>>3140

>>3141

Try investigating the path to the solution for every every multiple of 5. If b is odd, then the solution in (e,1) is two cells above the (e,1) for the starting cell (setting n=1 and generating a cell.)

Particularly for multiples of 5, you can deduce that b is odd because it ends in the digit 5.

Example:

145 = {1:61:12:11:1:145}—(1, 61, 6)

(1, 1, 6) = {1:1:72:11:61:85}

(1, 1, 4) = {1:1:32:7:25:41}

7 = x for 5*29

(1, 5, 4) = {1:5:12:7:5:29}

Example 2:

95 = {14:39:9:8:1:95}—(14, 39, 5)

(14, 1, 5) = {14:1:47:8:39:57}

(14, 1, 3) = {14:1:19:4:15:25}

4 = x for 5*19

(14, 3, 3) = {14:3:9:4:5:19}

I don't know how many cells above the (e,1) solution is for even b, for multiples of 5. Maybe you can tell me. Have to work.

>>3143

Try investigating it.

Try 10, 15, 20, 25, 30 etc

>>3145

It might look that way, but I haven't, I just found some things to investigate that are proving fruitful so far. Like the number systems crumbs. I'm starting to have some ideas on why he told me to look at it in binary. Binary is 1 and 0, the VQC is based on odd and even. And though not confirmed, every example I've tested has shown me that the solution cell for multiples of 5 is based on odd and even b.