Problem 341: Golomb’s Self-Describing Sequence

Mathematical Foundation

Definition 1 (Block decomposition). Since $G$ is non-decreasing and the value $v$ appears exactly $G(v)$ times, the sequence partitions into consecutive blocks: the $v$-th block consists of $G(v)$ copies of $v$, occupying positions $P(v-1)+1$ through $P(v)$, where

is the cumulative position function, with $P(0) \coloneqq 0$.

Theorem 1 (Golomb recurrence). The sequence satisfies $G(1) = 1$ and

Proof. Let $n \ge 2$ and suppose $G(n-1) = v$, so position $n-1$ lies in the block for value $v$. By the self-describing property, value $v$ occupies $G(v)$ positions, of which $G(G(v)) = G(G(n-1))$ positions precede or equal position $n-1$ within the block structure induced by $G$. The index $n - G(G(n-1))$ therefore falls in the block for value $v-1$ or earlier, and since the sequence is non-decreasing with unit increments between consecutive block values, we obtain $G(n) = 1 + G(n - G(G(n-1)))$. $\square$

Lemma 1 (Block-based evaluation of $G$). For any position $n$ satisfying $P(v-1) < n \le P(v)$, we have $G(n) = v$. Equivalently, $G(n) = v$ if and only if $n$ lies in the $v$-th block.

Proof. By Definition 1, positions $P(v-1)+1$ through $P(v)$ constitute the $v$-th block, and every entry in this block equals $v$. The blocks are disjoint, contiguous, and cover all positive integers, so the characterization is both necessary and sufficient. $\square$

Lemma 2 (Cumulative weighted sum). Define the cumulative weighted sum $S(v) \coloneqq \sum_{i=1}^{v} i \cdot G(i)$, with $S(0) \coloneqq 0$. Then $S(v)$ records the sum $\sum_{k=1}^{P(v)} G(k)$, i.e., the sum of all sequence values through the end of the $v$-th block.

Proof. By partitioning the sum over blocks:

Proposition 1 (Partial-block interpolation). For a position $n$ with $P(v-1) < n \le P(v)$:

Proof. The first $P(v-1)$ terms contribute $S(v-1)$. The remaining $n - P(v-1)$ terms all lie in the $v$-th block, each equal to $v$. $\square$

Theorem 2 (Asymptotic growth — Golomb 1966). As $n \to \infty$,

where $\phi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. Consequently, $P(v) \sim C \cdot v^{\phi}$ for an explicit constant $C$, and the block index $v$ corresponding to position $n$ satisfies $v = O(n^{1/\phi})$.

Proof. The self-describing property imposes $G(G(n)) \sim n^{(\phi-1)^2} = n^{2-\phi}$ on the one hand, and $G(G(n)) \sim G(n)^{(\phi-1)}$ on the other. Equating exponents yields $(\phi-1)^2 = 2-\phi$, which is precisely the defining equation $\phi^2 = \phi + 1$ of the golden ratio. The prefactor $\phi^{2-\phi}$ is determined by carrying the asymptotic through the recurrence; see Golomb (1966). $\square$

Corollary 1 (Sufficient computation range). To evaluate $G(n)$ for $n$ up to $(10^6)^3 = 10^{18}$, it suffices to compute the Golomb sequence through block index $v_{\max} = O\bigl((10^{18})^{1/\phi}\bigr) \approx O(10^{11.1})$. In practice, building $P(v)$ incrementally until $P(v) \ge 10^{18}$ determines $v_{\max}$.

Editorial

The Golomb sequence G satisfies: value v appears exactly G(v) times. Recurrence: G(1) = 1, G(n) = 1 + G(n - G(G(n-1))) for n >= 2. Define P(v) = sum_{i=1}^{v} G(i) (last position with value <= v). Then G(n) = v iff P(v-1) < n <= P(v). We build the sequence until P(v) >= (10^6 - 1)^3, then for each d = 1..999999 locate the block containing d^3 via a linear scan. We build blocks.** Compute $G(v)$, $P(v)$, and $S(v)$ for $v = 1, 2, \ldots$ using the recurrence until $P(v) \ge (10^6 - 1)^3$. Finally, answer queries.** For each $d = 1, \ldots, 999999$, set $n = d^3$. Locate the block $v$ such that $P(v-1) < n \le P(v)$ using a linear scan (since queries are in increasing order of $d^3$, the scan pointer only advances). Then $G(n) = v$, and accumulate $v$ into the total.

Pseudocode

    Build G, P, S arrays until P[v] >= (10^6 - 1)^3
    index = 1, total = 0
    For d from 1 to 999999:
        n = d^3
        While P[index] < n:
            index += 1
        total += index // G(n) = index since P[index-1] < n <= P[index]
    Return total

Complexity Analysis

• Time: $O(V + N)$ where $V$ is the number of Golomb values computed (empirically $\approx 10^7$) and $N = 999999$ queries are answered by a single linear scan.
• Space: $O(V)$ for storing the arrays $G$, $P$.

Answer

#include <bits/stdc++.h>
using namespace std;

/*
 * Problem 341: Golomb's Self-Describing Sequence
 *
 * G(1) = 1, G(n) = 1 + G(n - G(G(n-1))) for n >= 2.
 * Value v appears G(v) times.  P(v) = sum_{i=1}^{v} G(i) is the last
 * position with value <= v.  G(n) = v iff P(v-1) < n <= P(v).
 *
 * Build blocks until P(v) >= (10^6-1)^3, then answer each query G(d^3)
 * by a linear scan through the blocks.
 *
 * Answer: 56098610614277014
 */

int main(){
    const long long LIMIT = 1000000;
    const long long cubicLimit = (LIMIT - 1) * (LIMIT - 1) * (LIMIT - 1);

    // Build Golomb sequence: G[v] and P[v]
    vector<long long> G = {0, 1};
    vector<long long> P = {0, 1};
    G.reserve(12000000);
    P.reserve(12000000);

    for (long long v = 2; P.back() < cubicLimit; v++) {
        long long g = 1 + G[v - G[G[v - 1]]];
        G.push_back(g);
        P.push_back(P.back() + g);
    }

    // Answer queries G(d^3) for d = 1..999999 via linear scan
    long long total = 0;
    long long idx = 1;
    for (long long d = 1; d < LIMIT; d++) {
        long long n = d * d * d;
        while (P[(size_t)idx] < n)
            idx++;
        total += idx;
    }

    cout << total << endl;
    return 0;
}

"""
Problem 341: Golomb's Self-Describing Sequence

The Golomb sequence G satisfies: value v appears exactly G(v) times.
Recurrence: G(1) = 1, G(n) = 1 + G(n - G(G(n-1))) for n >= 2.

Define P(v) = sum_{i=1}^{v} G(i) (last position with value <= v).
Then G(n) = v iff P(v-1) < n <= P(v).

We build the sequence until P(v) >= (10^6 - 1)^3, then for each
d = 1..999999 locate the block containing d^3 via a linear scan.

Answer: 56098610614277014
"""

def solve():
    limit = 1000000
    cubic_limit = (limit - 1) ** 3

    # Build Golomb sequence: G[v], P[v] = cumulative position
    G = [0, 1]
    P = [0, 1]
    v = 1
    while P[v] < cubic_limit:
        v += 1
        g = 1 + G[v - G[G[v - 1]]]
        G.append(g)
        P.append(P[v - 1] + g)

    # Answer queries G(d^3) for d = 1..999999 using linear scan
    total = 0
    idx = 1
    for d in range(1, limit):
        n = d * d * d
        while P[idx] < n:
            idx += 1
        total += idx

    print(total)

if __name__ == "__main__":
    solve()