Featured

## You are the One! ❤️

Lord GOD loves you more than any other being including you.

Please remember that when you feel less of yourself or when you feel alone.

Or when you begin to dislike self.

And it matters most that Lord GOD loves you the most!

Beware of human beings succumbing to evil and deception.

Homeless, poor, despised, sad, sick, victimized, cheated, ashamed, afraid, abandoned, lost, hated, spurned, or even on death row, Lord GOD loves you when all others may not.

Please have the greatest faith in Lord GOD.

You are the one that Lord GOD loves in this life and beyond it.

Amen!

“Do not love the world or the things in the world. If anyone loves the world, the love of the Father is not in him.”  1John 2:15.

“The Prince”;

“Machiavelli’s The Prince: Still Relevant after All These Years”;

EU Set to Ban Surveillance, Start Fines Under New AI Rules;

UN rights chief urges moratorium on AI that risks rights abuses;

“In a democracy, everybody has the chance to be everything. There is no superior race and no inferior race. All of us have royal blood. We’re all God’s children. Everybody matters.”

Why did God create the integers?

Secrecy (2008 documentary)”;

“The Quest for Peace and Justice”;

“Love: Our True Destiny”;

###### 12 Attributes Of Above Average People — According To Philosophy;

Daily Devotionals;

When faced with the vicissitudes of life, one’s mind is unshaken, and it is free from sorrow, free from defilements, and free from insecurity. This is the highest blessing!

And having acted that way, everywhere undefeated and everywhere finding happiness. This is the highest blessing! Amen!

Featured

## World Peace ~ S.N. Goenka — Astitva Meditation

“There cannot be peace in the world when people have anger and hatred in their hearts. Only with love and compassion in the heart is world peace attainable.” ― S. N. Goenka

World Peace ~ S.N. Goenka — Astitva Meditation

“A film about a messed up world and how to fix it. Sprinkled with music, this documentary lyrically weaves a tapestry through 20 countries and is as magical as it is informative.” — One Peace at a Time;

“Instead of focusing on the negative effects of climate change, this documentary spotlights people suggesting solutions and their actions.” — Tomorrow.

“Welcome those big, sticky, complicated problems. In them are your most powerful opportunities.” — R. Marston.

Featured

## A Borrowed Life (Existence)

This life is not ours.

Though we claim it as ours.

And we are the masters of our fate too.

Right??

Wrong!!

This life, we claim as ours, is a precious gift from Lord GOD.

And for a while, we act, we grow, we enjoy, …, we cry, we succeed.

And we eventually die (we return the gift).

It’s all providence, the Lord’s Domain.

Amen!

Dave.

## Solve the following Diophantine equation for the integers, x and y.

Solve

$x^{7}+x^{6}y + x^{5}y^{2} + x^{4}y^{3} + x^{3}y^{4}+ x^{2}y^{5}+xy^{6} +y^{7} =234,567,890,122$

for integers, $x$ and $y$.

What types of Diophantine equations are unsolvable?

Good Luck!

Dave.

## Is P = NP?

Yes! P = NP = NP-Complete! Wow! (Please see the link above for proof.)

Thank Lord GOD! Praise Lord GOD! Amen! 😊

Go Blue! 👍✌

Dave.

## What could be an indirect proof of Goldbach’s Conjecture?

We can attempt a proof of Goldbach’s Conjecture via a proof by contradiction.

Suppose there exists a positive even integer, $e > 4 * 10^{18}$, that is not a sum of two odd primes.

We let e = p + nq (or equivalently, e ≡ p mod q) where odd primes, p and q, are less than e with odd integer, n > 1.

But there are k odd primes, $p_{i}$, less than e. Therefore, we generate a system of k Diophantine equations:

$e = 3 + n_{1}q_{1}$;

$e = 5 + n_{2}q_{2}$;

$e = 7 + n_{3}q_{3}$;

$e = p_{k} + n_{k}q_{k}$

where odd integer, $n_{i} > 1$ for $i \in$ {1, 2, 3,…, k}.

Remark: We must show that least one of those equations is false (contradicting our assumption that e is not a sum of two odd primes) to prove Goldbach’s Conjecture is true.

Claim: We strongly believe Goldbach’s Conjecture is provable with some more analysis and with the help of some tools/methods from prime number theory.

A Crucial Idea: The distribution of all odd primes less than e helps us to prove Goldbach’s Conjecture.

Good Luck! 🙂

Dave.

## Is the Euler–Mascheroni constant an algebraic number or a transcendental number?

Valid Assumption (Convergence): $1. \gamma = \frac{\sqrt{z^{2} - 1}}{2}$.

Remark: We tentatively assume $z$ is an algebraic number (rational number) where $1.52 < z < 1.53$.

We can apply Newton’s Method to equation one and study its convergence with regards to $\gamma$.

Warning: This may be a difficult problem…

Dave.

## “David Hilbert: The Architect of Modern Mathematics”

“In July 1915, Albert Einstein paid a visit to the University of Göttingen (Germany) on the invitation of the mathematician It was a fruitful encounter for both men that continued over the following months with an intense scientific correspondence. Einstein described that period as the most exhausting and stimulating of his entire life, the result of which was a series of studies and articles, authored by one or the other scientist, in which they formulated the equations of the gravitational field of the General Theory of Relativity (GTR)…

David Hilbert: The Architect of Modern Mathematics.

Hilbert’s Problems: 23 and Math.

## A Brief Analysis of the Collatz Conjecture

Algorithm for the Collatz Conjecture.

Remark: If n = 1, the algorithm terminates all computations.

Crucial Idea: R+ is a dense set.

Remark: R+ is the set of all positive real numbers.

We claim

1. $n_{t} = (\frac{3}{2} )^{\frac{t}{2}} * (\frac{3}{4} )^{\frac{t}{4}} * (\frac{3}{8} ) ^{\frac{t}{8}} * ... *( \frac{3}{2^{k}} ) ^{\frac{t}{2^{k}}} * r = 1$

over the Collatz sequence of positive odd integers, from $n_{0} > 1$ to $n_{t} = 1$ where the index, t, is the number of trials it takes the Collatz sequence of odd integers to converge to one.

Remark: There are no infinite (nontrivial) cycles of any length ( $(\frac{3}{2} )^{\frac{t}{2}} * (\frac{3}{4} )^{\frac{t}{4}} * (\frac{3}{8} ) ^{\frac{t}{8}} * ... *( \frac{3}{2^{k}} ) ^{\frac{t}{2^{k}}} \rightarrow 0$ and $r \rightarrow \infty$ as $t \rightarrow \infty$) in the Collatz sequence since the index, t, is clearly finite in equation one.

In addition, our claim is fundamentally based on the distribution of maximum divisors, $2^{i_{max}}$, for finite sets of consecutive positive even integers (e.g. {2, 4, 6, …, $e_{max}$}) where 2 is the minimum positive even integer and where $e_{max}$ is the maximum positive even integer belonging to those sets…

The positive real number, r, is determined by the algorithm for the Collatz conjecture. And therefore, its computation is generally complicated because we cannot easily compute the maximum positive even integer, $e_{max}$, in the Collatz sequence (orbit) for many odd positive integers, $n_{0} > 2^{68}$.

Question: For any odd positive integer, $n_{0} > 2^{68}$ , can we approximate the maximum positive even integer, $e_{max}$, in the Collatz sequence (orbit)?

The variable, k, is determined by the maximum positive even number, $e_{max}$, in the Collatz sequence: $k = \left \lfloor{\frac{log(e_{max })}{log(2)}}\right \rfloor$.

Can we refute our analysis?

We assume maximum divisors, $2^{i}$, for any positive even integer.

Remark: The Collatz conjecture is true for all positive integers, $n_{0} \le 2^{68}$.

For any positive odd integer, $n_{0} > 2^{68} -1$, we have

$n_{1} = \frac{3n_{0} +1}{2^{i_{1}}} = \frac{3}{2^{i_{1}}} * r_{1}$ where $r_{1} = n_{0} +\frac{1}{3}$.

$n_{2} = \frac{3(\frac{3n_{0} +1}{2^{i_{1}}}) + 1}{2^{i_{2}}} = \frac{3}{2^{i_{1}}} *\frac{3}{2^{i_{2}}} * r_{2}$

where $r_{2} = \frac{1}{9}(2^{i_{1}} + 9n_{0} +3)$.

$n_{3} = \frac{3(\frac{3(\frac{3n_{0} +1}{2^{i_{1}}}) + 1}{2^{i_{2}}}) + 1}{2^{i_{3}}} = \frac{3}{2^{i_{1}}} *\frac{3}{2^{i_{2}}} *\frac{3}{2^{i_{3}}} *r_{3}$

where $r_{3} = \frac{2^{i_{1}}}{27}*(2^{i_{2}} + 3) + n_{0} +\frac{1}{3}$.

$n_{t} = r_{t} * 3^{t}\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}})$ where $r_{t} > 1$ since $3^{t}\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}}) < 1$ and since $n_{t} \ge 1$.

Question: Is $3^{t}\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}}) > 1$ possible? No!

Why? Hint: $n_{0} > 1$.

Therefore, $r_{t} \ge 3^{-t}\frac{1}{\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}})}$

Thus, $r_{t} = 3^{-t}\frac{1}{\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}})}$ or $r_{t} > 3^{-t}\frac{1}{\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}})}$

Moreover, $n_{t} = r_{t} * 3^{t}\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}}) = r * (\frac{3}{2} )^{\frac{t}{2}} * (\frac{3}{4} )^{\frac{t}{4}} * (\frac{3}{8} ) ^{\frac{t}{8}} * ... *( \frac{3}{2^{k}} ) ^{\frac{t}{2^{k}}} \ge 1$ where $r_{t} > r$.

We conclude $r_{t} = 3^{-t}\frac{1}{\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}})}$ since $r_{t}$ is a positive rational number and since there are no infinite (nontrivial) cycles in any Collatz sequence. Hence $n_{t} = 1$.

Important Remark: The Collatz Conjecture is true! 🙂

Example: If we let $n_{0} = 57$, then we compute $e_{max} = 196$, $k = 7$, and $t = 10$.

Therefore, $r = r(57, 10) = 1/.0841 394 = 11.8850384$.

Dave’s Conjecture: $r = r(n_{0 }, t) =$ O(t) or $r = c_{t} * t$ for some real number, $c_{t}$, such that either $c_{t} > 1$ or $0 < c_{t} < 1$.

Example: If we have $t = 1$ for some $n_{0 }$, then as $k \rightarrow \infty$,

$r = \frac{1}{ \prod_{i=1}^{k }(\frac{3}{2^{i}})^{\frac{t}{2^{i}}}} \rightarrow \frac{4}{3 } = c_{1 }$.

Therefore, $r = r(n_{0 }, 1) \approx \frac{4}{3}$ for infinitely many positive odd integers, $n_{0} > 1$.

Remark: For our example, we compute $n_{0 } \in$ {5, 21, 85, 341, …, $\frac{2^{2j} - 1}{3}$, …} for all positive integers, $j > 1$.

Questions: What are the values (convergent) for $c_{2}, c_{3}, c_{4}, ...$?

Example: We compute $r = r(n_{0 }, 2) \approx 1 \frac{7}{9}$ for infinitely many positive odd integers, $n_{0} > 1$.

And therefore, $c_{2} = \frac{8}{9}$ where

$n_{0 } \in$ {7281, 29125, 116501, 466005,…, $4 *l_{j}+1$, …} for all positive integers, $j \ge 4$.

Remark: For our example, we assume $l_{1} = 7,281, l_{2} = 29,125, l_{3} = 116,501, l_{4} = 466,005$.

Old Example: If we let $n_{0} = 57$, then we compute $e_{max} = 196$, $k = 7$, and $t = 10$.

Therefore, $r = r(57, 10) = 1/.0841 394 = 11.8850384$. However,

$r = r(n_{0}, 10) \approx \frac{1}{ \prod_{i=1}^{\infty }(\frac{3}{2^{i}})^{\frac{10}{2^{i}}}} = \frac{1}{.0563135} \approx 17.75773127$

for infinitely many positive odd integers, $n_{0} > 1$.

And therefore, $c_{10} \approx 1.775773127$ where

$n_{0 } \in$ {57, 229, 917, 3669,…, $4 *l_{j}+1$, …} for all positive integers, $j \ge 4$.

Remark: For our old example, we assume $l_{1} = 57, l_{2} = 229, l_{3} = 917, l_{4} = 3669$.

For $n_{0} = 57$, we compute the following Collatz sequence where $t = 10$:

172

86

#1: 43

130

#2: 65

196

98

#3: 49

148

74

#4: 37

112

56

28

14

#5: 7

22

#6: 11

34

#7: 17

52

26

#8: 13

40

20

10

#9: 5

16

4

2

#10: 1

For $n_{0} = 917$, we compute the following Collatz sequence where $t = 10$:

2752

1376

688

344

172

86

#1: 43

130

#2: 65

196

98

#3: 49

148

74

#4: 37

112

56

28

14

#5: 7

22

#6: 11

34

#7: 17

52

26

#8: 13

40

20

10

#9: 5

16

8

4

2

#10: 1

Remark: The common Collatz sequence of positive odd integers for our old example is

{43, 65, 49, 37, 7, 11, 17, 13, 5, 1}.

Final Remark (Hmm): The $3n_{0} + 1$ Problem (Collatz Conjecture) has $n_{0}$$1$ (mod $4$) solutions for a given index, t. The solutions are not unique for a given index, t. We assume $n_{0} \gets 4n_{0} + 1$.

1. Wolfram Cloud (Mathematica Online);

The function, nvalue[t], computes a random positive odd integer, $21 \le n_{0} < 100,000$ (initially), for a small index, $1 \le t \le 10$ (number of trials for a Collatz sequence of odd integers to converge to one from the computed $n_{0}$).

Source Code:

nvalue[t_] := (
tt = t;
icnt = 0;
n = 2 * RandomInteger[{10, 50000}] + 1;
While[icnt != tt,
While[n != 1,
If[icnt == 0, nstart = n];
n = 3n + 1;
While[EvenQ[n], n = n/2];
icnt = icnt + 1;
If[icnt > tt, n = nstart + 2 *RandomInteger[{1,10}]]
If[icnt > tt, icnt = 0]]];
Return[{tt, nstart}])

__________________________________________________________

Some Examples:

nvalue[2] computes for t = 2 a random value, $n_{0} = 54, 613$;

nvalue[10] computes for t = 10 a random value, $n_{0} = 67, 077$;

nvalue[5] computes for t = 5 a random value, $n_{0} = 7,885$.

nvalue[6] computes for t = 6 a random value, $n_{0} = 82,485$.

nvalue[4] computes for t = 9 a random value, $n_{0} = 69,965$.

nvalue[1] computes for t = 1 a random value, $n_{0} = 87, 381$.

nvalue[41] computes for t = 41 a random value, $n_{0} = 29, 137$.

nvalue[75] computes for t = 75 a random value, $n_{0} = 57, 855$.

nvalue[75] computes for t = 75 a random value, $n_{0} = 15, 051$.

nvalue[150] computes for t = 150 a random value, $n_{0} = 2, 139, 623$.

nvalue[150] computes for t = 150 a random value, $n_{0} = 2, 143, 079$.

nvalue[200] computes for t = 200 a random value, $n_{0} = 6, 899, 327$.

nvalue[211] computes for t = 211 a random value, $n_{0} = 21, 069, 775$.

nvalue[250] computes for t = 250 a random value, $n_{0} = 46, 912, 975$.

nvalue[300] computes for t = 300 a random value, $n_{0} = 330, 552, 319$.

nvalue[351] computes for t = 351 a random value, $n_{0} = 1, 505, 110, 398, 721, 395$.

nvalue[351] computes for t = 351 a random value, $n_{0} = 1, 514, 992, 426, 990, 283$.

nvalue[408] computes for t = 408 a random value, $n_{0} = 319, 762, 236, 301, 419, 029$.

Go Blue! 👍✌

An Analysis of the Collatz Conjecture.

“Counting and ordering stuff (objects, sets, numbers, spaces, etc.) are fundamental.”

P.S. FIGHT RACISM IN THE SCIENCES INCLUDING MATHEMATICS! THANK YOU!

Oops! The Proceedings of the London Mathematical Society rejected the paper,  “A Brief Analysis of the Collatz Conjecture, for publication! Why?

We are confident our work is valid, and we suspect our work was rejected because of political reasons… It happens…

Reference Link: Two Important Properties of Convergent Collatz Sequences.

Bakuage Offers Prize of 120 Million JPY to Whoever Solves Collatz Conjecture, Math Problem Unsolved for 84 Years.

Dave.