I was looking at numbers where the digit-runs of n in one base, b₁, represented the digits of n in another base, b₂. To understand digit-runs, take the number 775555111 (in whatever base). The function digitrun(775555111) = 2,4,3, because reading 775555111 left-to-right you see two 7s, four 5s, three 1s. Here’s one pattern I found:

10000011111 in base 2 = 1055 in base 10. The digit runs of 10000011111 are 1,5,5 and 155₃₀ = 1055₁₀ = 10000011111₂.
10000022222 in base 3 = 59291 in base 10. The digit runs of 10000022222 are again 1,5,5 and 155₂₄₁ = 59291₁₀ = 10000022222₃.
10000099999₁₀ → 1,5,5 and 155₉₉₉₉₈ = 10000099999₁₀

The pattern isn’t difficult to understand — what are 30+2 and 99998+2? — and it’s easy to prove that it applies to all standard bases. But I wondered how AI would go about the proof. Here’s the result:

For base b ≥ 2, let the largest digit be d = b − 1. Then

100000ddddd_b

has digit runs

1, 5, 5

so the run-length encoding is always

155.

The value of 155 is interpreted in base

b⁵ − 2,

because

155_b⁵−2 = (b⁵−2)² + 5(b⁵−2) + 5.

Thus

100000ddddd_b = 155_b⁵−2.

Base b	n = 100000(b−1)(b−1)(b−1)(b−1)(b−1)b	Run Lengths	Compressed Form	Base-10 Value of n
2	100000111112	1,5,5	15530	1,055
3	100000222223	1,5,5	155241	59,291
4	100000333334	1,5,5	1551022	1,049,599
5	100000444445	1,5,5	1553123	9,765,624
6	100000555556	1,5,5	1557774	60,466,175
7	100000666667	1,5,5	15516805	282,475,248
8	100000777778	1,5,5	15532766	1,074,266,111
9	100000888889	1,5,5	15559047	3,486,270,320
10	1000009999910	1,5,5	15599998	10,000,099,999
11	100000AAAAA11	1,5,5	155161049	26,094,908,050
12	100000BBBBB12	1,5,5	155248830	62,191,972,031
13	100000CCCCC13	1,5,5	155371291	136,858,491,292
14	100000DDDDD14	1,5,5	155537822	281,793,968,543
15	100000EEEEE15	1,5,5	155759373	547,918,611,874
16	100000FFFFF16	1,5,5	1551048574	1,099,512,627,199
17	100000GGGGG17	1,5,5	1551419855	2,141,423,647,200
18	100000HHHHH18	1,5,5	1551889566	4,058,363,974,751
19	100000IIIII19	1,5,5	1552476097	7,502,570,010,432
20	100000JJJJJ20	1,5,5	1553199998	13,120,000,000,000

(Here A = 10, B = 11, …, J = 19.)

Proof for All b ≥ 2

The number

100000(b−1)(b−1)(b−1)(b−1)(b−1)_b

has value

b¹⁰ + (b−1)(b⁴ + b³ + b² + b + 1).

Since

b⁴ + b³ + b² + b + 1 = (b⁵ − 1)/(b − 1),

this becomes

b¹⁰ + (b⁵ − 1) = b¹⁰ + b⁵ − 1.

Now let

B = b⁵ − 2.

Then

155_B = B² + 5B + 5.

Substituting B = b⁵ − 2:

(B² + 5B + 5) = (b⁵−2)² + 5(b⁵−2) + 5

= b¹⁰−4b⁵+4+5b⁵−10+5

= b¹⁰ + b⁵−1.

Hence

b¹⁰ + b⁵ − 1 = (b⁵ − 2)²+5(b⁵ − 2)+5

for every b ≥ 2.

Therefore,

100000(b−1)⁵_b = 155_b⁵−2.

This is exactly the pattern observed in bases 2 to 20, and it holds for all bases.