Look-and-Say Sequence and Floor of (4+√11)^n: Weekend Math Challenges with Solutions

Two Challenges for the Weekend

N.1 For those who want a light warm-up
N.2 For those who want to think beautifully

Challenge N.1: The Look-and-Say Sequence

What is the next number in the sequence?
1, 11, 21, 1211, 111221, …

Challenge N.2: Floor of (4+√11)^n

Given a positive integer n, prove that the floor of (4+√11)^n is always an odd number.

Answer to the Look-and-Say Sequence

312211

Each term describes the previous one: “three ones, two twos, one one.” This is the classic look-and-say sequence — you read the digits of the previous number aloud and write down what you say.

Proof: The Floor of (4+√11)^n Is Always Odd

Step 1. Define the sequence.

Let:

Aₙ = (4+√11)ⁿ + (4−√11)ⁿ

The numbers 4+√11 and 4−√11 are roots of the equation x² − 8x + 5 = 0. Therefore, Aₙ satisfies the recurrence relation:

Aₙ₊₂ = 8Aₙ₊₁ − 5Aₙ

With initial values A₀ = 2 and A₁ = 8, one can show by induction that every Aₙ is an even integer.

Step 2. Estimate the conjugate term.

Since 4 − √11 ≈ 0.683 < 1, for any positive integer n:

0 < (4−√11)ⁿ < 1

This means the fractional part of (4+√11)ⁿ equals 1 − (4−√11)ⁿ, and therefore the floor of (4+√11)ⁿ is exactly Aₙ − 1.

Step 3. Conclude.

Since Aₙ is always even, the value Aₙ − 1 is always odd.

Conclusion: The floor of (4+√11)^n is always an odd number. ∎