Wasted a lot of time on (e,n).
I think that the solution is back on (e,1)
Exploring relation between (2c,1) and (2Nc,1)
I'm sure there is value in (e,N) but it take massive iteration to find records. I have not found a computational pattern.
Looking at VQC's hints they remain focused on (e,1) on (n,0) and on factor trees
a(t=1) equals c ….
for (2c, 1) and 2c-1,1)!!!!!!!!!!!!!!!!!!
.
The a(2c-1, t) factor tree has twice as many a,b as factors as does a(2c, 1, t) tree.
I'm doing side by side looking at different multiples of e = 2c and (2c-1)
Very impressive!
Deleted post for first time found myself in alternate universe full of deleted posts. Found a VQC post and got excited until I realized it was a deleted post!
I thought a lot about this hint. In row one n = 1 so na is just a. There was also a hint about subtract d from a to get n-1. Well at t=1 a = d and n=1 so we get a=d=n=0. If we subtract 2d we get a n = -1.
I was looking at some of the records you were generating. Didn't follow all of them but they tested. Do you think you give me a couple (30, 98) records
theoretically we know where the aa and bb records are. I've looked at odd next to even. I've looked at 2Nc versus 2c records but if you just know c the a and b remain hidden. I think Chris has some magic sauce that will seem easy when someone figures it out.
Ghost in our machine??
For e=2c-1 and factor tree A of (e,1, t) and c=a*b
Using A for value of a in tree because its bigger than c=a*b
a is factor of A when t/a is integer or has remainder 1
b is factor of A when t/b is integer or has remainder 1
c is factor of A when t/c is integer or has remainder 1
here's the weird part when a is a factor of t but b has a remainder of 1 (or vice versa) c is
still a factor so c is a factor more often than a or b
When we square c there are 5 possible factorizations. They all have different n's but d=c so f is the same for all 5 records including target record {0,n,c, c-aa, aa, bb}. We use f to move to between n's.
If we look at records with different f like moving down the (e,1) factor tree we have to be careful that we know how to get back to original f.
all the e's equal 1?
I have a couple of questions regarding your and other transformations I see all over this board.
Some review first.
We start with c Looking to find a and b such that a*b =c
Original record {e, n, d, x, a, b}
VQC suggested squaring c.
New record {0, (cc+1)/2-c, c, c-1, 1, cc}
Are a and b factors of new c?
Questions about your transformations
Is original c a factor of new c's?
Are both original a and b factors of the new c's?
Are we looking to determine a and b?
If a or b or both aren't factors of new c how do we find a and b????????????????
If c isn't factor of new c is transformation legit???????????????
>>2687 (You)
After testing it turns out that
(e = 2c, 1, 1) = {2c, 1, c, 0, c, (c+2)}
(e = 2c-1, 1, 1) = {2c-1, 1, c+1, 1, c, ((c+1)^2+2c-1)/c}
for both new records a of (e,1, 1) = c
So new.c = old.c * new.b
Is old.c a factor of new.c??
Are these transformations legit??
For (0, n) when n= 2mm for integer m there are multiple chains the rest of (0,n) you can develop from the first record.
These are records which a and b are perfect squares!
I tried your formula on some (e,1) records based on various combinations of c and N.
Unfortunately those records have already been solved. It just confirms what we already know. Your formula doesn't get us to the goal record. At least so far.
I can beat it. But I need to buy a computer and some software. Any recommendations for a windows fag.
Saw something in factor trees yesterday that wasn't there. Have to be careful doing math and weed together.
I thought factor trees worked because factors were seeded into the tree and would pile up at their appointed times . It doesn't work like that.
The algorithm for generating the factor trees is simple to follow. The initial record for the a b generating factor tree is {2c, 1, c, 0, c, c+2}
generating new records is simple
next.d = d+4t
next.x= x+2
next.a = b or next.d-next.x
next.b = (next.d*next.d + e=2c)/ next.x
Somehow this algorithm that knows nothing about a or b produces records containing a and b as factors without c many times more often than c records.
Somehow the knowledge to produce records with a and b factors exists in the initial record cause it isn't in the algorithm. Ghost in the machine!
Nice!
If I get any more [D]umb [M]oron ideas I'll be sure to share them!
Banged through a bunch of e=n records.
Discovery!
Guess z in c terms?
a of (z, z, 1) = c if c semi prime
Interesting records to investigate:
For c = a*b prime
(2c-1, 2c-1, 3c-1, 2c-1, c, b)
(2*c-1, 1, c+1, 1, c, b)
(2*c, 1, c, 0, c, b)
(0, 2c, 3c, 2*c, c, b)
For all t (0, 2*c, t) a and b are multiples of c
Yes its moving up the factor tree. There were a number of hints in that regard in terms of where factors would turn up. For example (2*c-1, 1, t) a and b turn up as factors almost twice as often as c but no way to find them without iteration. But it doesn't help move down from N.
I disagree with your t's.
What you call (0, 10, 1) isn't a valid record. If a = 0 everything is 0.
{0 10 15 10 5 45} is the first record.
Every positive integer t will exist for (0, 2d, t) and x/2*d will give you t.
This is (1, 1, t) first 10 records. Notice d pattern of two times perfect square starting with 1. t=1 d=2*tt. Lots of patterns for t math when you use this for
t = 1
t e n d x a b
1 1 1 2 1 1 5
2 1 1 8 3 5 13
3 1 1 18 5 13 25
4 1 1 32 7 25 41
5 1 1 50 9 41 61
6 1 1 72 11 61 85
7 1 1 98 13 85 113
8 1 1 128 15 113 145
9 1 1 162 17 145 181
10 1 1 200 19 181 221
I get your t math now it based on potential not valid x
Need some help!
I found two very interesting records in (e,n)
I was expecting a record at (2c,2c)
There isn't one.
But there are records at:
(2c, a,1) {2c: a: b: 0: b: b+2a}
(2c, b,1) {2c: b: a: 0: a: a+2b}
I [D]id [M]ention these to Chris a couple hours before the crumb yesterday.
Anyway was doing algebra on this mess and keep bouncing.
Did some geometry and found an even crazier record.
(2(c-aa), b-a, 1) {2(c-aa): b-a: a: 0: a: 2b-a}
I think this mess is solvable somehow.
Draw these suckers on graph paper! Very interesting.
Thoughts??
Apologies for autistic response.
If e is even then you get {e, 1, a, 0, a, b}
If e is odd you get get {e, 1, a+1, 1, a, b}
You know e and n you can solve for a
Even a=e/2
odd a = (e+1) /2
So you can solve for the t=1 record.
Then you have to figure out x.
We know d(t)- d(t=1) partly due to n(t)-n(t=1) and partly due to x(t) - x(t=1)
If we call original values E,N,T, D, X, A, B
Then a(E, 1, T) = NA
d(E, 1, T) = NA+X
For e even T=X/2+1 and e odd (X+1)/2
Take a record. Keep adding one to d. Keep a and b constant. For each increase in d n goes down by one and x goes up by one. But e is decreased by 2*d+1 each time.
The formula for changing e by an integer m:
e(t+m) = e(t) - ( d(t)*m+ mm)
If you can find m=n(t) then adding n(t) to d then n=o which means d+x = b.
xx=-e
So you are looking for a negative e that is a perfect square. If the square root is x then you are done.
For example for 5,23
{15, 4, 10, 5, 5, 23}
{-6, 3, 11, 6, 5,23}
{-29, 2, 12, 7, 5, 23}
{-54, 1, 13, 8, 5, 23}
{-81, 0, 14, 9, 5, 23}
-e is perfect square equal to x
Trying to figure out how to predict when n=0 or -e = perfect square
Q posted this today.
Election fraud fries me!
Can't wait til they do something about it
Think of the grid defining a space made up of perfect squares.
Then t equal d.
Think of n=0 records from baker. For an e perfect square there exists a record for every d positive integer where x = sqrt(abs(e))
Each increase of d by one is the next square.
Factor solution for c is n=0 then x tells you how far apart they are.
For each decrease in n by 1 d and x go up by 1 and a and b stay the same.
e(d+1) = e-(2d(d)+1)
Solution is -e=(n+x)^2
Solution is (-(n+x)^2, 0, d+n)
{-(n+x)^2: 0: d+n: x+n: a: b}
Notice d=long side x= short side
If you want to move multiple d at a time the formula is e+m = e-2dm-mm for increasing d
formula is e-m = e+2dm-mm for decreasing d
I just use m as integer variable for movement.
In this case the movement is between records describing the same a*b=c.
When you take a record (e, n, t) you can also describe it by changing d and e without changing a or b.
The relationship is d and x increase by m
n decrease by m
for e you use formula
Its the same record just a different way of describing it. However when you move n steps then n=0 and you have your solution.
Amazingly and frustratingly if you use algebra on the formula you just get (n+x)^2.
Whenever you use algebra to solve in the grid you get something you already know.
Think t = d
Grid is infinite series of squares with side = d
When you increase d by one subtract the difference between d and d +1 from remainder.
Its the same record.
You can do it multiple times it still describes a*b=c