Beal And Integer Quadrant Arcs

Consider a circle of circumference $4C^z$,

The circumference along arc EDF is of length $C^z$.

Let $p$ be a factor of $C^z$, we draw another smaller circle of circumference $\cfrac{4C^z}{p}$.  The quadrant arc is of length $\cfrac{C^z}{p}$.  This arc is of whole integer length as $p$ is a factor of $C^z$.  And this quadrant arc scales to $C^z$ by $p$.

A line OD is drawn to divide the smaller circle such that,

$m+n=\cfrac{C^z}{p}$

where both $m$ and $n$ and $\cfrac{C^z}{p}$ are integers.  Markings on the longer arc EDF is of integer width being divided by $C^z$.  Markings on arc GHI is also of integer width being divided by $\frac{C^2}{p}$.  The shorter arc GHI is scaled by $p$ to the longer EDF.  Line OD from O intersect a integer marking on arc GHI and then intersect another integer marking on the arc EDF.  The width of the markings on arc EDF is $p$ times smaller than the width on  arc GHI. One marking on GHI scale to $p$ markings on EDF.

This line divides $C^z$ into two integers, $K$ and $J$.

As the circumference $C^z$ was reduced by $p$,  The arc $K$ is, 

$K=mp$ 

and arc $J$ is,

$J=np$

Both K and J have a factor $p$ and are integers.  So,

$K+J=(m+n)p=\cfrac{C^z}{p}p=C^z$

and all $K$, $J$ and $C^z$ has a common factor $p$.  

Suppose $p=ab$ and we scale to $a$ now,

arc $G'H'I'=\cfrac{C^z}{a}=\cfrac{C^z}{p}b$.  When we divide arc G'H'I' into equally space markings, their width is, 

width of markings on arc G'H'I$=\cfrac{p}{bC^z}=\cfrac{1}{b}\times$width of markings on arc GHI.

The marking has scaled proportionately to accommodate line OD.  So, when there is a factor of $C^z$, all other factors of $C^z$ are also factors of $K$ and $J$.

If $p$ is prime, that $p=ab$, either $a=1$ and $b=p$, or $a=p$ and $b=1$, and $pa=b$, either $a=1$, $p=b$ or $b=1$ and there is no scaling, the arc remains on $C^z$. In this case of prime $p$, as the only (repeated) factor of $C^z$, all markings along quadrant arcs above and below $p$ will not accommodate line OD.  None of the markings generated above and below $p$ will intersect the line OD, because arc GHI drawn are not of integer length.

Both $K$ and $J$ share this prime factor.

This does not prove Beal Conjecture, but if given a plot of actual $A^x$, $B^y$ on a quadrant arc of length $C^z$,  and a line OD through center O dividing$C^z$ into $A^x$ and $B^y$, any quadrant arc that intersects OD at integer markings along the arc are factors of both $A^x$ and $B^y$ and so, also a factor of $C^z=A^x+B^y$.

As for the conjecture, if there is a factor of $C^z$, then both $A^x$ and $B^y$ share this factor.  Other factors that divide $C^z$ into integers are also possible on the same plot. These factor cannot be prime.  If $C^z$ has one prime factor then no other factor are possible on the plot.  Both $A^x$ and $B^y$ also have this prime factor.

Average Proof Pigeon Can Do It,

 Divide a unit circle into $n$ equal sector each of arc length $\cfrac{1}{n}$, $n$ runners occupy these $n$ sections.  On average, one runner is in each section, and so, their average and possible consecutive distance for two consecutive runners along the path is $\cfrac{1}{n}$.

This is true for all numbers of runners greater than $3$, with constant distinct speeds.

And the lonely runner conjecture is proved.  No pigeonholes needed.

Phase Important? Not Specified

 What all runners start at a common origin, their phase are the same.  When the phase of the system is not considered then the runners simply pepper themselves on the path at various time and start at various phase.  

This is the situation that the proof is for, where a snapshot was taken, that given a initial least distance the next least distance requirement is also satisfied.

 What else is a common origin for?  Does it matter if one guy is late.  Does the system forgets.  Does the reference for which this analogue is based show concern for phase and starting point?

Drop And Start Running.

Maybe the initial conjecture is formulated without a start point.  That the situation is simply $n$ runners running on a unit circle path each with distinct constant speeds.  It does not matter whether their paths trace back to a common start point.  They simply drop and start running.  

The need for such a common start point comes from computer simulation of the problem.

Rotational Invariant Path

With reference to the Post "Another Theorem By Changing The Question" dated 30 Dec 2022.  

Adding a reference runner with arbitrary speed is like rotating the path at that speed with all the runners on it.  As far as the runners are concern nothing changes.  But there is this strange guy by the track, seems to be going backwards.  Never seen before, must be new, a NPC!

Initiatively the proof is flawed.  It could be that this added $n+1$ runner does not start at the origin, but does that matter when his relative speed is zero.

This added runner must be half way between the lonely runner and the runner closes to him.  That position may not be the starting point.  But he is stationary always.  Can the whole system be rotated, rotational invariant?  The situation can stop and every body runs nothing should change.

What gives?  Starting position.

Another Theorem By Changing The Question

 Suppose it is established that $n$ runner running about a unit circle with arbitrary constant speeds, $v_i$, one just experienced a moment of loneliness, and is at least $\cfrac{1}{n}$ away from the closest runner.

We add the next $(n+1)$ runner, with relative speed zero, where all other runners take reference, right in between and achieve a least distance of $\cfrac{1}{2n}$.  This $(n+1)$ runner is at the origin zero (starting point).  The added $(n+1)$ runner still has arbitrary speed, distinct and constant.

This is the trivial case of $n=3$.

Since     $\cfrac{1}{2n}\lt\cfrac{1}{(n+1)}$

for $n\gt 2$

Adding the next $(n+1)$ runner becomes trivial, and with the trivial case of $n=3$, by induction the conjecture is true.  The constraint is still a least distance of $\cfrac{1}{n}$, it has not been changed to $\cfrac{1}{2n}$.

What?  Using relative speeds is not suppose to change any aspect of the analogy.

 Happy? 

Note: All speeds changed by a constant, but the transition of time has no bearing here. The origin is rotated by a constant.  The circle is rotating at the reference speed, speed of the $(n+1)$ runner.  Starting time is later for all.

Futile And Mistaken Lonely Runner

Continuing from the last post "Lonely Runner Not Lovely", 

Runners having the same origin is important.  They start at the same place, same time.  With that in mind,

Can the next runner added to any point on the unit circle, with arbitrary speed provide a lonely opportunity in the future?

Immediately, this runner cannot land in arc $\pm \cfrac{1}{n+1}$ around the $n$ runner, that is,

$v_{n+1}.t=m_{n+1}\pm\cfrac{1}{n+1}$

This forbidden region is smaller with large $n$.  

But can this runner, provide an lonely opportunity in the future?  The $n$ runner is the stationary runner at the origin.

Not necessarily.  Only the point placement at exactly at $\cfrac{1}{n+1}$, with making any other assumption.

Lonely Man Not Lovely

 The distance requirement is a distraction, because for all number $n$, if it is found that, in an instance (t), that any one of the runners are at least $\cfrac{1}{n}$ in front and back from other runners, then it is also true that in the distance from $\cfrac{1}{n+1}$ to $\cfrac{1}{n}$ taking reference from the same runner, that another runner be place there (either side of this reference runner) with any speed.   

Then the distance condition for $(n+1)$ runners is satisfied.  In fact for all $m\ge n$ this snapshot is true.

However this $n+1$ runner must be placed with speed $v_{n+1}$ such that all runners has a common starting point in time and space (started together (same time) at the same place, with arbitrary speed).  

Can speed $v_{n+1}$ be arbitrarily any speed, over a placement allowance of $d_{\epsilon}=\cfrac{1}{n}-\cfrac{1}{n+1}$? 

No, $v_{n+1}$ can only varies in around $\pm d_\epsilon+m_{n+1}$ circles.  Not the other way round.  $v_{n+1}$ is restricted but can be infinitely many.

Let turn back the clock anyway, given time $t$,  the duration that returns every runner to the starting point.

Consider, is it possible that,

$v_{n+1}.t=m_{n+1}-\cfrac{1}{n+1}-\epsilon$

where $m_{n+1}$ is any integer greater than $1$ (completing $m_{n+1}$ full circles) less $\cfrac{1}{n+1}+\epsilon$, $\epsilon$ is up to the length $d_{\epsilon}$ above.

$m_{n+1}$ is unbounded and the expression $\epsilon$ can be made proportionally infinitely, small but exact.

And we place the $n+1$ runner at a distance $\epsilon$ from the $\cfrac{1}{n+1}$ boundary away from the $n$ runner.

With this placement, $n$ lonely runners implies $n+1$ lonely runners and with an $n=n_o$ to start (done computationally or be found in any discussion of this question), we would have proven the Lonely Runner Conjecture.  

Unfortunately, runner $n+1$ added does not have arbitrary speed!  There's no proof.

Thinking About Plasma Condensing

 Plasma liquefies at high temperature; remember how high pressure is the result of particles coalescing under heat resulting in higher momentum exchanges at the containment surface, because of higher masses involved, and not due to velocities increase. 

For this reason, lower temperature and pressure when plasma liquefies and sticks, to prevent local scalding.  Do not increase temperature.  

Going Half Wave Standing

 If plasma is more wave than particles,

it is better to contain it as a standing wave.  Assuming radio frequencies, half wavelength is at a few meters.  Would energy be tapped from the side or top and bottom?

Even Expectation

 Every proven $n$ in Collatz Conjecture leads a path to $1$.  So. there can be a number $n_{max}$, below which all numbers have been proven true for Collatz Conjecture. (This can be created computationally).  The next immediate number after $n_{max}$, $n_{p}$ must be a odd integer, because an even integer will immediately be divided into $2$ and be send into the pack of proven numbers, as the conjecture dictate. 

This odd number $n_{p}$ will be move to bigger value to $3n_{p}+1$.  This number being even is divided by $2$, 

$n_e=\cfrac{3n_{p}+1}{2}$

this number can be odd or even.

Given that the probability of an even number by the operations in the Conjecture is $P(even)=\cfrac{3}{4}$ and the probability of an odd number after any operation is $P(odd)=\cfrac{1}{4}$.

The expected value for $n_e$ is then

$E(n_e)=\cfrac{3(3n_{p}+1)}{8}$     when $n_e$ is even

$E(n_e)=\cfrac{3n_{p}+1}{8}$     when $n_e$ is odd

These are expected value of $(n_e)$ given large numbers where the probability provide good approximation.

The next value of $n_e$ when it is even as we apply $\cfrac{n}{2}$ is,

$\cfrac{1}{2}\cfrac{3(3n_{p}+1}{8}=\cfrac{3(3n_{p}+1)}{16}=\cfrac{9n_p}{16}+\cfrac{3}{16}\lt n_p$

for large value of $n_{max}$, this number is below $n_p$ and so a proven number.

The next value of $n_e$ when it is odd, we apply $3n+1$

$3\cfrac{3n_{p}+1}{8}+1=\cfrac{9n_p}{8}+\cfrac{11}{8}$

as this is always s even number, we divide by $2$,

$\cfrac{1}{2}\left(\cfrac{9n_p}{8}+\cfrac{11}{8}\right)=\cfrac{9n_p}{16}+\cfrac{11}{16}\lt n_p$.

which is also in the pack below $n_p$

As such with a large continuous number of Collatz number, of maximum $n_{max}$, as starting point, $n_{max}$ implies the next integer $n_p$ is also a Collatz number, then by induction all integer $n\ge 2$ are Collatz numbers.

Happy?  Be Happy New Year.

Collatz Funnel

Given any number $n$,

$$n_{next} =\begin{cases} E(n)=n/2,  & \text{if $n$ is even} \\ O(n)=3n+1, & \text{if $n$ is odd}\end{cases}$$

The proof needed was that all numbers are reduced to $1$ using these mappings.

Given that a number, $n$ is odd, the operation,

$O(n)=3n+1$ 

will always provide a even number,

P(O(n)=even)=1

P(O(n)=odd)=0

Given that a number $n$ is even, the operation,

$E(n)=\cfrac{n}{2}$

can given either a even or odd number.

P(E(n)=odd)=1/2

P(E(n)=even)=1/2

The probability of any operation, producing an odd number is P(odd)=$\cfrac{1}{4}$ 

and the probability of any operation, producing an even number is P(even)=$\cfrac{3}{4}$

This system of equations is bias towards producing even number, and so tends to reduce $n$.

Given that a large number of lower numbers have already been found to reach $1$, this shows that the system has a tendency to reach $1$. This is because when a large unproven number is reduced to a smaller number proven to reach $1$ then this large number has just been proven to reach $1$ also.

Consider this,

This graph is symmetric about $tan(45^o)$ line through the origin.  The label 'odd' and 'even' show how any selected number enters into the 'funnel' and is reduced.  The red lines deflect all odd numbers to the black lines that divide the number.  At this point the resulting number can be odd or even.  If even, the number is further reduced by the other black line.  If odd, the number is deflected to the red lines.

Number caught between the black lines eventually reach $1$.  The red lines increase the numbers deflected off them and are always even.  These lines do not produce the needed $1$.

Does this prove the Collatz Conjecture.  No, just that, the more lower number proven, the more likely the conjecture is true.

Taking Out The Explosives First

  $$\sum^{\infty}_{n=1}{\cfrac{csc^2(\cfrac{n\pi}{5})}{n^3}}=\sum^\infty_{m=0}\sum^{10}_{n=1}{\cfrac{1}{(10m+n)^3}}csc^2(\cfrac{n\pi}{5})$$

As the sum at $n=5$ and $n=10$ are positive and negative infinite sum, they cancel and we are left with,

  $$\sum^{\infty}_{n=1}{\cfrac{csc^2(\cfrac{n\pi}{5})}{n^3}}=\sum^\infty_{m=0}\sum^{2,3,4,6,7,8,9}_{n=1}{\cfrac{1}{(10m+n)^3}}csc^2(\cfrac{n\pi}{5})$$

We have eventually,

$$\cfrac{\pi^2}{25}\left[\sum^{2,4,5}_{i=1}csc^2(\cfrac{(2i-1)\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(10m+(2i-1))^3} +\\ \cfrac{1}{8}\sum^{4}_{i=1}csc^2(\cfrac{2i\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(5m+i)^3}+\\ \sum^{2,3,4,6,7,8,9}_{n=1}{\cfrac{1}{n^3}}csc^2(\cfrac{n\pi}{5})\right]$$

This will converge and can be solved, numerically.

Changing Flint Miller Summation

 Consider a circle of perimeter,

$2\pi r=10$

a unit circle that has been scaled by 

$r=\cfrac{10}{2\pi}$.  

A scaling factor of $\cfrac{2\pi}{10}$ has to be applied to the results afterwards because of this scaling.

This way, as a integer $n$ run along the circumference, integers overlap with a previous integer (the last digit being the same).  All integer falls into $10$ points on the circumference.

The scaling factor is necessary because on a unit circle, one unit length along the circumference extends one radian (rad) at the center.  It is not true here.  Scaling the unit circle to radius $\cfrac{10}{2\pi}$, to accommodate $10$ points on the circumference, simplifies the analysis. 

Consider, 

$$(\cfrac{2\pi}{10})^2\sum^{\infty}_{n=1}{\cfrac{csc^2(n.\cfrac{2\pi}{10})}{n^3}}=\cfrac{\pi^2}{25}\sum^{\infty}_{n=1}{\cfrac{csc^2(\cfrac{n\pi}{5})}{n^3}}$$

where $(\cfrac{2\pi}{10})^2$ scales $(csc^2\,\,n)$ back to its length on a unit circle.

$$\sum^{\infty}_{n=1}{\cfrac{csc^2(\cfrac{n\pi}{5})}{n^3}}=\sum^\infty_{m=0}\sum^{10}_{n=1}{\cfrac{1}{(10m+n)^3}}csc^2(\cfrac{n\pi}{5})$$

because when $n=10$,  $\cfrac{n\pi}{5}=2\pi$ and the next $10$ addends scaled by $\cfrac{1}{n^3}$ repeats on the circle in the same position.  We extract the first $10$ terms for $m=0$,

$$\sum^{10}_{n=1}{\cfrac{1}{n^3}}csc^2(\cfrac{n\pi}{5})$$

We are then left with eq(1),

$$\sum^\infty_{m=1}\sum^{10}_{n=1}{\cfrac{1}{(10m+n)^3}}csc^2(\cfrac{n\pi}{5})$$

Consider, when $n=1$,

$$csc^2(\cfrac{\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(10m+1)^3}$$

Consider, when $n=2$,

$$csc^2(\cfrac{2\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(10m+2)^3}=csc^2(\cfrac{2\pi}{5})\sum^\infty_{m=1}\cfrac{1}{8}\cfrac{1}{(5m+1)^3}$$

which is for all even $n$,

$$\cfrac{1}{8}\sum^{5}_{i=1}csc^2(\cfrac{2i\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(5m+i)^3}$$

then (1) is then,

$$\sum^{5}_{i=1}csc^2(\cfrac{(2i-1)\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(10m+(2i-1))^3} +\\ \cfrac{1}{8}\sum^{5}_{i=1}csc^2(\cfrac{2i\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(5m+i)^3}$$

And the origin summation we started with is,

$$\cfrac{\pi^2}{25}\left[\sum^{5}_{i=1}csc^2(\cfrac{(2i-1)\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(10m+(2i-1))^3} +\\ \cfrac{1}{8}\sum^{5}_{i=1}csc^2(\cfrac{2i\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(5m+i)^3}+\\ \sum^{10}_{n=1}{\cfrac{1}{n^3}}csc^2(\cfrac{n\pi}{5})\right]$$

Convergent but...NOT Quite!

Addends at $\pi$ and $2\pi$ should cancels leaving $8$ points on the circle, but it is not satisfying.

Note:

Flint Miller Series is $$\sum^{\infty}_{n=1}{\cfrac{csc^2(n)}{n^3}}$$

Stepping Out Odd And Stepping Out Even

 Stepping out odd steps and stepping out even steps are different,

Odd and even steps already have one common point at the origin.  They will not meet again unless they have at least one odd common factor in their divisors that created those step sizes.  Since only even numbers can have a odd factor (and at least one other even factor), and a odd number has only odd factors (an even factor changes it to a even number).  This common odd factors are in the even number divisor that creates a even step.

A new step everyday.

Erdős–Straus Conjecture In Odd Steps

 This is the reason why a half step was take,

a old number of steps is required to reach length $1$ at $\cfrac{\pi}{2}$ further down.  Without the half steps the formulation of $L$ will scream for an even factor.

$M=4\cfrac{N}{n}$ is an even partition such that integer steps can be taken to $N$.  $a$ is odd because $N$ is odd due to $n$ being odd.  $b$ is also odd for the same reason.  So $c$ is even such that $a+b+c$ fills up $M$ an even number.  But full steps in $c$ cannot reach $\cfrac{\pi}{2}$.  $L$ is in odd steps only due to $N$ and $n$ being odd.  So, a half step is taken to find the step count and $c$ is recovered by doubling the half step count ( times $2$).

It is possible instead to step all the way to $\cfrac{3\pi}{2}$ and then divide the count obtained by $3$?  In this case the expression for length $\bar{c}$ is$\cfrac{3}{L}$ which is the reciprocal of an integer only if $L$ has a factor $3$.  This is not the case for prime numbers other then $3$, and the prove will fail.

We start not with $n$ dividing a circle, we start with $4$ dividing the circle. 

Goodnight.

Plasma Deflected

 Knowledge is often unquantifiable, math brings the details and provides the next small step.  For example plasma, as it was dealt with here, plasma was electromagnetic not particles.  So, to deflect plasma, total internal reflection,

reflective index is important not magnetic field.  As wave plasma has long wavelength, alternating electric and magnetic fields out of phase means the fields can pass through the material meters ahead before the fields are at the correct angles to interact with the confining magnetic fluxes.

Board direction changes attention from particles to waves; not the mathematical details, which nudge the next small step forward.

Big Fuck Deflated

 In the prove for Erdős–Straus Conjecture, for even $n$, the expression,

$\cfrac{1}{N}+\cfrac{2}{N}+\cfrac{1}{N}=\cfrac{4}{n}$

is equivalent to

$\cfrac{1}{N}+\cfrac{1}{N_1}+\cfrac{1}{N}=\cfrac{4}{n}$

because $N=n=2N_1$ is even, so the $2$ in numerator cancel with $N$ to give an integer $N_1$.

If we state,

$\cfrac{1}{A}+\cfrac{1}{B}+\cfrac{2}{C}=\cfrac{4}{n}$

where $n\ge 2$ and $A$, $B$, $C$ are integers, it is for all $n$ odd and even.

If we state,

$\cfrac{1}{A}+\cfrac{1}{B}+\cfrac{1}{C}=\cfrac{4}{n}$

then it is true, only for even integer $n\ge 2$.

Thank you.

ChanHL Theorem Big Fuck

  Consider a circle of perimeter length $4$,

we divide this circle into $n$ sector.  We divide one of these sectors of angle $\cfrac{2\pi}{n}$ further into $M$ sectors such that,

$M=4*\cfrac{N}{n}$, 

$M\ge 3$, both $M$ and $N$ are integer

This way,

$\cfrac{1}{M}*\cfrac{2\pi}{n}*N=\cfrac{n}{4N}\cfrac{2\pi}{n}*N=\cfrac{\pi}{2}$

That is we can walk $N$ step of $(\cfrac{1}{M}*\cfrac{2\pi}{n})$ to $\cfrac{\pi}{2}$ along the circumference for a total distance of $1$.

On the sector with $M$ markings, we divide it with two radial lines drawn through the markings into three sectors.  Obviously,

$a+b+c=M$

If $a$ is on the first marking, $N$ steps will take it  down $(\cfrac{\pi}{2})$ to $1$, ie. the length of $a$

$\bar{a}=\cfrac{1}{N}$

Then the next line divides the remaining $M-1$ into $b$ and $c$, such that both $b$ and $c$ divides $N$ (The proper choice of $M$ and $N$ will make this possible and easy).  We formulate,

$b.B=N$ and $b.C=N$

then in $B$ steps $b$ will travel through $\cfrac{\pi}{2}$ for a distance of $1$ along the circumference.  Similarly, in $C$ steps $c$ will travel through $\cfrac{\pi}{2}$ for a distance of $1$ along the circumference.  That is to say,

$\bar{b}=\cfrac{1}{B}$   and 

$\bar{c}=\cfrac{1}{C}$

Since, the size of this sector is $\cfrac{4}{n}$, we have

$\bar{a}+\bar{b}+\bar{c}=\cfrac{1}{N}+\cfrac{1}{B}+\cfrac{1}{C}=\cfrac{4}{n}$ for any $n\ge 2$ and $N$, $B$, $C$

all integer.  Provided $B$ and $C$ can be found.

For the proper choice of $M$ and $N$;

When $n$ is even we chose $N=n$ then $M=4$.

 $a$ is found as above, one marking wide, and takes $N$ step to reach $1$ along the circumference

$\bar{a}=\cfrac{1}{N}$

$b$ is set to $2$ markings wide such that $b$ will take $\cfrac{N}{2}$ steps to travel a distance of $1$ so,

$\bar{b}=\cfrac{2}{N}$

since $N=n$ is even, $\cfrac{N}{2}$ is an integer.

After taking step $a$, one step,

$b+c=M-1=4-1=3$

After taking step $b$, two steps,

$2+c=3$

then only one step $c=1$ remains and so,

$\bar{c}=\cfrac{1}{N}$

since $a$, $b$ and $c$ make up the sector of length $\cfrac{4}{n}$

$\cfrac{1}{N}+\cfrac{2}{N}+\cfrac{1}{N}=\cfrac{4}{n}$

we have proven Erdős–Straus conjecture for even numbers.

When $n$ is odd, it has at least one odd factor $f_o$.  We let

$n=f_of_1$  where $f_1$ is another odd factor and may be one.

 Let

$N=\cfrac{f_1}{2}(n+f_o)(f_o-1)=\cfrac{f_1}{2}(f_1f_o+f_o)(f_o-1)$

$N=\cfrac{f_of_1(f_1+1)(f_o-1)}{2}=\cfrac{n}{2}(f_1+1)(f_o-1)$ 

since $(f_1+1)$ $(f_o-1)$ are even, $N$ is an integer.

So, $M=4\cfrac{N}{n}=\cfrac{2n(f_1+1)(f_o-1)}{n}=2(f_1+1)(f_o-1)$

We let

$a=f_1(f_o-1)$

the first step we take is $f_1(f_o-1)$

$\bar{a}=\cfrac{1}{K}$

$K=\cfrac{f_of_1(f_1+1)(f_o-1)}{2f_1(f_o-1)}$ since $(f_o-1)$ and $(f_1+1)$ are even, $K$ is an integer.

Let,

$b=f_o-1$     $\bar{b}=\cfrac{1}{J}$

$J=\cfrac{f_of_1(f_1+1)(f_o-1)}{2(f_o-1)}$ $J$ is an integer. 

and the length of this step is $\bar{b}=\cfrac{1}{J}$

After step $b$,

$c=M-a-b$

$c=2(f_1+1)(f_o-1)-f_1(f_o-1)-(f_o-1)$

$c=2f_of_1+2f_o-2f_1-2-f_of_1+f_1-f_o+1=f_of_1+f_o-f_1-1$ 

$c=f_1(f_o-1)+(f_o-1)=(f_o-1)(f_1+1)$ 

But we take half step instead to go further to $\cfrac{\pi}{2}$ away, when the total length travelled by $c$ is $1$.

$c=\cfrac{(f_o-1)(f_1+1)}{2}$ 

since $(f_o-1)$ and $(f_1+1)$ are even, $c$ is an integer.

$\bar{c}=\cfrac{2}{L}$

$L=\cfrac{2f_of_1(f_1+1)(f_o-1)}{2(f_o-1)(f_1+1)}$

and $L$ is an integer.

The total length of these steps $a$, $b$ and $c$ is $\cfrac{4}{n}$, we have,

$\cfrac{1}{K}+\cfrac{1}{J}+\cfrac{2}{L}=\cfrac{4}{n}$

where $K$, $J$ and $L$ are integer.

and we have ChanHL Theorem for all odd $n$, $n\gt 2$, that corrects Erdős–Straus Conjecture.

Erdős–Straus Conjecture Not But ChanHL Theorem

 It seems that to take half steps at $c$ does not resolve the problem,

$c=\cfrac{(f_o-1)(f_1+1)}{2}$ 

since $(f_o-1)$ and $(f_1+1)$ are even, $c$ is an integer.

$\bar{c}=\cfrac{2}{L}$

$L=\cfrac{2f_of_1(f_1+1)(f_o-1)}{2(f_o-1)(f_1+1)}$

and $L$ is an integer.

The total length of these steps $a$, $b$ and $c$ is $\cfrac{4}{n}$, we have,

$\cfrac{1}{K}+\cfrac{1}{J}+\cfrac{2}{L}=\cfrac{4}{n}$

where $K$, $J$ and $L$ are integer.

This will not prove Erdős–Straus Conjecture but this is now my very own ChanHL Theorem.

Thank you.

Erdős–Straus Conjecture

 Consider a circle of perimeter length $4$,

we divide this circle into $n$ sector.  We divide one of these sectors of angle $\cfrac{2\pi}{n}$ further into $M$ sectors such that,

$M=4*\cfrac{N}{n}$, 

$M\ge 3$, both $M$ and $N$ are integer

This way,

$\cfrac{1}{M}*\cfrac{2\pi}{n}*N=\cfrac{n}{4N}\cfrac{2\pi}{n}*N=\cfrac{\pi}{2}$

That is we can walk $N$ step of $(\cfrac{1}{M}*\cfrac{2\pi}{n})$ to $\cfrac{\pi}{2}$ along the circumference for a total distance of $1$.

On the sector with $M$ markings, we divide it with two radial lines drawn through the markings into three sectors.  Obviously,

$a+b+c=M$

If $a$ is on the first marking, $N$ steps will take it  down $(\cfrac{\pi}{2})$ to $1$, ie. the length of $a$

$\bar{a}=\cfrac{1}{N}$

Then the next line divides the remaining $M-1$ into $b$ and $c$, such that both $b$ and $c$ divides $N$ (The proper choice of $M$ and $N$ will make this possible and easy).  We formulate,

$b.B=N$ and $b.C=N$

then in $B$ steps $b$ will travel through $\cfrac{\pi}{2}$ for a distance of $1$ along the circumference.  Similarly, in $C$ steps $c$ will travel through $\cfrac{\pi}{2}$ for a distance of $1$ along the circumference.  That is to say,

$\bar{b}=\cfrac{1}{B}$   and 

$\bar{c}=\cfrac{1}{C}$

Since, the size of this sector is $\cfrac{4}{n}$, we have

$\bar{a}+\bar{b}+\bar{c}=\cfrac{1}{N}+\cfrac{1}{B}+\cfrac{1}{C}=\cfrac{4}{n}$ for any $n\ge 2$ and $N$, $B$, $C$

all integer.  Provided $B$ and $C$ can be found.

For the proper choice of $M$ and $N$;

When $n$ is even we chose $N=n$ then $M=4$.

 $a$ is found as above, one marking wide, and takes $N$ step to reach $1$ along the circumference

$\bar{a}=\cfrac{1}{N}$

$b$ is set to $2$ markings wide such that $b$ will take $\cfrac{N}{2}$ steps to travel a distance of $1$ so,

$\bar{b}=\cfrac{2}{N}$

since $N=n$ is even, $\cfrac{N}{2}$ is an integer.

After taking step $a$, one step,

$b+c=M-1=4-1=3$

After taking step $b$, two steps,

$2+c=3$

then only one step $c=1$ remains and so,

$\bar{c}=\cfrac{1}{N}$

since $a$, $b$ and $c$ make up the sector of length $\cfrac{4}{n}$

$\cfrac{1}{N}+\cfrac{2}{N}+\cfrac{1}{N}=\cfrac{4}{n}$

we have proven Erdős–Straus conjecture for even numbers.

When $n$ is odd, it has at least one odd factor $f_o$.  We let

$n=f_of_1$  where $f_1$ is another odd factor and may be one.

 Let

$N=\cfrac{f_1}{2}(n+f_o)(f_o-1)=\cfrac{f_1}{2}(f_1f_o+f_o)(f_o-1)$

$N=\cfrac{f_of_1(f_1+1)(f_o-1)}{2}=\cfrac{n}{2}(f_1+1)(f_o-1)$ 

since $(f_1+1)$ $(f_o-1)$ are even, $N$ is an integer.

So, $M=4\cfrac{N}{n}=\cfrac{2n(f_1+1)(f_o-1)}{n}=2(f_1+1)(f_o-1)$

We let

$a=f_1(f_o-1)$

the first step we take is $f_1(f_o-1)$

$\bar{a}=\cfrac{1}{K}$

$K=\cfrac{f_of_1(f_1+1)(f_o-1)}{2f_1(f_o-1)}$ since $(f_o-1)$ and $(f_1+1)$ are even, $K$ is an integer.

Let,

$b=f_o-1$     $\bar{b}=\cfrac{1}{J}$

$J=\cfrac{f_of_1(f_1+1)(f_o-1)}{2(f_o-1)}$ $J$ is an integer. 

and the length of this step is $\bar{b}=\cfrac{1}{J}$

After step $b$,

$c=2(f_1+1)(f_o-1)-f_1(f_o-1)-(f_o-1)$

$c=2f_of_1+2f_o-2f_1-2-f_of_1+f_1-f_o+1=f_of_1+f_o-f_1-1$ 

$c=f_1(f_o-1)+(f_o-1)=(f_o-1)(f_1+1)$ 

But we take half step instead,

$c=\cfrac{(f_o-1)(f_1+1)}{2}$ 

since $(f_o-1)$ and $(f_1+1)$ are even, $c$ is an integer.

$\bar{c}=\cfrac{1}{L}$

$L=\cfrac{2f_of_1(f_1+1)(f_o-1)}{2(f_o-1)(f_1+1)}$

and $L$ is an integer.

The total length of these steps $a$, $b$ and $c$ is $\cfrac{4}{n}$, we have,

$\cfrac{1}{K}+\cfrac{1}{J}+\cfrac{1}{L}=\cfrac{4}{n}$

where $K$, $J$ and $L$ are integer.

and we have proven Erdős–Straus Conjecture for all $n$, $n\ge 2$

Good morning.

Notes:

The steps, $a$, $b$ and $c$ are sectors in units of markings provided by dividing the sector $\cfrac{4}{n}$ by $M$.  These steps size are chosen such that,

$a$.$A$=$N$,     $b$.$B$=$N$ and     $c$.$C$=$N$

Given step size $a$, it will take $A$ step to reach $N$,  where $\cfrac{\pi}{2}$ is through a lenght of $1$.  So, the length of $a$ is,

$\bar{a}=\cfrac{1}{A}$

similarly for $b$ and $c$.  $M$ goes down to $\cfrac{4}{n}$, $N$ goes down to $1$.

Bertrand–Chebyshev Step Forward

 Consider, the last prime, $p$ for which

$p+q=2n$

for $n\gt 2$ where $q \le p$ is also prime.  

In addition, $S_n: p+q=2n,\, p\ne q,\,n\gt 2$ is all true up to $n$, that is to say $S_n$ is true for all even numbers below and including $2n$.

This starting condition is easily satisfied by enumerating the first $S_n$.  Then for the interval $2n$ to $4n$ in which a prime exist by Bertrand–Chebyshev Theorem, this prime is the next prime, $p^{next}$ in succession.  Since $S_{n}$ exist for all $n$.  By replacing one of the prime pair with $p^{last}$ or $p^{next}$ the interval $(0,2n]$ will translate to $(2n,4n]$ and so, $S_{n}$ exist for all $n$ up to $4n$.

Then we move to the next interval $n$ up to $8n$ still by using the fact that Bertrand–Chebyshev Theorem guarantee a prime number in the interval $4n$ to $8n$.

ad infinitum.

Roughly Goldbach's Conjecture is again proved.

Mind The Gap, Goldbach

 The weak point in the proof of Goldbach's Conjecture is the insistent that all skipped indexes are reachable by selecting the next prime number (backward or forward) and reach out with known primes.

Skipped index, in $S_{n}$ is the direct consequence of gap between consecutive primes.  Such gaps are small compared to the primes numbers that mark the gap.  We know from Bertrand–Chebyshev Theorem that there is a prime between $n$ and $2n$ (inclusive).  So the widest prime gap in this interval is $n$, in which case the last prime is $2n$. On average the gap is $\cfrac{n}{2}$.  

Given a prime $n$, the next prime is at most $n$ away at $2n$, from Bertrand–Chebyshev Theorem.  But since the average prime gap over $n$ ratio, tends towards $\cfrac{1}{log(n)}$(and is asymptotically zero) by the Prime Number Theorem.  The gap is smaller than the last prime $n$, in the worst case scenario.  Please refer to the post "Squeeze Goldbach" dated 24 Dec 2022, also.

The last prime (at $2n$) after a widest gap of at most $n$ remains reachable from the last but one prime $p=n$.

ie., with $p=n$ and $q=n$

$p+q=2n$

which is $S_{2n}$.  In cases when the last prime is $p\lt 2n$ we will use this prime, and all of any one of the previous prime to reach beyond $2(n+p)$ and $2n$. 

$p+q=2n_i$  where $2n_i\lt 2n$

$q=3,5,...q_i\t p$

Goldbach Taking Prime Hops

Let,

$S_n:\,\, p+q=2n$ $p\neq q$ and $p$ and $q$ are both prime and $n\gt 2$

Consider $n$ such that $p=2n-1$ is prime and so,

$2n-1+3=p+q=2n+2=2(n+1)$

with $q=3$,  such that $S_{n+1}$ is true.

Consider further that, at $n$ we take a step $q=5$, down the number line to $2(n+2)$, obviously,

$p+5=2n-1+5=2(n+2)$

as such $S_{n+2}$ is true.

Again from $n$, we take a step $q=7$,

$p+7=2n-1+7=2(n+3)$ 

as such $S_{n+3}$ is true

and with a step $q=11$

$p+11=2n-1+11=2(n+5)$ 

as such for any prime $p=2n-1$, $S_{n+1}$, $S_{n+2}$, $S_{n+3}$, $S_{n+5}$ are true.

In fact all steps of known prime are possible resulting in $S_{n+a}$, where $a=\left\lfloor\cfrac{p}{2}\right\rfloor$, being proven true.

We have then, $S_{n+a}$ where $a=\left\lfloor\cfrac{q}{2}\right\rfloor$ for all know prime $p=2n-1$.

We starts $S_{n}$ with $n=2$ where $p=2(2)-1=3$ is prime

$2(2)-1+3=2(3)=p+q=2(n+1)\,\,\,\implies\,\,S_{3}$

from previously results, we have also $S_{4}$, $S_{5}$, $S_{7}$

When we starts $S_{n}$ with $n=3$ where $p=2(3)-1=5$ is prime

$2(3)-1+3=2(4)=p+q=2(n+1)\,\,\,\implies\,\,S_{4}$

from previously, we have also $S_{5}$, $S_{6}$, $S_{8}$

The skipped index in $S_n$, are recovered when we start $S_{n}$ with a later prime.  Because for any prime $p=2n-1$, $S_{n+1}$, $S_{n+2}$, $S_{n+3}$, $S_{n+5}$ are true.

The next consecutive odd (also a prime) gives $S_{n+2}$, $S_{n+3}$, $S_{n+4}$, $S_{n+6}$.  From which we obtain $S_{n+4}$.  The rest of the indexes not not relevant, only the skipped index and the prime from which to reach the skipped index.

These skipped indexes are the result of odd numbers not being prime along the number line.

The size, $i$ of the series $S_{n}$....$S_{n+i}$ is the number of primes used to hop forward, which is the number of primes already discovered.  Given the large size of this number, and the relatively small number of consecutive odd between consecutive primes, all skipped index can be recovered.

This means all $S_{n}$ for $n\gt 2$ can be proven true.  That is to say,

Given $p\neq q$ and, $p$ and $q$ are both prime 

$p+q=2n$ 

is true for all $n\gt 2$.

Goldbach's Conjecture proved!

Enhanced Double Slits

 More than one double double slits,

narrower as they are placed further from the center hole, such that their first dark band overlap at the center hole.

Merry Christmas...

First Thing First Goldbach

For, $p$ and $q$ both prime, is

$p+q=2n$, $n\gt 2,\,\in\mathbb{Z}^+$

true for all $n$.

if,

$p=q$

$p+q=2p=2n$

$p=n$

but not all $n$ are prime as such $p\neq q$.  

So previously where $n_o=n_q$, cannot hold.

What For? Negative Energy

 If in a double slit experiment, the dark bands are that pronounced,

then between two double double slits, we might just observe a negative energy field.

Negative what energy?  What particle source again?

Easy Goldbach Bigger

 Consider again,

since the distance between two consecutive primes is roughly $log(n)$ and the ratio of this distance to $n$ is asymptotically zero by the Prime Number Theorem, the distance between the last prime and next prime as illustrated is also very narrow.  And by Bertrand–Chebyshev Theorem a prime exist between $n$ and $2n$.  We have,

the space below the last prime $\gt n$, is full of prime numbers.  So, as $n\rightarrow\infty$, it is easier to find $p+q=2n$, although not necessarily certain.

The point is as $n\rightarrow\infty$, Goldbach's Conjecture is easier to satisfy.

Squeeze Goldbach

 How does $p+q=2n$ fails? If,

$\cfrac{p-q}{4}\longrightarrow\cfrac{n+1}{2}$ as primes$\longrightarrow$large and $n\longrightarrow\infty$

that insist on last admitted odd number, $2n-1$ be prime is the only way to fail, then we may have a proof in the previous post "Silly Me, Goldbach" dated 24 Dec 2022, because the Prime Number Theorem indicates that it is not possible to fail that way.  

Consider, from previously,

$n_g+n_o=n+1$  

from the same post limits

$n_o$ to about $[1,\cfrac{n+1}{2})$.

but does not guarantee that both $p$ and $q$ exist for any $n\gt 2$, in

$p+q=2n$

$p=2n-2n_o+1$ and $q=2n-2n_g+1$

$p$ and $q$ must be prime.

If we substitute for $n_o$ in $p$ for $n_g$

$p=2n-2(n+1-n_g)+1$

$p=2n_g-1$ and similarly,

$q=2n_o-1$ 

Hey, Hey, Hey...a simpler way to find $p$ and $q$ and more importantly,

$p$ and $q$ are more relevant to the last odd number admitted as $n$ increments.  As,

$\cfrac{p-q}{4}\longrightarrow\cfrac{n+1}{2}$, $n_g\longrightarrow n$ as primes$\longrightarrow$large and $n\longrightarrow\infty$.

In the worst case, where $p$ and $q$ are consecutive primes of increasing gap, ie. $p-q$ increases most as $n$ increases, Prime Number Theorem shows that $p-q$ is roughly $log(n)$ in the first $n$ integers.  That means,

$p=2n_g-1\lt\lt 2n-1$

where $2n-1$ is the last odd number added. $p$ will not fail because the last odd number added when $2n$ increases to $2(n+1)$ is not a prime number.  $p$ is prime number located away from this boundary as $n$ increases.  $n_g$ and $n_o$ are located in a small gap about $\cfrac{n+1}{2}$.

Can $p$ and $q$ always be found?  Is the only way to fail when $p$ and $q$ becomes too wide?

Maybe,

But PNT says average $p-q$ when $p$ and $q$ are consecutive is of order $log(n)$ as $n$ increases and so strictly, $p-q\lt n$.

Average!  If the average fails then $(p-q)_{max}$ will fail.  But not vice versa.

Goldbach's Conjecture will fail when the last consecutive primes $p$ and $q$ are greater than $n$ apart.

$p-q\gt n$

not the average gap between consecutive primes but the largest.  Unfortunately, Bertrand–Chebyshev theorem prove this wrong, there is a prime between $n$ and $2n$.

Silly Me, Goldbach

 From, 

$p+q=2n$

$p_1+q_1=2n+2$

$p+q-(p_1+q_1)=2$

This is true for all consecutive $p_i$ and $q_i$ pair,

$p_i+q_i-(p_{i+1}+q_{i+1})=2$

This is a simple predictor for $p_{i+1}$ and $q_{i+1}$, given $p_{i}$ and $q_{i}$ and a list of primes $p\lt 2(n+i+1)$.

Alternatively, if we use the formulation,

$p=2n-2n_o+1$

$q=2n-2n_g+1$

$\cfrac{p-q}{2}=n_g-n_o$ --- (1)

but

$p+q=2n=4n+2-2(n_o+n_g)$

$n_g+n_o=n+1$   --- (2)

We add (1) and (2),

$2n_g=\cfrac{p-q}{2}+n+1$

$n_g=\cfrac{p-q}{4}+\cfrac{n+1}{2}$

Similarly,

$n_o=\cfrac{q-p}{4}+\cfrac{n+1}{2}$

We can find $n_g$ and $n_o$ exactly.

For the next pair of primes,

$p_1+q_1=2(n+1)$

from (2),

$p_1+q_1=2(n+1)=2(n_g+n_o)$ 

Graphically,

The constrain seems to be on $n_g$ and $n_o$ growing in opposite directions about $\cfrac{n+1}{2}$. $n_o$ is confined to about $[1,\cfrac{n+1}{2})$.

When $n_o=1$

$p=2n-2(1)+1=2n-1$

$q=1$

that means the last number introduce to the set of primes available to construct $p+q$ is $2n-1$.  $2n-1$ must be a prime.

$n_g$ is confined to about $(\cfrac{n+1}{2},n]$, 

When $n_o=n$

$q=2n-2(n)+1=1$

$p=2n-1$

Equivalently,  $2n-1$ must be a prime.

We can squeeze $n$ such that $n_o$ and $n_g$ must be on the boundary that insist on $2n-1$ being prime that contradicts with not all odd numbers are prime.  That happens for $n$ small, but those values has been enumerated.  Squeezing $n$ will not disprove the conjecture.

However if $p$ and $q$ diverges quickly, where $\cfrac{p-q}{2}$ forces $n_g$ and $n_o$ to their boundary values that insist on $2n-1$ being prime and since not all odd number ($2n-1$) admitted as $2(n-1)$ moves to $2n$ is prime, then we have a contradiction.

Does $p$ and $q$ diverges quickly?  Where,

$\cfrac{p-q}{4}\longrightarrow\cfrac{n+1}{2}$ as primes$\longrightarrow$large and $n\longrightarrow\infty$

and the Prime Number Theorem says?  The average gap between consecutive prime numbers among the first $n$ integers is roughly $log(n)$. 

No.  PNT does not serve to contradict the conjecture.  In the worst scenario, when $p$ and $q$ are consecutive with increasing gap,

$\cfrac{p-q}{2}$ does not squeeze close to $\cfrac{n+1}{2}$ and insist that the last odd number just admitted into $2n$ is prime.

PNT does not guarantee the existent of $n_g$ and $n_o$ either.

Pulling Hair Off Goldbach

From the previous post "More Prime Spiral Goldbach" dated 24 Dec 2022,

 $n_{gi+1}+n_{oi+1}=n_{g}+n_{o}+i+1$            $i$ is odd, 1, 3, 5...

predicts the next pair $n_{gi+1}$ and $n_{oi+1}$ from an odd position $i$ odd.

$n_{gi+1}+n_{oi+1}=(2n+(i+1)-2)-n_{g}-n_{o}$  $i$ is even, 2, 4, 6

predicts the next pair $n_{gi+1}$ and $n_{oi+1}$ from a even position $i$ even.

Since $n_{gi+1}$ and $n_{oi+1}$ is not guaranteed to ensure,

$2(n+i+1)-2n_{oi+1}+1=p_{i+1}$ and $2(n+i+1)-2n_{gi+1}+1=q_{i+1}$

where both $p_{i+1}$ and $q_{i+1}$ are prime and,

$p_{i+1} +q_{i+1}=2(n+i+1)$

Goldbach's Conjecture not proven here.

But take any pair of primes, $p$ and $q$ and calculate,

$n=\cfrac{p+q}{2}$

$n_o=\cfrac{2n-p+1}{2}$ and

$n_p=\cfrac{2n-q+1}{2}$ 

starting with the odd position prediction find candidates for $n_{g1}$ and $n_{o1}$ under the constrain,

$p_{1} +q_{1}=2(n+1)$

then use the even position prediction to find candidates for $n_{g2}$ and $n_{o2}$ under the constrain,

$p_{2} +q_{2}=2(n+2)$

repeat till no possible candidates for $n_{gi}$ and $n_{oi}$ can be found.  The series ends.

Does Goldbach has long hairs?  Spirally long hairs?

More Prime Spiral Goldbach

 Consider,

 $2n=p+q$

where $p$ and $q$ are primes and $n$ an integer greater than 2.

$2n-q=p$

let $q=\cfrac{a}{b}$

$2n-\cfrac{a}{b}=p$ ---(*)

$\cfrac{2nb-a}{b}=p$

If p is prime,

$b=2nb-a$ or

$b=1$

When $b=1$,

$2n-a=p$ is prime

we have instead,

$2n=p+a$

so,

$a=q$

interesting result, $\rightarrow$备用

or from $b=2n_ob-a$

$a=2n_ob-b=b(2n_o-1)$

$\cfrac{a}{b}=2n_o-1$

for a specific number $n_o$ and $b$ not zero.

sub. into (*) we have,

$2n-2n_o+1=p$

$2(n-n_o)+1$ is prime as $p$ is prime and $n\gt n_o$ as primes are positive.

Let $2(n-n_{oo})+1=q_i$ where $n\gt n_{oo}$ for positive prime.

where $n$ is paired with $n_{oo}$ to give $q_{i}$ as $n$ and $n_o$ is paired to give $p$.  Setting $n_i$ instead of $n$ is not necessary.

In which case, we formulate,

$q_i+p=2(n-n_o)+1 + 2(n-n_{oo})+1$

and so,

$q_i+p=2(2n-n_o-n_{oo})+2=2n_j$

where $n_j=2n-n_o-n_{oo}+1$ which is just another integer and both $q_i$ and $p$ are prime. And when we set $2n=n_o+n_{oo}$ we start from $n_j=1$.

Similarly,

$p_i+q=2(n-n_g)+1 + 2(n-n_{gg})+1$

and so,

$p_i+q=2(2n-n_g-n_{gg})+2=2n_j$

where $n_j=2n-n_g-n_{gg}+1$ which is just another integer and both $p_i$ and $q$ are prime.

Is $n_j$ strictly non negative?  Yes. And When we set $2n=n_g+n_{gg}$ we start from $n_j=1$.

And we add the mess together,

$p_i+q_i+p+q=2(2n-n_g-n_{gg})+2(2n-n_o-n_{oo})$

since $p+q=2n$

$p_i+q_i=2(3n-n_g-n_{gg}-n_o-n_{oo})$

in order that 

$p_i+q_i=2(n+1)$

we set, 

$2(n+1)=2(3n-n_g-n_{gg}-n_o-n_{oo})$

by choosing appropriate values for $n_g$, $n_{gg}$, $n_o$ and $n_{oo}$.

$2n=n_g+n_{gg}+n_o+n_{oo}+1$

So, $p_i=2(n-n_{gg})+1$ and $q_1=2(n-n_{oo})+1$

where given $n$, $n_{gg}$ and $n_{oo}$ are chosen such that $p_i$ and $q_i$ are primes.

$n_g$ and $n_o$ are derived from the starting/previous primes $p$, $q$ and $n$.

$n_g$, $n_{gg}$, $n_o$ and $n_{oo}$ are not necessarily prime just integers greater than 2.

One way to simplify is to set $p=q$, but this series is troubled.

Given, $2n=n_g+n_{gg}+n_o+n_{oo}+1$,The set $[$ $n_g$, $n_{gg}$, $n_o$, $n_{oo}$$]$ has one odd or 3 odd numbers.  

If we go on further, the next equation is,

$2(n+1)=n_{gg}+n_{ggg}+n_{oo}+n_{ooo}+1$

$2n+1=n_{gg}+n_{ggg}+n_{oo}+n_{ooo}$

we have to introduce $n_{ggg}$ and $n_{ooo}$ and the set $[$ $n_{ggg}$, $n_{gggg}$, $n_{ooo}$, $n_{oooo}$$]$ has one odd or 3 odd numbers, as before.

and the next in the series after $n_{ggg}$ and $n_{ooo}$

$2(n+2)=n_{ggg}+n_{gggg}+n_{ooo}+n_{oooo}+1$

$2n+3=n_{ggg}+n_{gggg}+n_{oo}+n_{oooo}$

We gather the equations with a change in notation,

$2n-1=n_g+n_{g1}+n_o+n_{o1}$ 

$2n+1=n_{g1}+n_{g2}+n_{o1}+n_{o2}$

$2n+3=n_{g2}+n_{g3}+n_{o2}+n_{o3}$ 

$2n+5=n_{g3}+n_{g4}+n_{o3}+n_{o4}$ 

$2n+7=n_{g4}+n_{g5}+n_{o4}+n_{o5}$ 

and in general,

$2n+(2i-1)=n_{gi}+n_{gi+1}+n_{oi}+n_{oi+1}$ --- (*)

where $i$ is an integer starting at 0,  $n_{g0}=n_{g}$ and $n_{o0}=n_{o}$, given a start,

$2n=n_g+n_{g1}+n_o+n_{o1}+1$  from $p+q=2n$ and $p_i+q_i=2(n+1)$.

If we take the difference of the expressions to retain the pair $n_g$ and $n_o$ and the last pair $n_{gi}$ and $n_{oi}$

We have two series depending on $i$ being odd or even.

For $i=1$,

$n_{g2}+n_{o2}=n_g+n_o+2$

For $i=2$ substitute away $n_{g2}$ and $n_{o2}$

$2n+1=n_{o}+n_{g}+n_{o3}+n_{g3}$ 

which involves $n$.

For $i=3$, 

$n_{g4}+n_{o4}=n_{g}+n_{o}+4$ 

For $i=4$, substitute away $n_{g4}$ and $n_{o4}$

$2n+3=n_{g}+n_{o}+n_{g5}+n_{o5}$ 

Looking forward, when $i$ is odd,

$n_{g2}+n_{o2}=n_g+n_o+2$  for $i=1$

$n_{g4}+n_{o4}=n_{g}+n_{o}+4$  for $i=3$

So, in general,

$n_{gi+1}+n_{oi+1}=n_{g}+n_{o}+(i+1)$  when $i$ is odd.

When $i$ is even,

$n_{g3}+n_{o3}=(2n+1)-n_{g}-n_{o}$ for $i=2$ 

$n_{g5}+n_{o5}=(2n+3)-n_{g}-n_{o}$ for $i=4$

In general,

$n_{gi+1}+n_{oi+1}=(2n+(i-1))-n_{g}-n_{o}$  when $i$ is even

In summary,

$n_{gi+1}+n_{oi+1}=n_{g}+n_{o}+i+1$            $i$ is odd, 1, 3, 5...

predicts the next pair $n_{gi+1}$ and $n_{oi+1}$ from an odd position $i$ odd.

$n_{gi+1}+n_{oi+1}=(2n+i-1)-n_{g}-n_{o}$  $i$ is even, 2, 4, 6

predicts the next pair $n_{gi+1}$ and $n_{oi+1}$ from a even position $i$ even.

Consider starting from a odd position,

$n_{gi}+n_{oi}=n_{g}+n_{o}+i$   and the next equation,

$n_{gi+1}+n_{oi+1}=(2n+(i+1)-2)-n_{g}-n_{o}$ 

where next $i$ is $i+1$.

$n_{gi+1}+n_{oi+1}+n_{gi}+n_{oi}=(2n+(i+1)-2)+i=2n+2i-1$  

which is the expression (*).

Consider starting from a even position,

$n_{gi}+n_{oi}=(2n+i-2)-n_{g}-n_{o}$ and the next equation,

$n_{gi+1}+n_{oi+1}=n_{g}+n_{o}+i+1$  

where next $i$ is $i+1$.

$n_{gi+1}+n_{oi+1}+n_{gi}+n_{oi}=(2n+i-2)+i+1=2n+2i-1$  

which is again the expression (*).

This checks the expressions for predicting $n_{gi}$ and $n_{oi}$.

This is not proof of Goldbach's Conjecture but provide guidelines to predict $n_{gi+1}$ and $n_{oi+1}$ from $n_{gi}$ and $n_{oi}$ provided with a start $n_{g}$ and $n_{o}$ and some integer $n$ greater than 2.  

$n_{gi+1}$ and $n_{oi+1}$ gives an expression

$p_{i+1}+q_{i+1}=2(n+i+1)$

where $2(n+i+1)-2n_{oi+1}+1=p_{i+1}$ and $2(n+i+1)-2n_{gi+1}+1=q_{i+1}$

starting at $p+q=2n$ with $i=0$, $p$ and $q$ being prime and $n$ is an integer greater than 2.

$n_{g}$ and $n_{o}$ are derived from the expressions,

$2n-2n_o+1=p$ and $2n-2n_g+1=q$

where $n_{g}$ and $n_{o}$  is chosen such that $p$ and $q$ are primes.

And so primes can be analyzed using an expression,

$p_n=2n-2n_p+1$ where $n\in\mathbb{Z},\,\,n\gt 2$.  $n$ group primes into $n_p\le n$, given $n$, $n_p\gt 2$.

We can restrict membership in $p_n$ to $3\le p_n\le 2n-1$.

Just Messing With Goldbach

Consider,

 $2n=p+q$

where $p$ and $q$ are both primes and $n$ is greater than $2$.

$2n-q=p$

let $q=\cfrac{a}{b}$

$2n-\cfrac{a}{b}=p$ ---(*)

$\cfrac{2nb-a}{b}=p$

When p is prime,

$b=2nb-a$ or

$b=1$

When $b=1$,

$2n-a=p$ is prime

we have instead,

$2n=p+a$

so,

$a=q$

interesting result, $\rightarrow$备用

or from $b=2n_ob-a$

$a=2n_ob-b=b(2n_o-1)$

$\cfrac{a}{b}=2n_o-1$

for a specific number $n_o$ and $b$ not zero.

sub. into (*) we have,

$2n-2n_o+1=p$

$p=2(n-n_o)+1$

$2(n-n_o)+1$ is prime as $p$ is prime and $n\gt n_o$ as primes are positive.

Also,

$q=2(n-n_{oo})+1$ 

where $n\gt n_{oo}$ for positive prime.  This is because if we start over and swap $p$ and $q$ and arrived at an expression for $q$ in the form,

$q=2(n-n_o)+1$

Both $n_o$ and $n_{oo}$ are chosen such that $p$ and $q$ are primes given $n$.

And so,

$q+p=2(n+n-n_o-n_{oo})+2=2(n_j+1)$

where $n_j=2n-n_o-n_{oo}$ and both $q$ and $p$ are prime.

Consider,

$n_j=2n-n_o-n_{oo}=n$

$n=n_o+n_{oo}$

From previously,

$p=2(n-n_{o})+1$

$q=2(n-n_{oo})+1$

 where $n_{o}$ and $n_{oo}$ are selected to make $p$ and $q$ prime, and initially,

$p+q=2n$

 If we apply the condition,

$n=n_o+n_{oo}$

to $p$ and $q$, substituting values for $n$ into the expressions for $p$ and $q$ and create new numbers,

$p_1=2(n-n_{o})+1=2n_{oo}+1$

$q_1=2(n-n_{oo})+1=2n_o+1$

and so,

$p_1+q_1=2(n_{o}+n_{oo})+1=2(n+1)$ 

But are $p_1$ and $q_1$ prime?

Specifically, given,

$p=2(n-n_{o})+1$ is prime,

is $q_1=2n_o+1$ also prime?

Obviously,

$p+q_1=2(n+1)=2n_1$   where $n+1=n_1$

since we started with $p+q=2n$ where $p$ and $q$ are both prime and $n\gt 2$.  $q_1$ can be prime.

Equivalently, given

$q=2(n-n_{oo})+1$

$p_1=2n_{oo}+1=2(n+1)-q$

$p_1+q=2n_1$

so $p_1$ can also be prime, because we started with $p+q=2n$ where $p$ and $q$ are both prime and $n\gt 2$.

More importantly, $n_o$ and $n_{oo}$ are chosen such that, $p_1$ and $q_1$ are primes as are $p$ and $q$.

And so assuming $p+q=2n\implies p_1+q_1=2(n+1)$ where $p$, $q$, $p_1$ and $q_1$ are primes and $n\gt 2$. By induction $p+q=2n$ is true when we can start with,

$6=3+3$ where $n_o=n_{oo}=5$, $p_1=7$ and $q_1=7$

We start with $n=3$ and so $2n=6$.

This proof, does not suggest a way to find the next pair of primes such that,

$p_1+q_1=2(n+1)$

but it proves that such a pair exist provided $n_{}o$ and $n_{oo}$ can be found.

Unfortunately, the proof does show a way and it is haywire afterwards!  The expression, $p_1+q=2n_1$ suggest that one of the prime is retained to move on to $n+1$.  But as stated, $p+q=2n$ does not differential between $p$ and $q$.  Easily, we end up with $p=q$ and so, $n_{o}=n_{oo}$.

Do we spiral out of control?  If this prove is true, hopefully, depending on where the start is, to infinite prime!

Good morning.  

Unfortunately Goldbach's Conjecture

 The previous post on Goldbach's Conjecture is unfortunately falsehood, a simplified 'proof' but on the same line of argument is,

All primes are odd, let,

$2n_1+1=p$ 

and

$2n_2+1=q$

so $p+q=2(n_1+n_2+1)=2n$

where $n=n_1+n_2+1$.

The actual proof needed is that $n$ spans all integer greater than $2$.

哥德巴赫想太多

Consider,

 $2n=p+q$

$2n-q=p$

let $q=\cfrac{a}{b}$

$2n-\cfrac{a}{b}=p$ ---(*)

$\cfrac{2nb-a}{b}=p$

If p is prime,

$b=2nb-a$ or

$b=1$

When $b=1$,

$2n-a=p$ is prime

we have instead,

$2n=p+a$

so,

$a=q$

interesting result, $\rightarrow$备用

or from $b=2n_ob-a$

$a=2n_ob-b=b(2n_o-1)$

$\cfrac{a}{b}=2n_o-1$

for a specific number $n_o$ and $b$ not zero.

sub. into (*) we have,

$2n-2n_o+1=p$

$2(n-n_o)+1$ is prime as $p$ is prime and $n\gt n_o$ as primes are positive.

Let $2(n_i-n_{oo})+1=q_i$ where $n_i\gt n_{oo}$ for positive prime.

where $n_i$ is paired with $n_{oo}$ to give $q_{i}$ as $n$ and $n_o$ is paired to give $p$.

In which case, we formulate,

$q_i+p=2(n-n_o)+1 + 2(n_i-n_{oo})+1$

and so,

$q_i+p=2(n+n_i-n_o-n_{oo})+2=2n_j$

where $n_j=n+n_i-n_o-n_{oo}+1$ which is just another integer and both $q_i$ and $p$ are prime.

Is $n_j$ strictly non negative?  Yes. And When we set $n+n_i=n_o+n_{oo}$ we start from $n_j=1$.

What conjecture?  Goldbach's.  All yours.  This is not prove of the conjecture.  It proves that the sum of two primes are even.

Note:  The trick is to realize that $n_o$ is not a running variable as $n$.

Volume Pixel Projecting Space

 Making it small,

If the coil at 141988 Hz acting as a variable-focus lens, squeezes the beam to nearer or further focus, then a single pixel can be projected into space at a specific distance.

First a line, then a circle, a disc, and finally a hollow sphere.  3D image immediately.

Bertrand Walking Along A Short Radius

 Walking along a radial line is shorter than walking along the circumference on which the chords originates.  This error is more pronounce where the curve has a high gradient.  This happens at the end of the radius near the circumference where the shorter cords are.  

The situation is the same as the radius rotates, undercounting at each turn does not make up for anything.  As the radius makes a full turn, undercounting all along the way, the final result is still less. Undercounting these short cords results in a higher probability for long chords. 

Ok?

Bertrand Paradox Not

 There is no paradox, any methods conceived to generate chords must generate all chords possible in the circle.  Otherwise a subset of the infinite set of chords possible will give a wrong answer.

Methods that are off the circle center cannot use symmetry to generate all chords with the same required condition, $\bar{c}\gt\bar{a}$.  Focusing on the sides and vertices of the triangle does not provide the symmetry that generate all possible chords as the condition 'any random chords, c' requires.  The triangle has to be rotated also but it does not have circular symmetry.   Choosing any sides or vertices introduces conditional probabilities that must be resolved.

Given any point in the circle, infinite number of chords through it is possible.  Methods that draw only one chord through each point have ignored many others through the same point that meet the required condition, $\bar{c}\gt\bar{a}$.  Setting the chosen point as midpoint of one chord added extra restrain/condition to the problem and reduced the set of chords considered.  

In particular, the center of the circle can have more than one chord through it that is longer than the side of the triangle.  Missing out on these chords, undercounts the long chords and so gives a lower probability for long chords.

OK?

Note:  The rest of the triangle does not matter, only the length $a$ and the parameter that defines $a$; $120^o$ at the center of the circle.

Amplify Space

 What if?

instead of amplifying photons at 15.088 kHz, we amplify space at 141.988 kHz?

Projected Space Hologram

 Maybe space and photons can both be projected,

The end of the space beam is imbued with photons from the screen image on the diaphragm and is a flat free standing image.

If the beam can be precisely controlled,

and only if space carries with it photons from a screen with a transparent vibrating diaphragm, enlarged and focused with a dense transparent lens.

Does space move with photons in them?

Focus, Focus , Focus

 If it is possible to slow the beam using dense materials, maybe it is possible to focus the effects of this longitudinal wave using a lens.  Assuming that the wave fronts, transverse to  its direction of travel, remain continuous lines,

 it is possible to focus the space beam using a dense matter concave lens.

And, using total internal refraction,

due to the dense space layer just above the vibrating diaphragm, a free standing virtual image from a screen image.

Maybe...

Lift Above The CG

 An UFO lifting itself by itself,

An aperture opens below the vibrating piezo-diaphragm to steal cows.

How's that for alien technology.

No Need For High Spin

 Speeding up $\beta$ decay,

using the region of high gravity created using space resonance at $141.988\,kHz$ played over a piezo-electric diaphragm.

Try Not To Kill Yourselves

 Here they are sticky speakers,

SpaceResonance 141987-697 Square Wave 14198-7697 Hz

141987-697 Hz is beyond the audio range but as a 14.1987697 kHz square wave it has integer multiple harmonics including 141987-697 Hz.

SpaceResonance141987-697 14198-7697 Hz

SpaceResonance141987-697 1419-87 Hz

Do not use headphones to listen to these frequencies.  There are strictly for coils, heating elements, antennae and maybe speakers only.

Do not puncture your ear drum.  Do not send a beam into your ears.  Do not point the beam at shelves, walls, trees, etc. Otherwise, have fun.

四大皆空

 Because all these will trigger resonance in space,

electric, thermal, gravitational AND magnetic.

Which might suggest thermal and magnetic to be fundamentally different.

Maybe each possible Platonic solids, with vertices equally spaced in a bounding sphere, corresponds to a type of energy with a characteristic basic particle of a matching $\psi$.

By equating the volume density equations in space and in time, a resonance frequency for the matching, but yet unknown $\psi$ can be obtained.

There can be more fundamental forces.

Space Full Of It, Purge

 How then to lower down?  Only when the field is first established as it extends slowly towards the ground.

Once the beam touches ground, its direction is up; all objects in the beam is attracted up.

If another resonator at $f_{res}=1.41988e5\,Hz$ but at phase $\pi$ (at anti-phase), is on the object, maybe this will cancel the beam locally and so drop the object down the beam.

What goes up must come down SLOWLY...

Good morning.

Space Full Of It

 In the post "Crop Circles and Gravity Wave Antenna" dated 31 Jul 2014, a resonance frequency for space was calculated by considering a black hole made by shrinking Earth,

$f_{res}=1.41988e5\,Hz$

This seems to be a very dangerous number!  It is so because, space is everywhere.  

What if energy is added to space incessantly, unceasingly?  A space beam for?

Such a beam will always have higher space density at the source and so greater gravity there.  Subjected to such a beam, objects will be moved towards the source.

A tractor beam!  Steal cows anyone?

No Need To Run

 If the collapsing field carries with it all matters in it,

then Bon Voyage.

Rat Bite Fever And Myopia

Also from rats,

Rat Bite Fever 920 Hz

Rat Bite Fever 722-53 Hz

Rat Bite Fever 697-50 Hz

Rat Bite Fever 562-50 Hz

Rat Bite Fever 137-50 Hz

Rat Bite Fever 85-31 Hz

Rat Bite Fever And Breaking SS 85-31 177-979 1779-79 Hz

Rat Bite Fever And Breaking SS 85-31 137-50 177-979 1779-79 Hz

and additional frequencies,

Rat Bite Fever 8531 Hz

Rat Bite Fever 853-10 Hz

Rat Bite Fever 3512 Hz

Rat Bite Fever 351-20 Hz

Rat Bite Fever 950 Hz

Rat Bite Fever 550 Hz

Rat Bite Fever 170 Hz

How common is rat bite?