>The best and most pragmatic advice I can offer is to use known numbers at scale as well as what you are doing.

>Reconcile an earlier post with a HUGE drop in BTC.

>Everything is connected.

And theeen you see that we're looking for the element in (e,1) where x + (e a value) - (-f a value) = original d value.

for c145 here's the calc:

original d value = 12

(e na transform) x value = 11

at x = 7, we have the following calc: x= 7 + (25) - (20) = 12 = original d value.

which gives us d = x + a, 12 = 7 + 5

So my idea is to use the original d value as the target that we're trying to make x+(e a)-(-f a) add up to. Have we already done this lads? Just thinking the clue through again and working it through.

>>7915

Nice work on this diagram, Anon!

Alright, I have an idea lads.

In the linked c6107 diagram, (n-1) = 35

Let's derive the staircase numbers, as I explained here: >>7881

(f-1)/8 = 16, T2 = 4 , T1= 3, T1+T2 = 7

The 2d(n-1) rectangles are 2730 + 2730, or 2 * 78 * 35

35 = (n-1)

T1 + T2 * iterate = correct (n-1)

7 * 5 = 35

So (f-1)/8 gives us the staircase triangle number bases to add together, and then multiply out looking for a lock. It will scale properly upward, even with huge c values.

Obviously we need to test some other examples, but deriving the T2 + T1 staircase numbers from f , and then using it to iterate/solve for (n-1) works in this c6107 example.