# Collatz Conjecture/Supposed Proof

## Conjecture

Let $f: \N \to \N$ be the mapping defined on the natural numbers as follows:

$\forall n \in \N: \map f n = \begin{cases} n / 2 & : n \text { even} \\ 3 n + 1 & : n \text { odd} \end{cases}$

For any given value of $n$, let the sequence $\sequence {S_k}$ be defined:

$\forall k \in \N: S_k = \begin{cases} n & : k = 0 \\ \map f {S_{k - 1} } & : k > 0 \end{cases}$

The Collatz conjecture is:

For all $n \in N$, there exists some $i \in \N$ such that sequence $S_i = 1$.

## Supposed Proof

As the author, I do not discourage deletion. Senojesse (talk) 20:47, 8 February 2021 (UTC)

The Collatz sequence consists of instances of a single odd term interspersed by one or more even terms.

The sequence grows if and only if there is only a single even term between two successive odd terms.

Consider the following sequences:

 $\ds$  $\ds 31 \to 94 \to 47$ $\ds 2051 \to 6154 \to 3077$ The ratio between the last and first terms is $1.5$ $\ds$  $\ds 41 \to 124 \to 62 \to 31$ $\ds 233 \to 700 \to 350 \to 175$ Here the ratio is $0.75$ $\ds$  $\ds$ $\ds 2429 \to 7288 \to 3644 \to 1822 \to 911$ Now it is $0.3785$

For a sufficiently large set of $3$ term sequences, approximately $50 \%$ will have one even term and $50 \%$ will have two or more even terms.

Let N = 4k+1 Then 3*n + 1 = 12*k + 4 This is always divisible by 4 Let N = 4k+3 Then 3*n + 1 = 12*k + 10. This is never divisible 4. The 4*k + 1 term can possibly be divided by any power of two. Then, a priori the last statement logical. There is no feature in the construction of a Collatz Sequence that favors 4*k + 3 over 4*k + 1.

Consider the following equations

  (3*N + 1)/2 = 1.5*N + 1/2
(3*N + 1)/4 = 0.75*N + 1/4
(3*N + 1)/8 = 0.3875*N + 1/8
(3*N + 1)/16 = 0.19375*N + 1/16
"           "         "
"            "         "
"             "          "
(3*N + 1)/(2^x) = (3*N)/(2^x + 1/(2^x)


This shows the rate of gain/loss based on the divsibility of the 3n+1 term.

A priori, 4*k+1 & 4*k+3 are equally probable.

The net result for a partial or completed sequence ts;

last odd term = first odd term *1.5^A*0.75^B*0.3875^C*0.1875^D* ...

  A, B, C, D represent the number of occurrences of each divisibility


This equation has very many terms,but for the purposes of this discussion; only three or four are required.

QUESTION: What is the relative ratio between A, B, C & D necessary to achieve minimal growth?

 If B = 200, C = 100, D = 50; what must A be?


1.5 A = 591 1.1747192942496588E+104* 0.75 B = 200 1.0286145857915894E-25* 0.375 C = 100 2.5300364191868604E-43* 0.1875 D = 50 4.4674901453521685E-37= 1.3657687020517226

Ans. A = 591

Even Terms Ratio No. of Terms Net Gain or Loss
$1$ $1.5$ $100$ $4.065611775352152 \times 10^{17}$
$2$ $0.75$ $50$ $5.663216564269376 \times 10^{-7}$
$3$ $0.3785$ $50$ $8.003762476146784 \times 10^{-22}$

Total loss: $1.8428214850660864 \times 10^{-10}$

Perhaps $200$ terms is not a "sufficiently large" set. It could be $200$ googols. Then the ratio would be $10^{-20}$ googols.