/vqc/ - Virtual Quantum Computer

However, we do have e = 121, which is 10 * 2n away from e=1 ??

correction.

12 * 2n away from e=1

delta is 120, = 10 * 12

>>12823

>For the Second constructed Element in the nN cell, a'=c.

Note it's the second Element for that series in the cell, not necessarily 2nd element. This is reason for the 'k' index for the SubSeries associated with a starting 'c' and then 2n growth in x for each element.

>>12827

Following!

Can you please elaborate on your notation for k?

My intuition is guiding me toward the matching values in (-f,1).

So we would end up with a chart or Grid of valid e and n values for the positive (e) side of the Grid that represent valid e and n pairings.

And we would also end up with a chart or Grid of valid e and (n-1) values for the (-f) side of the Grid that represent valid e and (n-1) pairings.

>And we would also end up with a chart or Grid of valid e and (n-1) values for the (-f) side of the Grid that represent valid e and (n-1) pairings.

Correction:

And we would also end up with a chart or Grid of valid e and (n-1) values for the (-f) side of the Grid that represent valid (-f) and (n-1) pairings.

The real question is, does that list exceed O(log n)?

>>6185

Excellent crumb, ctrl F works wonders.

>>12828

>Can you please elaborate on your notation for k?

'k' is the Index of the Construction.

Start from initial element, k=0.

1st Element k=1, 2nd k=2, ..

The Elements in a cell that are related, grow in 'x' by 2n. The 'k' index is this sequence.

The "ab_shift" have mentioned before directly relates to this. For n=1, the ab_shift is always '1' (meaning the 'b' value in t, is the 'a' value in t+1). All row elements captured in single sequence (e.g. k=t).

Consider (23,24), using handy grid output you can quickly see it has an ab_shift of '8'.

And (23,26) has ab_shift of '4'.

So in (23,24), each incremental 'k' element is 8 Grid 't' values apart, and similarly there are 4 sub-sequences for (23,26).

>>12831

Thank you MM!

Very clear explanation.

Nice job on figuring out the construction of new elements!

Ok, so this whole swapping e and n thing has me fascinated.

For example, with c145 for ease of understanding…

if we swap bigN = 61 up to the e column, we actually have a match for where one of the e+2n locations would be.

e = 1 ===>>> e = 61

delta is 60, which is 6 * 2(5) = 60

so, 6 steps forward.

new a should be 5 + 6 = 11

new b should be 29+ 6 = 35

new d should be 12 + 6 = 18

new c = 385

{61:5:18:7:11:35}

e = 61

n = 5

d = 18

x = 17

a = 11

b = 35

(x+n) remains 12 = 7 + 5

So quick recap.

We swap bigN up to the e column

and it appears that the value of b + 6 = 35 is right at the top at both t[1] and t[2].

This is the value of b that we need to solve the 2n jump idea (i believe)

also, a GCD could be a possible solution.

(61, 1, 1) is 31 * 35 = 1085

GCD 1085 and 145 is 5, which is solution (a) and (n)

Bottom line: the number we need (35) is right there.

Flipping N and e seems pretty cool to me.

Very intuitive and yet, very counter intuitive.

Simple.

I like simple explanations for complex ideas.

What a long, strange journey we've been on, Anons.

Hopefully this new idea can finally bear fruit.

Reviewing old breads this evening, found this gem that PMA wrote. Unfortunately a lot of the images aren't loading, but oh well.

>>5851

>(n-1) balances out the f center square.

Mannnnnn, I just want to say.

I fucking love all of you so much.

It's ridiculous how much I love and appreciate you all. But you're, we are, some of the most amazing human beings to ever Be Here Now.

Make sure all of your loved ones know you love and appreciate them today.

Blessssss.

Deus Vult

Just realized that (n-1) itself could be a construction of two polite triangle numbers.

Visualize with me…

At the center of the (x+n)^2 area…

We leave an open space for (n-1) in the middle

f (or f-1) surrounds the (n-1) space.

4 * T(3) + 4 * T(2) - 1 = (n-1)

4 * 6 + 4 * 3 - 1 = 35

Total inner portion is a perfect square,

134 + 35 = 169

Feels like the bookshelf in Interstellar. Something about Love (which is also Truth, who together with Truth is G-d) is the only thing that transcends both time & space (simultaneously).

Something about a trip, and a vision, and a crystal globe, and a rose bush named Love in the midst of a Fallen Pine.

All while a good friend stepped on a peace of glass.

Blessssss

Today is the day.

Multiply c by 4.

Take the new e',1 at 4 x c.

That cell at new e',1 will have a value that is double a value at e,1.

The value at e,1 is na or (x^2+e)/2 and the value in e',1 is 2na.

If we have two cells at e,1 and e',1 how do we find values that are twice the other?

>>12843

>>12844

Have fun.

If you need any help with finding the double value between cells, I'll be back later.

Let's do the biggest RSA number together?

End of an era, beginning of a new one.

As my brother from another mother once said, and still says…

When you gon' drop that verse? Nigga, you taking long"

WHEN U GON' DROP THAT EQUATION? NIGGA U TAKIN LONG.

So now I'm back spittin' that heat, could pass a polygraph

That Reverend Run rockin' Adidas out on Hollis Ave

That FOI, Marcus Garvey, Nikky Tesla

I shock you like an eel, electric feel, Jay Electra…

[Verse 3: Jay Electronica & Just Blaze]

They call me Jay Electronica—fuck that

Call me Jay Elec-Hanukkah, Jay Elec-Yarmulke

Jay Elect-Ramadan, Muhammad as-salaam-alaikum

RasoulAllah Subhanahu wa ta'ala through your monitor (Ooh!)

My Uzi still weigh a ton; check the barometer

I'm hotter than the motherfucking sun; check the thermometer

I'm bringing ancient mathematics back to modern man

WE BRINGIN' ANCIENT MATHEMATICS BACK TO MODERN MAN

My mama told me, "Never throw a stone and hide your hand"

I got a lot of family, you got a lot of fans

That's why the people got my back like the Verizon man

I play the back and fade to black, and then devise a plan

Out in London, smoking, vibing while I ride the tram

Giving out that raw food to lions disguised as lambs

And—by the time they get they seats hot

And deploy all they henchmen to come at me from the treetops

I'm chillin' out at Tweetstock, building by the millions

My light is brilliant..

OUR LIGHT IS BRILLIANT.

Outro: Just Blaze]

Woooo!

I rest my case

'09, Act 3

First chapter of the end

The last chapter of a new beginning

If it's so

The things we do without even trying…

>>12846

First Chapter of The E.N.D.

Last Chapter of a New Beginning.

Just Blaze.

Sheeeeeeit, baked.

Blessssss 🙏

>>12849

Blessings to you & your mind, and your family. Let us take care of those we love and make sure they know that we cherish them and hold them near to our hearts.

Blessings to All. Past, Present, & Future.

And now, as our savior taught us, we are bold to say:

Our Father, who art in Heaven,

Hallowed be Thy Name…

>>12851

I only thought to check variables, now I'm checking gaps between variables within elements and between adjacent elements. It does exist as a gap between adjacent d values here:

(27,1,32) = {27:1:2061:63:1998:2126}, f=-4096, c=4247748, u=32, i=2062, j=64

(27,1,33) = {27:1:2191:65:2126:2258}, f=-4356, c=4800508, u=33, i=2192, j=66

2191-2061=130, which is 2na. The higher of the two x values is na. I don't know if that actually matters though because for every element in (e,1) the gap between two adjacent d values is two times the higher element's x value, so that's implied algebraically. I still don't see anything useful about these elements that would help us find them, and all the patterns that involve gaps between d values in (e,1) either involves taking away our original d value from d[t] or d fitting between the two d values, so I don't know if any of this is useful. Also you would need to know unknowns to find these elements regardless of anything else.

The only way I can make sense of this post is if the 2na he's talking about in (e',1) is the gap between d values. Can't find any use for this or patterns that could be used to find these elements though.

>>12852

By the way, pretty sure today is the 7th anniversary of the "my hand is forced" post, what better time to finally post the fucking solution already

>>12855

Should clarify, in that first linked post he specifically mentions "gaps between the square remainders" (i.e. square remainders are e so maybe relevant to gaps between d values, which is the only place 2na exists in (e',1) that I could find other than forcing it as a t value or something) and the fractal thing is relevant given the fractal nature of one element's squares being made up of double another's variables and so on upwards as you multiply by 4. I think these latest two posts are related to the n0 triangles.

>>12857

Thanks AA, reviewed your output in detail.

{30:5:23:10:13:43} c=559

I think the only advantage of multiplying c x 4 is that we get more n locations?

>Still doesn't really seem like anything is obviously relevant.

Yeah, not sure exactly how this helps.

Doing work here is a welcome daily break when IRL duties are done, so I'll just keep thinking and working. Nice to have a hobby I enjoy, and friends to work on it with.

Huh.

I don't know what to make of the new crumb.

At face value, it doesn't take any advantage of the Offset between (-f,1) and (e,1).

That's fine, since I love being here and this is my happy place lol.

Also, a couple finds…

I found 4(an) but not 2(an)

For 4 x c(6107), the 4(an) element is located at (92,1,48) with 4(an) being 4464 = 1116 * 4

I also found 1/2 * (an) = 558 at the location (92, ,1, 17)

Not sure what this means.

Just another day tending the rows and columns at the VQC farm.

"In all labor, there is profit. But idle chatter leads only to poverty."

Proverbs 14:23

AA you lurking?

Did we actually find solutions for every semiprime c up to 1M?

I recall that you said we did, using the binary idea.

For example, if the a[t] values in (e,1) are even, we moved to (-f,1) to use the odd a[t] values to look for the binary tag.

Same with this multiply c x 4, 9, 16 idea.

We need a column where a[t] is odd.

For example, in c6107 the (e,1) column is even a[t] values

But when we move to (-f,1) then we have odd a[t] values to examine.

So we need movement to the place where a[t] values are odd.

>>12856

> gap between squares and remainders

xx+e / 2 = an

(square + remainder) / 2 = an

This is why there is a flow, or repeating pattern of remainders in the a[t] column for any (c) value.

The squares cycle, but the remainder is fixed based on (e).

A wave pattern.

That is embedded in the data.

Perhaps by moving to 4c we can observe this wave pattern more easily.

I haven't tried a binary analysis on 4c yet.

There should be 4x as many binary tags for (a)

There should also be 4x as many (n) locations

The key is that we need a multiple that delivers odd a{t] values, as that is where our prime value for (a) will be found.

Whoops. Copied part of my last post, here's the new part.

The wave pattern is (x squared + e) / 2

It's the part that doesn't fit that's left over.

For example, 2975

101110000000 = 2944

011111 = 35

101110011111 = 2975

So what happens with this binary stuff when we multiply to a column that has MORE of the (a) value we need?

It just has to be a (c) multiple that gives us odd a[t] values I think.

>>12860

>Did we actually find solutions for every semiprime c up to 1M?

>I recall that you said we did, using the binary idea.

I don't remember where it stopped (the fact that it stopped alone shows that it's not the solution) but it was inconsistent and it would be unpredictable which variable appeared where. A year or two later (I've lost track of time in this context) I wrote an entirely new program that checked the same thing except I tried to see if every possible combination of parities among the variables and other types of groupings brought out any patterns and there weren't any. If the solution is based on finding an unknown in the binary of a known, I've exhausted every possible way of doing that that I could and I didn't find a way to do it that wouldn't require waiting thousands of years for bigger numbers (since it would be O(n) where n is the length of each variable in bits, rather than O(log n) (and even then it wouldn't find a solution in all cases)).

>For example, if the a[t] values in (e,1) are even, we moved to (-f,1) to use the odd a[t] values to look for the binary tag.

I was already just generating every known and every unknown and checking all the unknowns (including recursively through the factors of d, e and f) to see if they existed in the binary of the knowns, so I already did that.

>>12866

>Glad to be here with you all.

Just checked in - guess will work a bit this eve.

>>12853

>7th anniversary of the "my hand is forced" post, ..

and just over 6 yrs:

>>8176 (VQC in #14)

>Last post.

ALL indications are that was the last post by Chris.

>>12864

>Hello AA. I appreciate all the effort you've put into our quest. Thank you.

SAME!

>False.

>(upon retrospection)

Same same.

>>12859

>I found 4(an) but not 2(an)

>For 4 x c(6107), the 4(an) element is located at (92,1,48) with 4(an) being 4464 = 1116 * 4

>I also found 1/2 * (an) = 558 at the location (92, ,1, 17)

Confirmed those. Interesting relationship.

>>12855

>Multiplying c by four is the same as multiplying a by 2 and b by 2

Right, and for 9c, a, b * by 3 and 3.

So the "trivial" "Big N" (as '1' will produce half-integer n with 0.5 remainders for 4c, ..) for c9 would use a=2, ..

One thing will be reviewing more, is the difference in n=1 between the "d_min" rows, above and below.

To review "d_min" is the minimum 'd' value for a given 'e' column. As e grows, so does d_min.

Consider e=23. The minimum d=12.

d^2+e = c = 144+23 = 167

Record (23,1,6) {23:1:83:11:72:96:6912:84:12}

Here starts perhaps a unique set of BigN, where each an relates to an N where a=1.

For t=1:5, the 'c' are valid, but the a,b are not of simplest form.

So t=5:

(23,1,5) {23:1:61:9:52:72:3744:62:10}

a=52, b=72

=2^2*26*36 = 2^4 * 13 * 18

In 'f', 'c' and 'i' again positive at t=6.

We look at the "related" 'c' values in the 'e' column at d_min and above.

d increases by 2 for each successive row. For e=23, d is even.

Also noted that the 'd' for the related c = 2*t.

Consider trivial case for c=6107.

t=39: *2=78 = d.

for the 6107 solution row, related c=2327:

t=24, related c is 2327 with d=48.

It has a BigN: (2327+1)/2-48 = 1116

BigX "happens" to be 47 (solution x for 6107).

(23,1,24) {23:1:1163:47:1116:1212:1352592:1164:48}

(23,1116,0) {23:1116:48:47:1:2327:2327:1164:1163}

And using that element as the BASE for the sub-series Generator, and looking at gcd patterns for 6107 (instead of 2327 as would typ do), 31 and 197 emerge:

b for:

(23,1116,-2) {23:1116:4324:-4417:8741:2139:18696999:5440:-3301}

= 2139 = a for:

(23,1116,-1) {23:1116:-46:-2185:2139:1:2139:1070:-1069}

gcd = 31.

at t=14, d and i have gcd = 31

(23,1116,14) {23:1116:470084:31295:438789:503611:220978967079:471200:32411}

Which is right in middle of repeated t=-2 and t=-1 pattern:

(23,1116,29) {23:1116:1944614:64775:1879839:2011621:3781523609019:1945730:65891}

(23,1116,30) {23:1116:2078628:67007:2011621:2147867:4320694362407:2079744:68123}

gcd(2011621, 6107) = 31

the 197 first appears at t=51 (b) and t=52 (a):

(23,1116,51) {23:1116:5924106:113879:5810227:6040217:35095031899259:5925222:114995}

(23,1116,52) {23:1116:6156328:116111:6040217:6274671:37900374443607:6157444:117227}

where gcd(6040217, 6107) = 197

also at 52, mod 'x':

mod(116111, 6107) = 78

>>12869

>What was interesting, is the 4*c is another way to quickly ID the d_min

Incorrect statement. Meant to simply show with the base calcs (c*1) that e=23 starting with t=6 (d_min = 12, c=167)

>>12868

>So the "trivial" "Big N" (as '1' will produce half-integer n with 0.5 remainders for 4c, ..) for c9 would use a=2, ..

just read that, which states the a for c0 a=2, should state 3, as a=2 is for c*4

In case c*9 a=3, c*16 a = 4, …

*bantz lol.

Too tired from all full day of work in the field.

Autocorrect obviously couldn't help.

*a full day

Pretty beat over here.

Glad to check in with you my bruddahs.

Going to read some crumbs.

I'll check the board while I work if anyone wants to hang out.