The n0 base is a way to calculate through iterations the correct value of n or n-1. The iterations required are no bigger in complexity than Big Oh for the root of c.
A staircase number is a number that can be expressed as the addition of two or more consecutive numbers (e.g. 3+4+5, 8+9+10+11+12+13, 22+23). To find one configuration of a staircase number for any odd number, we add one to the value then divide it by two. Then the number is equal to the result added to the result minus one (e.g. 67 = (67+1)/2 + ((67+1)/2)-1 = 33+34).
One place in the grid in which we find two consecutive numbers like this which can be added together to find a staircase number is the x values in (e,1) and (f,1). Two elements from two (e,1) and (f,1) cells with the same t value will have consecutive x values. For example, (-17,1,6) has an x value of 11, but (30,1,6) has an x value of 10. These are consecutive numbers, so they produce the staircase number of 10+11=21.
Something Chris has suggested that we look into is where the x values of two paired (e,1) and (f,1) elements at the same t add together to produce na or n-1 (depending on which one is odd).
>a from (e,1,1) = c at (2c,1,1) and (2c-1,1,1)
We can find c at t=1 in both of these cells. We can also find c at t=2, t=3 etc. Since all a values in (e,1) are twice a square or the sum of consecutive squares (based around t) plus either e/2 or (e-1)/2, but we're choosing t, we can just take 2(t-1)(t-1) and (t-1)(t-1)+tt away from c to find e/2 or (e-1)/2 and therefore calculate the columns in which c will appear as an a value in (e,1). So we can find where c appears in (e,1) at t=1, t=2, t=3 etc through calculation. There is a pattern to the columns where c will appear at each t value. Where c145 appears as a[t] in (e,1) with t increasing from 1 upwards, to begin with for even e is 290, 286, 274, 254, 226, 190, etc. -4, -12, -20, -28 etc. The gap between cells increases by 8 each time for even e. For odd e, 289, 281, 265, 241, 209, 169, etc. -8, -16, -24, -32, etc. The gap between cells also increases by 8 each time, but the first started at 4 and this one starts at 8.
In all of these columns, there will be a cell where n equals our a and one of the a values equals our b, as well as where n is our b and one of the a values is our a. This is because each a[t] represents na for some c. All of these columns will have this in common. When Chris posted about this, he said "how many columns do you need to create a lookup?" This implies that having a certain number of these columns calculated will give us enough information to create a lookup. Nobody seems to have figured out what information is specifically important in this context so far.
"Values of t that are derived from imaginary numbers can be thought of as orthogonal to the grid." This could relate to when Chris told us to consider the idea of elements in the grid coming out towards you. He never elaborated about what imaginary numbers these t values need to be derived from. Obviously if we're taking a square root in order to produce imaginary numbers this would imply he's talking about negative c values. Taking sqrt(-c), we would end up with di, which, squared, would be a negative square. For example, if we take c=-559, d=23i, and dd=-529. e = c-dd, so e here would be -30. This is e for c=559 but negative. f=e-(2d+1), so here f = -31-46i. I never did imaginary numbers in school so I'm not sure where you'd go from there, but it would appear that using negative c produces a new set of cells and numbers that may be worth analyzing. This is obviously not a completely-understood pattern, but this paragraph at the very least potentially explains the opening statement from a post by Chris, which can be expanded upon when understood further.
If c has two prime factors (and is therefore semiprime), it'll have two na values (including BigN). If c has three prime factors, it'll have four na values. If c has four prime factors, it'll have eight na values. The number of na values a given c has increases exponentially with its total number of prime factors. So within (e,1) and (f,1), there will be more values of na and (n-1)a if c has a higher number of factors. This also means the number of overall factors (prime and/or composite) and the number of valid n values for a given c increases exponentially with the number of factors it has.
Chris said that we're meant to make use of this concept by multiplying c by another number q. q is meant to be the product of several small known prime numbers (e.g. 51317), and is meant to be slightly bigger than d. There was a vague post in RSA #15 that led us to believe that these small primes are meant to be odd sums of two squares, but Chris never confirmed. We're meant to multiply c by q to give us a new number for which there are more na values, giving us more information. With a semiprime c, since we will know these small prime numbers, we will know all of the corresponding na values for each known factor. So, for example, if we were to multiply 817 (1943) by q=513=65 to get qc=53105, we still wouldn't know 19 and 43 are factors of 817, but we would know that 5, 13, 65 and 817 are factors of 53105. This would give us eight known na values, thereby increasing the amount of information we know within the (e',1) and (f',1) cells. It also obviously means qc will turn up in more cells within the e' and f' columns than c appears in the e and f columns. According to Chris, "there is a point where the number of factors gives up the answer." Also according to Chris, our original a must appear twice in the first a+1 [e',1,t] elements of [e',1], and "the two values we are forcing with the grid" are the "factors [that] will also make up the two values of t where a is a factor". This statement is a bit vague. He refers to these two numbers we're forcing with the grid as factors, which would imply that we're looking for a[t]/a for the two elements in (e',1) where a appears as a factor between 1 and a+1, but he specifically says "values of t", which specifically refers to t, and not to a[t]. Either way, we are apparently looking for two values that both directly relate to two elements in (e',1) where a[t] is divisible by our unknown a and where t is between 1 and a+1 (inclusive).
This process is also allegedly meant to increase the smoothness of BigN-n (i.e. the highest prime factor of BigN'-n' should be lower than the highest prime factor of BigN-n). Chris said "if you multiply c by small primes, the smoothness of BigN-n increase". This does not appear to be the case, however, as I showed >>9049 here. It's worth mentioning in Grid Patterns given Chris said it. Maybe he'll come back and elaborate on this point eventually and I can update this post.
There are three types of prime or rather there is a way to separate primes.
Those in column [1,n], those in column [2,n] and those that are not.
Those in column 1 are the sum of two squares.
Those in column 2 are the sum of three squares.
c%d is congruent to e%d
Column (-1,n) contains an entry in every cell, like column (0,n). In (-1,1), the series of d values (1, 7, 31, 49, etc) are each two times a square minus one (so they're also the n values in (0,n) where a and b are both squares minus 1). Also in (-1,1), the series of a values (4, 12, 24, 40, etc) contain within them as factors every single prime number. They appear in a pattern, shown >>8883 here. Where a prime number appears as a factor of a[t] in (-1,1), it will also appear at this t+1. The first two appearances of primes in a cell at [e,1] have the property where their values of their position t, summed is the value of the prime plus one. Five will appear twice within the first six elements, seven twice within eight, etc. This series is meant to help our lookup somehow.
The product of a and some values of q will appear before N'c' in [e',1] with one of the n' values.
There's meant to be a trivial lookup based on everything we already know and a non-trivial lookup that relies on the results of the trivial lookup. The trivial lookup returns t, e and f (we don't know if that t value is the correct one but given there's a separate non-trivial lookup it doesn't seem likely), and the non-trivial lookup returns n (-1 for prime, 0 for square c). Chris didn't give us enough information to infer anything about how the non-trivial lookup works yet (aside from explaining various patterns that are apparently relevant), but he did say that we would use columns -f,0,1 and e; rows 1, (n-1), n, X, Y and C. It depends at least on odd e/even e (i.e. it'll work differently for either one) at a minimum (meaning it could potentially be optimized by going further with things like odd x+n, even n, etc). Column 0 contains the square of c. X and Y are the positions of n between 1 and the square of c (this is extremely vague but it's a direct quote). X and Y will not exist for primes (that depends on the value of f and d - again, a pattern we haven't necessarily understood yet). The work we did with triangles will show which integers are primes. It also pays to mention that a certain notorious weirdo on /qresearch/ once said out of the blue "use X on Y at C", which could potentially be relevant (but possibly not).
In the (e,1) and (f,1) cells corresponding to our given c, from t=1 upwards, we see an increasing infinite series of d values. There are two elements in (e,1) and two elements in (f,1) where the d from our given c fits between the d[t] values. Something about this is meant to be important.
Here are a few more things to add to >>7982 that (as far as I could tell) nobody has looked into yet or are not fully understood.
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The non-trivial method uses a pattern you might not have seen yet but has been discussed.
You know how to look up a number c from column 0 or 1. One way is to take a[t] from c in [0,1] or [1,1], divide the remainder by two and add that to the column to get c in another column.
The value c can be looked up in many columns this way.
This gives enough information to find a, x and n.
In other words.
The columns that c appears in, determines it's composition.
That is the power behind the grid.
Try it with both primes and composite numbers. Anything strike you? Once you c it…
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Here is a tip, if i(t) has not been done before.
Take products, c, that have odd x+n. And even n.
Take one from x+n.
8 triangles.
Create a table.
What does f and 2(n-1) contribute?
Because the tips of all 8 triangles are from (n-1)(n-1) - 1.
Find the pattern from large to small.
There are h families!
Avoid the 13th step of AA.
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All values in the cells at n=1 are products where you add a small square to e to make a square with c.
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There always exist another multiple of a that has a modular congruence with c. (it was from the c%d stuff)
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sqrt(a) is our long side length of the triangle, with t and t + 1 being the other two sides.
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(In relation to qc)
By controlling the low prime numbers as a product with, an increasing (non-linear) amount of information is forced upon [e',1]. Each of those factors have to appear early and forever after once they appear in the grid. The sequential probability of consecutive primes being in a random value of [e',1] are low and place huge restrictions on the other primes, including a and including the values of x.
The low primes act to filter out the numbers that a cannot be based on what we know in [e,1] and, importantly in [-f,1]. Because one is odd, and one is even, the overlap and distribution of odd numbers in one of these columns and e' is such that a lookup becomes possible. BigN' comes into play.
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