How many 8-digit positive integers consist only of the digits 1 and 3, and contain at least one pair of consecutive 1s? - inBeat

Understanding the Context

Why This Pattern Is Gaining Attention

Across the US, especially among tech-savvy readers, there’s rising interest in how patterns emerge from simple rules—much like the way social media algorithms decrypt user behavior, or how coders build apps using binary logic. The constraints of using only 1s and 3s—without digits like 0, 2, 4, 5, etc.—create a controlled space for exploring combinations. When added with the condition of at least one pair of consecutive 1s, it becomes a familiar challenge in sequence analysis—relevant in everything from cryptography to digital art and generative design.

This isn’t just a side math question—it’s a gateway into learning about combinatorics, recursive patterns, and digital sequences, all increasingly valued in fields like computer science, data analysis, and creative coding.

How Many 8-Digit Numbers Use Only 1s and 3s with Consecutive 1s?

Key Insights

To find how many 8-digit numbers made only of 1 and 3 contain at least one pair of consecutive 1s, we approach it through combinatorics. The total number of such 8-digit sequences is simple: since each digit has 2 choices (1 or 3), the total count is 2⁸ = 256.

But instead of counting all valid sequences outright, we use complementary counting: find how many 8-digit sequences use only 1s and 3s without any consecutive 1s, then subtract from 256.

Let’s define aₙ as the number of n-digit sequences of 1s and 3s with no two consecutive 1s. This follows a recurrence relation:
a₁ = 2 (either 1 or 3)
a₂ = 3 (13, 31, 33; exclude 11)
For n ≥ 3:
aₙ = aₙ₋₁ + aₙ₋₂

This mimics the Fibonacci sequence—because a valid sequence of length n ends either with 3 (followed by any valid sequence of n–1) or 1 (followed by a 3 and any valid sequence of n–2).

Using this:
a₁ = 2
a₂ = 3
a₃ = 2 + 3 = 5
a₄ = 3 + 5 = 8
a₅ = 5 + 8 = 13
a₆ = 8 + 13 = 21
a₇ = 13 + 21 = 34
a₈ = 21 + 34 = 55

Final Thoughts

So, 55