That should be 2d-1. It also appears to be the case for all numbers regardless of the number of factors.
What does "too small" mean? The point of this thread is to get all the patterns we completely understand in one place, so you'll need to figure specifically what types of n values this works for and specifically what elements they appear in in (e,1) and (f,1).
Enumeration so far
(e,1) and (f,1) elements of note
>In (e,1) we have elements where a[t]=an and a[t]=bn. In (f,1) we have elements where a[t]=a(n-1) and a[t]=b(n-1). These elements are n and n-1 apart in terms of t respectively.
>Since an and bn appear in (e,1) and a(n-1) and b(n-1) appear in (f,1), the equivalent elements appear for c=1c. In (e,1) we can find elements where a[t]=BigN and cBigN. In (f,1) we can find elements where a[t]=BigN-1 and c(BigN-1).
>In (e,1) where x=c-d and where x=c+d+1, a[t]=cBigN and a[t]=c(c-BigN+1) respectively, and in (f,1) we have equivalent elements where a[t]=c(BigN-1) and a[t]=c(c-BigN+2). In both positive and negative space, these cells are 2d+1 apart in terms of t (the d in this gap being the d of the c we're trying to factor)
>i[t]=i occasionally pops up in the elements in (e,1) and (f,1) where d is between the d[t] values. This is also true for j[t] where c falls between the i[t] values (which also means c falls between the d[t] values).
(e,1) (f,1) concepts
>If an integer p is a factor of a[t], then p will be a factor of a[p+1-t] for ALL cells in row n=1 (na and nb are n apart (at a[t] and a[t+n])
>If p is a factor in a[t] then there exists (e,p)
>If a number at position t has a factor s, then s is a factor at (t+s), (t+2s) and so on for a at (e,1).
>Also, if a number at position t has a factor s at (e+1), then s is a factor at (s+1-t), (2s+1-t), etc for a at (e,1).
>In row 1 (where n=1) in positive space (so not the (f,n) cells), f=(x+n)(x+n)
>(1,1) contains as values for a and b the values of two consecutive squares added together.
>The n values in (0,n) where a and b are squares appear as the d values in (1,1). They also appear as the a values in (0,1).
>In (e,1) and (f,1), there's meant to be an asymmetrical pattern of n-1 as a factor of (f,1)'s d[t]-d compared with n as a factor of (e,1)'s a[t].
>Since d[t]-d values have the opposite parity (odd or even) of the a[t] values, then that means that for any e, we already know the parity of the d[t]-d values.
>Where a[t] = bn, d[t-1]-d = b(n-1)
>Between the starting prime solution cell and the a[t]=na cell in (e,1), the difference between the a value from the first to the second is a(n-1). This is obvious since na-a = (n-1)a. To add to this, though, the difference between the d values in these cells is also a(n-1).
>Take a cell in (e,1). This holds for any cell in (e,1). If it's even, its b values will appear as the d values in (e+1,1), and the a values from (e-1,1) will appear as its d values. If it's odd, its a values will appear as the d values in (e+1,1), and the b values from (e-1,1) will appear as its d values.
>We can find elements in (e,1) and (f,1) where a[t]=c. 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).
(0,1) and (1,1) concepts
>(1,1) contains as values for a and b the values of two consecutive squares added together.
>The n values in (0,n) where a and b are squares appear as the d values in (1,1). They also appear as the a values in (0,1). These are two times the square numbers.
>The values of a[t] at (0,1) are twice the square numbers. The values of d[t] at (0,1) are 4 multiples by the triangular numbers.
>In (0,1) we know a[t] = tt2. We also know how to "move" up and down (0,1) whereby t=p+t. This means that if we try to find all the a[t] that are divisible by 3, we can simply list: t = 3, 6, 9, 12, etc.
>By multiplying c with 4 we're creating a new record and distributing the 4 (2x2) to both a and b, giving us 2a, 2b. This also gives us an even square to work with 2(x+n)^2. Odd (x+n)^2 = 8(Tu) + 1. Even (x+n)^2 = 4(Tu + T(u-1)). We now have two triangle problems to solve. For an even square, half of the triangles is one unit longer than the other half. The even square can be expressed as two d values from (0,1) added together, d[u+1] + d[u].
>The a[t] values in (1,1) are all odd sums of two squares (although not every odd sum of two squares appears as an a[t] value). The a[t] values in (1,1) are also only divisible by numbers which are also odd sums of two squares.
(0,n) cells of note
>In (0,0) every a and b are equal and every c is a perfect square.
>If c is a semiprime, it'll show up where [a,b] = [1,c^2], [a,ab^2], [b,a^2b], [a^2,b^2] and [c,c]
>Root of d = {0, 2xd, 3xd, 2xd, d, 9xd}. We don't know what this is used for but it's meant to be important.
>There is a cell in (0,n) where a=(x+n)(x+n) and b=(d+n)(d+n).
>In (0,n), if you set x=2c (and therefore with semiprimes t=c+1), each valid n value is divisible by either a or b (or both).
(0,n) concepts
>Squares only ever appear in a and b in (0,n) where n is two times a square, so 2, 8, 18, 32, 50, 72, etc
>The n values in (0,n) where a and b are squares appear as the d values in (1,1). They also appear as the a values in (0,1).
>In column zero, x is a multiple of n
Relationships between variables
>Based on the parity of d and e, we can also find the parity of x, a, b, f and c.
>The parity of BigN is the same as the parity of n.
>2(BigN-n) = (a-1)(b-1), and we can find the cell where a=a-1 and b=b-1 (or where a=a+1 and b=b+1, for which our current unknown a and b represent 2(BigN-n)) by finding the cell (e-2n,n) (or (e+2n,n) for +1s).
>When you decrease e by 1 but increase n by 1, the increase in f is equal to 2d-1. When you increase e and n by 1, at the same t, the increase in f is equal to 2*(the increase in d)+1.
>ab can be represented as the sum of consecutive odd numbers where a is the number of terms and b is the midpoint: 529=145=25+27+29+31+33=145
>If a cell contains an element c, another element in the cell can be constructed from it. e'=e, n'=n, x'=x+2n, a'=b, d'=a'+x', b'=a'+2x'+2n
>By multiplying c with 4 we're creating a new record and distributing the 4 (2x2) to both a and b, giving us 2a, 2b. This also gives us an even square to work with 2(x+n)^2. Odd (x+n)^2 = 8(Tu) + 1. Even (x+n)^2 = 4(Tu + T(u-1)). We now have two triangle problems to solve. For an even square, half of the triangles is one unit longer than the other half. The even square can be expressed as two d values from (0,1) added together, d[u+1] + d[u].
>For each odd (x+n)(x+n), there is a finite set of cells where n!=1, and an infinite set where n==1.
>c%(BigN-1) = 2d-1
>c%d is congruent to e%d
>(x+n) = (b-a)/2
Relationships between cells
>If a cell has values at (e,n) then there will be values in a cell at (e+2n,n). This is how you get the horizontal pattern of grid cells.
>The values of a,b and d each increase by ONE 2n cells to the right. They decrease by one 2n cells to the left. This also means that one cell's c is equal to the next one to the right's 2(BigN-n).
>From (-f,n-1) = c, the value of a,b and d increase by ONE every 2(n-1) cells, as you move from left to right in the grid.
>From (e,n) = c, the value of a,b and d decrease by ONE every 2n cells, as you move from right to left in the grid.
>Choose any cell (e,n). Take a look at the corresponding cell (e, a). The a value of the 2nd cell = the n value of the 1st cell.
>Take a cell in (e,1). This holds for any cell in (e,1). If it's even, its b values will appear as the d values in (e+1,1), and the a values from (e-1,1) will appear as its d values. If it's odd, its a values will appear as the d values in (e+1,1), and the b values from (e-1,1) will appear as its d values.
>At any (e,n,t), there will be another record at (e,n,t+n) where that record's b is equal to the original record's a, and the new x is equal to the old one plus 2n.
>You can create any negative x record by substituting a and b. (1,5,4) = {1:5:12:7:5:29} = 145 becomes (1,5,-8) = {1:5:12:-17:29:5} = 145
>The a values in (-1,2) are equal to 12, 23, 34, 45, 56, 67, 7*8 etc. So our n(n-1) will always appear as an a (and technically also a b) in (-1,2).
Bigger overarching concepts that we were told were meant to be used to directly solve the thing
>There's some kind of tree thing we're meant to use. It starts of as a decision tree which terminates if we can immediately solve (e.g. if gcd(d,e)>1). There's then a proper tree diagram in which we find d and e of c, then we find d and e of d and d and e of e, and so on until reaching 1, and also dividing by 2 if we get an even number until it's odd. This is meant to precede whatever concepts we use in either (e,1) or (0,n) to solve, although once we learned to make the tree it was never brought up again and no links were made to any other concepts. Summing up all of the leaves of the number tree diagram gets us a number that is relatively close to one of the factors. "The factor tree is used to factor d and e", apparently, even though that hasn't been shown. "Factoring these (and down the tree) allow for the factoring of c" too apparently. Another claim is that "the tree solution finds x+n or x."
>We're meant to be able to use (x+n)(x+n)=nn+2d(n-1)+f-1 to find the solution. This involves patterns that were looked for but never found, and a variable n0 which is meant as a guess base for each triangle that makes up (x+n)(x+n) and then eventually finds the right triangle base in some way we were never shown.
>Polite/staircase numbers are numbers that can be expressed as the sum of consecutive integers. There are many different variables which pop up as expressions of polite numbers, such as 2d+1. We also find staircase numbers when we add the x values from an (e,1) element and an (f,1) element (this was specifically pointed out), although nothing useful has come from analysis of this either. "The solution to this problem introduces a new form of algebra where two concurrent forms of equations run side by side and then merge. The two sets of equations take the problem and simplify it. Together they handle the "lock and key" nature of the problem/solution, particularly when c is divisible by 1, c, and two other prime numbers, such as in RSA. The two forms of equations that merge together handle staircase numbers where the base of one staircase, is one unit longer than the other." According to this statement, staircase numbers directly relate to the solution, but they have only ever come up in the explanation of the triangle method, which, as stated, we don't know how to use.
>Multiplying c by known small primes is meant to help us to find the unknown factors by increasing the frequency of their appearance. We multiply c by another variable q, which is the product of those small known primes (which for some reason are meant to all end in 01 in binary). We are then meant to multiply qc by another variable v, which will transform it into (0,n). We haven't found anything useful through analysis of qc alone, and it would appear to not currently be possible to directly calculate v. Chris also said that multiplying c by q would increase the smoothness of BigN-n (in other words, decrease its highest prime factor). We discovered that this is actually incorrect, and that the highest prime factor of BigN-n actually increases in every test case of qc.
>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 ebot on /qresearch/ once said out of the blue "use X on Y at C", which could potentially be relevant (but possibly not).
>There are meant to be separate ways to solving using either (0,n) with X, Y and C (possibly including (0,e) and (0,f), or they might be their own separate solution path), (e,1) and (f,1), row 2, and (1,1).
Things we don't know what to do with
>Root of d = {0, 2xd, 3xd, 2xd, d, 9xd}. We don't know what this is used for but it's meant to be important.
>In (e,1) and (f,1), there's meant to be an asymmetrical pattern of n-1 as a factor of (f,1)'s d[t]-d compared with n as a factor of (e,1)'s a[t].
>Apparently there's a useful values in the cell in (e,1) where x=f or f-1.
>There's meant to be a "lookup x" for our given c that "gives you na or (n-1)". There's some x value somewhere in the grid that can applied to c in some way to give us na or n-1. Since there are two possibilities, it's probably based on e and f parities, given all of the patterns based around e and f parities. Also, since the x value in (e,1) where a[t] = na will always be equal to the x value in the prime solution record, finding that x value would bypass (e,1) and na completely, since we wouldn't need to factor na to know what a and b are (since we'd have the prime solution record). So this x value is most likely not the x value from the (e,1) record where a[t] = na.
>Apparently aan(n-1) is meant to be helpful. We don't know how to find it or what to do with it.
<There are more things in this thread that I left out because either they were extremely vague, we looked into them and didn't find anything, or they're interesting concepts but they've never been brought up by anyone other than the person who discovered them and don't seem to have much use. I may include these in this section later on if anyone wants.