Code shared from the Rust Playground

playground.rs

use num_traits::Float;

/// Calculates the lengths of the hypotenuses in the Spiral of Theodorus,
/// which are the square roots of integers starting from sqrt(2).
fn calculate_theodorus_hypotenuses<F: Float>(count: u32) -> Vec<F> {
    // We want roots from sqrt(2) up to sqrt(count + 1).
    // The number of triangles is 'count'.
    let mut lengths = Vec::with_capacity(count as usize);

    // The length of the first hypotenuse (the base of the spiral) is sqrt(1^2 + 1^2) = sqrt(2).
    let mut current_hypotenuse = F::from(1.0).unwrap().sqrt(); // Start with sqrt(1) = 1, will be squared in the loop
    let one = F::from(1.0).unwrap();

    // The spiral starts with the hypotenuse sqrt(2)
    lengths.push(one.hypot(one)); // sqrt(1^2 + 1^2) = sqrt(2)

    // Calculate the remaining hypotenuses up to sqrt(17)
    // The previous hypotenuse 'h_n' and the constant side '1' form the legs of the
    // next triangle, with the next hypotenuse 'h_{n+1}' = sqrt(h_n^2 + 1^2).
    // Since h_n = sqrt(n), h_n^2 = n, so h_{n+1} = sqrt(n + 1).
    for n in 2..=(count as usize) {
        // The square of the previous hypotenuse is 'n-1'.
        let prev_hypotenuse_squared = F::from(n as f64 - 1.0).unwrap();
        
        // The next hypotenuse is sqrt((n-1) + 1) = sqrt(n).
        // For the n-th iteration (starting at n=2), we calculate sqrt(n).
        let next_hypotenuse = (prev_hypotenuse_squared + one).sqrt();
        lengths.push(next_hypotenuse);
    }

    lengths
}

fn main() {
    // The image shows hypotenuses from sqrt(2) up to sqrt(17).
    // This requires 16 triangles (n=2 to n=17).
    const NUM_TRIANGLES: u32 = 16;
    
    // We'll use f64 for precision.
    let hypotenuse_lengths: Vec<f64> = calculate_theodorus_hypotenuses(NUM_TRIANGLES);

    println!("📐 Spiral of Theodorus Hypotenuse Lengths (The square roots):");
    println!("---");
    
    // The lengths correspond to sqrt(2), sqrt(3), ..., sqrt(17).
    for (i, length) in hypotenuse_lengths.iter().enumerate() {
        // i starts at 0, representing sqrt(2). So we use i + 2 for the root number.
        let n = i + 2; 
        
        println!("√{n:<2} = {length:.8}");
        
        // Add a blank line for sqrt(4), sqrt(9), and sqrt(16) for visual grouping 
        // as they represent integer lengths in the spiral.
        if n == 4 || n == 9 || n == 16 {
            println!("---");
        }
    }
}