Verify a complex pattern for a mathematical function's sequence generation using floor division and pattern recognition | Step-by-Step Solution

What We're Solving:

We need to prove that the formula f(n) = ⌊(n+1)/2⌋ × ⌊(n+2)/2⌋ correctly generates the sequence 1, 2, 4, 6, 9, 12, 16, 20, 25, 30, 36... This is a beautiful pattern recognition and proof problem!

The Approach:

To prove this relationship, we'll use two strategies:

• 1. Direct verification - Test the formula with several values to see if it matches our sequence
• 2. Pattern analysis - Understand WHY this formula works by examining how the sequence behaves for odd and even values of n

The key insight is that floor functions create different behaviors for odd and even inputs, which is exactly what we need for this alternating-growth pattern!

Step-by-Step Solution:

Step 1: Verify the formula with concrete examples

• For n=1: f(1) = ⌊2/2⌋ × ⌊3/2⌋ = 1 × 1 = 1 ✓
• For n=2: f(2) = ⌊3/2⌋ × ⌊4/2⌋ = 1 × 2 = 2 ✓
• For n=3: f(3) = ⌊4/2⌋ × ⌊5/2⌋ = 2 × 2 = 4 ✓
• For n=4: f(4) = ⌊5/2⌋ × ⌊6/2⌋ = 2 × 3 = 6 ✓

Step 2: Analyze the floor function behavior Let's see what ⌊(n+1)/2⌋ and ⌊(n+2)/2⌋ equal for different values:

For odd n (n = 2k+1):

• ⌊(2k+2)/2⌋ = ⌊k+1⌋ = k+1
• ⌊(2k+3)/2⌋ = ⌊k+1.5⌋ = k+1
• So f(2k+1) = (k+1)(k+1) = (k+1)²

For even n (n = 2k):

• ⌊(2k+1)/2⌋ = ⌊k+0.5⌋ = k
• ⌊(2k+2)/2⌋ = ⌊k+1⌋ = k+1
• So f(2k) = k(k+1)

Step 3: Verify this pattern matches our sequence

• n=1 (odd, k=0): f(1) = (0+1)² = 1 ✓
• n=2 (even, k=1): f(2) = 1×2 = 2 ✓
• n=3 (odd, k=1): f(3) = (1+1)² = 4 ✓
• n=4 (even, k=2): f(4) = 2×3 = 6 ✓
• n=5 (odd, k=2): f(5) = (2+1)² = 9 ✓

Step 4: Understand the sequence structure The sequence alternates between:

• Perfect squares: 1, 4, 9, 16, 25, 36... (at odd positions)
• Products of consecutive integers: 2, 6, 12, 20, 30... (at even positions)

This is exactly what our formula produces!

The Answer:

The relationship f(n) = ⌊(n+1)/2⌋ × ⌊(n+2)/2⌋ is proven correct.

The formula works because:

• For odd n, both floor expressions equal the same value, creating perfect squares
• For even n, the floor expressions differ by 1, creating products of consecutive integers
• This alternating pattern perfectly generates the given sequence

Memory Tip:

Think of this as "two stairs going up at slightly different rates" - when n is odd, both stairs are at the same level (making squares), and when n is even, one stair is one step ahead (making rectangular products). The floor functions are what create this beautiful "stair-step" behavior!

You've tackled a really elegant problem that combines number theory, pattern recognition, and proof techniques. Well done working through this! 🌟