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.

(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.