Solution

Answer: 16475640049

Let $F(n)$ denote the number of ways to fill a row of length $n$ using:

grey squares of length $1$,
red blocks of length at least $3$,
with at least one grey square between any two red blocks.

We are asked to compute $F(50)$.

Mathematical analysis

We derive a recurrence from first principles.

Step 1: Understand the structure

Consider the leftmost position of a row of length $n$.

There are two possibilities:

Case 1 — First square is grey

Then the remaining $n-1$ positions can be filled in $F(n-1)$ ways.

Contribution:

$$F(n-1)$$

Case 2 — First piece is a red block

Suppose the first red block has length $k$, where

$$k \ge 3$$

There are two subcases.

Subcase A: The block reaches the end

If $k=n$, this contributes exactly one arrangement.

Subcase B: Space remains after the block

If $k<n$, then the next square must be grey (separator), consuming one extra unit.

So after placing:

red block of length $k$,
one mandatory grey square,

the remaining length is

$$n-k-1$$

which can be filled in

$$F(n-k-1)$$

ways.

Thus for fixed $k<n$, contribution is:

$$F(n-k-1)$$

Step 2: Build recurrence

Combining all cases:

$$F(n) = F(n-1) + 1 + \sum_{k=3}^{n-1} F(n-k-1)$$

The $+1$ corresponds to the arrangement where a red block of length exactly $n$ fills the row.

Step 3: Initial conditions

If the row length is less than $3$, no red block can fit.

So only the all-grey arrangement exists:

$$F(0)=F(1)=F(2)=1$$

The problem statement says for $n=7$ there are $17$ ways, so we can use that to verify.

Python implementation

def count_ways(n):
    # F[i] = number of valid fillings for length i
    F = [0] * (n + 1)

    # Base cases
    F[0] = 1
    if n >= 1:
        F[1] = 1
    if n >= 2:
        F[2] = 1

    # Compute values bottom-up
    for length in range(3, n + 1):

        # Case 1: first square is grey
        total = F[length - 1]

        # Case 2: first block is red
        for block_len in range(3, length + 1):

            if block_len == length:
                # Red block fills the entire row
                total += 1
            else:
                # Need one grey separator after the block
                total += F[length - block_len - 1]

        F[length] = total

    return F[n]

# Verify small example
print(count_ways(7))   # should be 17

# Compute required answer
print(count_ways(50))

Code walkthrough

Array setup

F = [0] * (n + 1)

Stores the number of valid arrangements for each row length.

Base cases

F[0] = F[1] = F[2] = 1

For lengths below $3$, only the all-grey arrangement exists.

Main DP loop

for length in range(3, n + 1):

We compute answers incrementally from smaller lengths.

Grey-first case

total = F[length - 1]

Place one grey square first, leaving a row of size $length-1$.

Red-first cases

for block_len in range(3, length + 1):

Try every allowable red block length.

Entire row filled

if block_len == length:
    total += 1

Exactly one arrangement.

Space remains

total += F[length - block_len - 1]

One separator grey square is mandatory.

Verification on the sample

For $n=7$:

count_ways(7) = 17

which matches the problem statement exactly.

Computing $F(50)$

Evaluating the recurrence gives:

$$F(50)=16475640049$$

This magnitude is reasonable because the count grows exponentially but not explosively fast.

Answer: 16475640049

Project Euler Problem 114