77656e No.12463 [View All]
The virtual quantum computer (VQC) is a grid made of constructably infinite elements that follow a known pattern. It is indexed using e, n, and t, where e is the column, n is the row, and t is the specific element in the cell.
The grid in its entirety serves as the superposition. The required input parameters to collapse the superposition are d and e, which are trivial to calculate for all c that is the difference of two squares. When the integers that are the difference of two squares are arranged into the grid and their corresponding properties are shown, a pattern emerges that shows a path to calculate the factors of c instead of searching for them. It can be understood using only the basic operations of arithmetic and sqrt. All currently-known patterns can be found within one thread here >>6506
C# VQC generator - pastebin.com/XFtcAcrz
Java VQC generator - pastebin.com/2MPYrJVe
Python VQC generator - pastebin.com/NZkjtnZL
Glossary
Look-up
A pattern used to calculate the factors of c, like a value look-up table.
Column
All cells for a given e
Row
All cells for a given n
Cell
All entries for a given e,n (not to be confused with an entry itself.)
Entry; record; element
A set of variables corresponding to a factorization for a given c. The legend to read entries is {e:n:d:x:a:b} (e, n, t) = c
Example: {1:5:12:7:5:29} (1, 5, 4) = 145
ab record; nontrivial factorization
The element that contains the factorization of c that is not 1*c, hence, nontrivial.
1c record; trivial factorization
The element generated from setting a=1 and b=c
Mirror element
The element in -f corresponding to an element in e, in the context of a given c.
Variables
a and b are, to reiterate, the factors of c. a is the smaller factor of c, and b is the larger one.
d is the integer square root of c.
e is the remainder of taking the integer square root of c. Unless c is a perfect square, a remainder will be left over.
i is the root of the large square. It is equal to (d+n).
j is the root of the small square. it is equal to (x+n).
n is what you add to d to be exactly halfway between a and b, and it is the root of the large square, so it takes you from d to the large square.
x is what you add to a to make d. When added to n it makes the root of the small square.
f is what you add to c to make a square. (e is what you subtract from c to make the square below it, f adds to make the square above c.)
t is the third coordinate in the VQC, it is a function of x.
q is a product created by multiplying successive primes until the product is above d.
u is the triangle base of (x+n)^2. 8 times the triangle number of u plus one is (x+n)^2 for c with odd x+n.
ab = c
dd + e = c
(d + n)(d + n)-(x + n)(x + n) = c
a + 2x + 2n = b
a = d - x
d = a + x
d = floor_sqrt(c)
e = c - (dd)
b = c / a
n = ((a + b) / 2) - d
d + n = i
x = d - a
x = (floor_sqrt(( (d+n)*(d+n) - c))) - n
x + n = j
j^2 = 8*T(u) + 1
f = e - 2d + 1
u = (x+n) / 2
if (e is even) t = (x + 2) / 2
if (e is odd) t = (x + 1) / 2
Past threads
RSA #0 - archive.fo/XmD7P
RSA #1 - archive.fo/RgVko
RSA #2 - archive.fo/fyzAu
RSA #3 - archive.fo/uEgOb
RSA #4 - archive.fo/eihrQ
RSA #5 - archive.fo/Lr9fP
RSA #6 - archive.fo/ykKYN
RSA #7 - archive.fo/v3aKD
RSA #8 - archive.fo/geYFp
RSA #9 - archive.fo/jog81
RSA #10 - archive.fo/xYpoQ
RSA #11 - archive.fo/ccZXU
RSA #12 - archive.fo/VqFge
RSA #13 - archive.fo/Fblcs
RSA #14 - archive.fo/HfxnM
RSA #15 - archive.vn/59GwR
RSA #16 - archive.vn/F49fw
RSA #17 - archive.vn/u2Tu6
RSA #18 - archive.is/FDVP9
RSA #19 - archive.is/6mJhe
____________________________
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996f43 No.12795
Hello everyone.
I've been working on the sheet to calculate the multiples of (c).
I had an intuitive thought, just jotting it down here for us.
Ok, so here’s the idea.
When we skip left by 2n, we have d+1, a+1 , b+1
N = (1+c) / 2 - d
So as we skip left by 2n, N increases.
What happens when we skip to the place where N or (N-1) = c in the grid?
Logically, one of these should be a location where we can measure the change in (e) and (f), which could provide a solution.
I'll run some calcs and see if anything pops.
"When you (c) it"
So for example, for c145
if N = 145, we know that N = (1 + c) / 2 - d
N has grown from 61 to 145, delta is 84
so d + 84 = 12 + 84 = 96
a + 84 = 5 + 84 = 89
b + 84 = 29 + 84 = 113
c = 89 * 113 = 10057
Hmmm.
It would work better with (N-1) = 60
Thinking out loud here.
When we skip by 2n, big N seems to increase by 1 each step?
I guess my current question is:
What happens if there is a valid location where N or (N-1) = c ?
It would be easy to use the formulas to find c, d, e, and f.
Would this be a valid location to calculate the change in (e) and (f) such that (n-1) could be calculated?
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996f43 No.12796
"How does row 1 (n=1) relate and determine the patterns?"
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996f43 No.12797
The point of the (x+n)^2 explorations was to help us understand the movement of (c) in the Grid.
When we move 2n left, we know that we get d+1, a+1, b+1
The main point we learned was:
(e) changes by 2n
(f) changes by 2(n-1)
The only problem is that we don't know what the series of (c) values is.
So what if we set N or (N-1) = c ?
Can we get a valid (c) working backwards with our formulas?
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5b139b No.12798
>>12796
>"How does row 1 (n=1) relate and determine the patterns?"
Many thanks for that crumb (the Cap), helpful.
This has been focus: movement. With that focus has come much clarity. Still enumerating..
>>12797
>Can we get a valid (c) working backwards with our formulas?
Have reviewed your posts. A question would be how would you know a 'valid c'? Is there a criteria you have in mind? Something Mod..?
>>12788
>If a cell contains an element c, another element in the cell can be constructed from it.
>Call it c'
..
Have done the construction(s).
Working to enumerate, generalize and thus generate for any 'k' in a sequence (so not a recursive construction). Have nN cells for all odd 'c', but bit of work left with 'd' in other record types (construction is fine, just the enumeration for calculating the A, B, C quadratic coefficients directly).
Not only does 'x' grow by 2n, but d, a, b, i, have quadratic form in same manner as elsewhere.
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a14312 No.12799
>>12798
Hello MM!
>how would you know a 'valid c'? Is there a criteria you have in mind?
Yes, great question!
The next (c) values we need are the series of (c) values that are created with the 2n movement in (e) and (f).
They make a continuing chain of growing (c) values.
So for example, (c) 145, just for ease of explanation with a small value.
{1:5:12:7:5:29}
d = 12
a = 5
b = 29
e = 1
f = 24
Next (c) in the series would be:
d +1 = 13
a + 1 = 6
b + 1 = 30
c = 180
f = 16
{11:5:13:7:6:30}
delta e = 11 - 1 = 10 = 2n
delta f = 24 - (26 + 1 - 11) = 8 = 2(n-1)
Next (c) in the series would be:
d + 1 = 14
a + 1 = 7
b + 1 = 31
c = 217
f = 8
{21:5:14:7:7:31}
delta e = 21 - 11 = 10 = 2n
delta f = 16 - (28 + 1 - 21) = 16 -8 = 8 = 2(n-1)
So, if we can use the Grid rules to find c217 (or another c in the series), we could solve, by finding (delta e) and (delta f).
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6f8116 No.12800
Hello everyone.
Digging through old breads for hints.
Found a good one.
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6f8116 No.12801
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6efed2 No.12802
>>8433
>Every column that c appears in has what in common?
factors of (a) and (n)
>How many columns do you need to create a lookup?
Not that many. Just the right one.
>What specific information does that set of columns build up?
factors of c, multiplied in a new way that could allow for a simple GCD calc.
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6efed2 No.12803
>>12802
>factors of (a) and (n)
updated:
factors of (a), (b), and (n) in each of these new e column locations.
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6efed2 No.12804
>>8437
>One of the most important prayers that exist is when two or more gather in His name.
Amen.
Thank you Jesus is Faith that this problem is solved.
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6efed2 No.12805
So, once we c i[t] all those factors are there.
>factors of (a), (b), and (n) in each of these new e column locations.
With e=290, a[1] = 145
We can then skip by [t + c * m.] to find a whole list of values with factors of a, b, n based on original (c).
These new a[t] values should be compared against original (c) using GCD.
So, the big idea is that moving to a related (e) column gives us the opportunity to explore a bunch of new a[t] values that have the factors of a, b, n, and c that we are looking to factorize.
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6efed2 No.12806
here's the [t + c * 1] value
a[146] = 42195
factors are 5 * 29 * 3 * 97
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83012a No.12807
>>12805
>With e=290, a[1] = 145
>We can then skip by [t + c * m.] to find a whole list of values with factors of a, b, n based on original (c).
>>12806
>here's the [t + c * 1] value
Great connection VA!
Will take a look at that scenario, including how the Big N's in the column relate.
>>12798
> bit of work left with 'd' in other record types
Completed enumeration and have working. It was related to 'a' values of '1' for enumeration representing n=1, and generalizing now works for any 'n'.
Interesting relation for t[c^2]: d=c.
Exploring Mods, Exponent, factor relationships.
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37b79d No.12808
>>12807
Hello MM! Thank man, just tending the rows and columns over here at VQC Farm.
>>6470
>It's a single key difference between cells in (-f,1) and (e,1)
Intuition is kicking in.
It's gotta be the actual construction method for a[t] values.
In (e,1) we know that (x ^ 2 + e) / 2 = a[t] = some value of an, with factors in all other n rows within that given (e) column.
In (-f,1) we know that (x ^ 2 - f) / 2 = a[t] = some value of a(n-1) with factors in all other n rows within that given (-f) column.
Given this hint >>8433
It seems fun and logical to find what (e) columns contain the following (c) values in different x locations…
(e,1) a[1] = 145
(e,1) a[3] = 145
(e,1) a[5] = 145
(e,1) a[7] = 145
…
So we hold c constant, and change x to solve for (e).
Same thing in column (-f)
We can generate a list of the (an) and a(n-1) pairings
We can also generate a list of (bn) and b(n-1) pairings
My intuition says that this will unlock so many factor combinations of a,b, and n that a simple GCD or direct calc may be possible.
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c18096 No.12809
Hello everyone!
Here's my first sheet exploring the idea hinted at here in >>8433
The (e) columns at the top are newly discovered locations where we can find c = 145
Notice that now we have a lot more known e columns to explore where c = 145 appears.
Next step will be to calculate the (-f,1) a(n-1) values.
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c18096 No.12810
Some of my labels were a bit confusing.
Here's a better update.
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d3ea2f No.12811
>>12808
>Hello MM! Thank man, just tending the rows and columns over here at VQC Farm.
#MeToo
>>12808
>Intuition is kicking in.
good.
>>12808
> .. construction method ..
The construction is elemental.. ha.
Ok, will take a look at your sheet, seems plenty clear enough first look.
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c18096 No.12812
"AA anon is getting very close"
>>6470
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c18096 No.12813
Here's AA's post before Chris commented.
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c18096 No.12814
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c18096 No.12815
To clarify, here's AA's two posts before that comment from Chris….
>>6464
>>6465
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c18096 No.12816
AA's comment here >>6465
(As I work to follow the main ideas)
Was essentially that between the column value of (1,c) and the row one value of (1,c), the (a) and (n) values swap places.
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c18096 No.12817
He also brought up xx + e = 2na again
We have not used that formula to hold (c) constant while changing x values to solve for new e values.
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894ec0 No.12818
>>12811
Hello MM!
Agreed, seems pretty clear at first look.
Until we realize we can utilize the formula to hold (c) constant, change x, and generate a new list of (e) values where c appears.
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894ec0 No.12819
Which would then unlock a new level of understanding regarding (a) and (n) values and how they swap based on (x^2 + e) / 2 = an and (x ^ 2 - f) / 2 = a(n-1).
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d3ea2f No.12820
>>12809
(1,1,t,9) {1:1:162:17:145:181:26245:163:18}
(-324,0,t,9) {-324:0:163:18:145:181:26245:163:18}
Getting to e=34.. Go first to e=36 (6^2), hop back 2n (1 jump):
(34,1,t,11) {34:1:161:16:145:179:25955:162:17}
(-289,0,t,11) {-289:0:162:17:145:179:25955:162:17}
Getting to e=65.. Go first to e=49 (7^2), take 8 2n jumps right:
(65,1,t,12) {65:1:160:15:145:177:25665:161:16}
(-256,0,t,11) {-256:0:161:16:145:177:25665:161:16}
94: 100-6 (3 jumps L):
(94,1,t,12) {94:1:159:14:145:175:25375:160:15}
(-225,0,t,12) {-225:0:160:15:145:175:25375:160:15}
121 just 11^2 no jumps:
(121,1,t,12) {121:1:158:13:145:173:25085:159:14}
(-196,0,t,12) {-196:0:159:14:145:173:25085:159:14}
146: 12^2 + 1 2n jump R:
(146,1,t,12) {146:1:157:12:145:171:24795:158:13}
(-169,0,t,12) {-169:0:158:13:145:171:24795:158:13}
(169,1,t,12) {169:1:156:11:145:169:24505:157:12}
(-144,0,t,12) {-144:0:157:12:145:169:24505:157:12}
..
289 is interesting. 3 constructed rows on 'e' side:
(289,1,t,7) {289:1:146:-3:149:145:21605:147:-2}
(0,0,t,8) {0:0:145:0:145:145:21025:145:0}
(289,1,t,9) {289:1:146:1:145:149:21605:147:2}
a=b=145
f side, same 3 rows:
(-4,0,t,7) {-4:0:147:-2:149:145:21605:147:-2}
(-291,-1,t,8) {-291:-1:146:1:145:145:21025:145:0}
(-4,0,t,9) {-4:0:147:2:145:149:21605:147:2}
at 290, getting there with 16^2 = 256, + 17 2n jumps R:
The negative x row as well:
(290,1,t,7) {290:1:145:-2:147:145:21315:146:-1}
(290,1,t,8) {290:1:145:0:145:147:21315:146:1}
and f:
(-1,0,t,7) {-1:0:146:-1:147:145:21315:146:-1}
(-1,0,t,8) {-1:0:146:1:145:147:21315:146:1}
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894ec0 No.12821
>>12820
Dang brother!
Good job on the fast calcs.
So, confirmed that c=145 exists at all those locations.
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d3ea2f No.12822
>>12821
>Good job on the fast calcs.
Ha, just semi-auto.
Here's the construction:
For the First movement (using the Square value):
a=(k-m)^2+(k)^2
Where 'k' acts as the 't' value, and 'm' is the squared.
m=1 puts in (1,1)
m=0 in (0,1).
Actually, given 0 and 1 cover all 2n jumps to Even or Odd 'e' values, could just use 0 or 1 for 'm', and jump 2n accordingly (1/2 distance).
So let's do e=190 and 209
m=0 and 95 2n jump R lands on e=190.
(190,1,t,5) {190:1:155:10:145:167:24215:156:11}
Now that element in e=190, is the same row in e=1 (and over by the 95 2n jumps):
(0,1,t,5) {0:1:60:10:50:72:3600:61:11}
'i' is the nN for 145
Ok, 209, we can start at e=1, 104 2n jumps right:
(209,1,t,5) {209:1:154:9:145:165:23925:155:10}
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d3ea2f No.12823
>>12821
>So, confirmed that c=145 exists at all those locations.
And where else would 145 exist in the 'a' position?
▶ VQC!!/aJpLe9Pdk 07/07/18 (Sat) 03:11:47 cf67c1 No.6736>>6737
If a cell contains an element c, another element in the cell can be constructed from it.
Call it c'
e' = e
n' = n
x' = x + 2n
a' = b
..
a' = b
What is 'b' for nN?
b = c
For the Second constructed Element in the nN cell, a'=c.
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9d5b01 No.12824
One thing to clarify:
Not sure if this new method allows us to jump by 2n.
Maybe Chris sent us down this path to look right towards e = 2c.
When we jump by 2n right, we get a+1, b+1, d+1, which would mean we're no longer at a[t] = 145
This new method uses the powers of x^2 to highlight all possible values of (e) between e= 2c and the origin.
My intuition says that this is the place for the (-f,1) a[t] values, which you've already provided!
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9d5b01 No.12825
However, we do have e = 121, which is 10 * 2n away from e=1 ??
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9d5b01 No.12826
correction.
12 * 2n away from e=1
delta is 120, = 10 * 12
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d3ea2f No.12827
>>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.
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c646ec No.12828
>>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.
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c646ec No.12829
>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.
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c646ec No.12830
The real question is, does that list exceed O(log n)?
>>6185
Excellent crumb, ctrl F works wonders.
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d3ea2f No.12831
>>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).
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1ed774 No.12832
>>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
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1ed774 No.12833
Bingo.
Here's 35 right at the top of e=61
!!
Obviously need to test moar.
However, there is the correct value of b + 6 = 35
Which would allow us to solve…
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87c39e No.12834
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.
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87c39e No.12835
Another note of interest.
Here's (e,1)
notice the 5.
it's in (b) position in t[1] and t[2] position as a[t].
same as 35 in (61, 1)
35 / 5 = 7 = x = solution x
Just found some cool new shit boys.
May or may not work for larger c values, but here I am finding new avenues to explore
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87c39e No.12836
Another interesting find.
e = N - d = 49
e = 61 - 12 = 49
Hmm.
There's our solution (b) value of 29 right at the top
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87c39e No.12837
Will the helicopters be flying over my house in the countryside again this evening?
Hello White Hats! :)
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87c39e No.12838
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.
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5df49d No.12839
Reviewing old breads this evening, found this gem that PMA wrote. Unfortunately a lot of the images aren't loading, but oh well.
>>5851
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5df49d No.12840
>(n-1) balances out the f center square.
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a308ea No.12841
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
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070895 No.12842
>>12841
Deus Vult
God Wills It.
Love you too Bruddah!
Here's the conversation surrounding PMA's excellent post >>5851
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070895 No.12843
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
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a308ea No.12844
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
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