Project Euler Problem 341

Solution

Answer: 56098610614277014

Let

$G(n)=1+G\bigl(n-G(G(n-1))\bigr)$

be the standard recurrence for the Golomb sequence (valid for $n>1$), with $G(1)=1$.

We want to compute

$$\sum_{1 \le n < 10^6} G(n^3).$$

The challenge is that $n^3$ reaches nearly $10^{18}$, so directly generating the sequence up to that index is impossible.

1. Mathematical analysis

Step 1: Understanding Golomb's sequence

The defining property says:

the integer $k$ appears exactly $G(k)$ times.

So if $G(k)=4$, then the value $k$ occurs four times in the sequence.

Define the cumulative endpoint function

$$S(k)=\sum_{i=1}^{k} G(i).$$

Then:

$k$ occupies positions

$$S(k-1)+1,\dots,S(k),$$

therefore

$$G(n)=k \quad\Longleftrightarrow\quad S(k-1)<n\le S(k).$$

So computing $G(n^3)$ reduces to finding the first $k$ with

$$S(k)\ge n^3.$$

Step 2: Avoid generating up to $10^{18}$

A naïve approach would compute all $S(k)$ until $10^{18}$, but that would require about $10^{11}$ values — impossible.

The key observation is:

If $G(r)=L$, then the value $r$ occurs exactly $L$ times among the indices $k$.

That means there is a whole block of $L$ consecutive $k$-values for which

$$G(k)=r.$$

Inside that block:

$S(k)$ increases by exactly $r$ each step,
so the cumulative values form an arithmetic progression.

Hence we can process entire blocks at once rather than one $k$ at a time.

Since $G(r)$ only grows to around a few million, generating $G(r)$ up to the necessary size is completely feasible.

Step 3: Efficient block processing

Suppose we are in a block where:

$G(k)=r$,
block length is $L=G(r)$.

Then:

$$S(k)$$

advances by increments of $r$.

If we already know the cumulative sum before the block, say $S_0$, then the end of the block is

$$S_0 + Lr.$$

For a cube $c=n^3$ inside this range, the correct index is

$$k = k_0 + \left\lceil \frac{c-S_0}{r} \right\rceil,$$

where $k_0$ is the index before the block starts.

Since there are only $999{,}999$ cubes, we can sweep through them once.

2. Python implementation

def solve(limit=10**6):
    """
    Compute:
        sum(G(n^3)) for 1 <= n < limit
    where G is Golomb's self-describing sequence.
    """

    # Golomb values, 1-indexed
    G = [0, 1]  # G(1) = 1

    total = 0

    # Current cube pointer: n^3
    n = 1
    cube = 1

    # We have already processed k = 1
    # S(1) = 1
    current_k = 1
    current_S = 1

    # Handle cubes <= S(1)
    while n < limit and cube <= 1:
        total += 1
        n += 1
        cube = n**3

    r = 2

    while n < limit:
        # Generate G(r)
        g_r = 1 + G[r - G[G[r - 1]]]
        G.append(g_r)

        # This block has length G(r)
        block_length = g_r

        # End of cumulative range for this block
        block_end_S = current_S + block_length * r

        # Process all cubes that fall in this block
        while n < limit and cube <= block_end_S:
            # Ceiling division:
            # j = ceil((cube - current_S) / r)
            j = (cube - current_S + r - 1) // r

            # Corresponding k-value
            total += current_k + j

            n += 1
            cube = n**3

        # Move to next block
        current_k += block_length
        current_S = block_end_S
        r += 1

    return total

# Verification checks
print(solve(1000))     # should be 153506976
print(solve(10**6))

3. Code walkthrough

Building Golomb values

g_r = 1 + G[r - G[G[r - 1]]]

This is the classic Golomb recurrence:

$$G(r) = 1 + G(r - G(G(r-1))).$$

We only generate as many terms as needed.

Processing blocks

If $G(r)=L$, then the value $r$ repeats $L$ times among consecutive indices $k$.

Instead of iterating over every $k$, we process the whole block:

block_end_S = current_S + block_length * r

because each step increases $S(k)$ by exactly $r$.

Finding where a cube lands

If $c=n^3$ lies in the current block:

j = (cube - current_S + r - 1) // r

computes

$$\left\lceil \frac{c-S_0}{r} \right\rceil,$$

the offset inside the block.

Then:

total += current_k + j

adds the correct Golomb value.

4. Verification

The problem provides:

$$\sum_{1\le n<10^3} G(n^3)=153506976.$$

Running the algorithm for limit=1000 gives exactly:

$$153506976,$$

so the implementation matches the official check.

The final computation for $10^6$ completes quickly and produces a value of the expected magnitude ($\sim 10^{16}$).

Answer: 56098610614277014