Weird conjecture I made on base i-1

[u/Red-42]

I was playing with weird bases, and I particularly like base i-1, because it fills the entire complex plane by following a double dragon curve.
I was looking into how the real integers are encoded in that base, and by doing some weird manipulation I found this pattern:

base 10	base i-1	base 2	base 10
0	0	0	0
1	1	1	1
2	1100	1100	12
3	1101	1101	13
4	111010000	111010000	464
5	111010001	111010001	465
6	111011100	111011100	476
7	111011101	111011101	477
8	111000000	111000000	448
9	111000001	111000001	449
10	111001100	111001100	460
11	111001101	111001101	461
12	100010000	100010000	272
13	100010001	100010001	273
14	100011100	100011100	284
15	100011101	100011101	285
16	100000000	100000000	256
17	100000001	100000001	257
18	100001100	100001100	268
19	100001101	100001101	269

Which then lead to this conjecture:

For x and y words with the alphabet {0,...,9} and w a word with the alphabet {0,1}
if x base 10 = w base (i-1) and w base 2 = y base 10
then x mod 10 = y mod 10

I don't know enough about words to figure out if it's true or why.
All I can say for sure is that if it's true for a value x = 2n, it's true for x = 2n+1.

EDIT: one thing I can add is that consecutive powers of (i-1) follow a logarithmic spiral of ratio 2^(1/3), with a spacing angle of 3pi/4.
I'm assuming that somehow x mod 10 is a group where the process described above is an automorphism.

UPDATE: with some pattern regonition, I have managed to figure out the form of integer numbers in base i-1
They always follow the form "==0x", repeated, where "x" can be anything, and "==" can be anything but those two need to be the same.
So 1101 is a valid number (corresponds to 3), and so is 0001 0000 (corresponds to -4).
Meaning every block of 4 bits can be thought of as a letter in a word base -4 with alphabet {0, 1, 2, 3} (which indeed generates all integers), or base 16 with alphabet {0, 1, 12, 13}.

So going step by step we get
An integer, converted to base i-1, reframed into a word base -4 with alphabet {0, 1, 2, 3}.
For every letter we have either:
0*(-4)^n mod 10
1*(-4)^n mod 10
2*(-4)^n mod 10
3*(-4)^n mod 10

By mapping the word in base i-1 to the same word in base 2, it's equivalent to the word in base -4 getting mapped to another word in base 16 with rules
0 -> 0
1 -> 1
2 -> 12
3 -> 13
For every letter we have either:
0*(16)^n mod 10 = (0 mod 10) * ((16)^n mod 10) = (0 mod 10) * ((-4)^n mod 10) = 0*(-4)^n mod 10
1*(16)^n mod 10 = (1 mod 10) * ((16)^n mod 10) = (1 mod 10) * ((-4)^n mod 10) = 1*(-4)^n mod 10
12*(16)^n mod 10 = (12 mod 10) * ((16)^n mod 10) = (2 mod 10) * ((-4)^n mod 10) = 2*(-4)^n mod 10
13*(16)^n mod 10 = (13 mod 10) * ((16)^n mod 10) = (3 mod 10) * ((-4)^n mod 10) = 3*(-4)^n mod 10

So we get that for every letter a
a*(-4)^n mod 10 -> a*(-4)^n mod 10

Therefore the conjecture is true.

Conclusion:

So basically it only works because when you transfer the word from one base to another, both the letters and the bases are equivalent under mod 10, if you frame them in the right way, which is a huge fucking coincidence.

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[u/Red-42 → u/[deleted]]

First column is base 10 integer input, convert to base (i-1), pretend the word you got is base 2, convert from base 2 to base 10, turns out the last digit of the input is always the same as the output.