1)/2}/2^{D+T}= 3^{O}/2^{D}. 2.(3/4)^{(T+1)/2}if T is odd

Usage: Make a directory. Unzip in that directory and Run the 'exe'.