Project Euler

Siegbert and Jo

Siegbert and Jo play a Fibonacci Nim variant. Define: H(N) = the smallest Fibonacci number in the Zeckendorf representation of N. G(n) = sum_(k=1)^n H(k). Given that G(10) = 22 and G(1000) = 396673...

Source sync Apr 19, 2026

Problem #0692

Level Level 12

Solved By 1,663

Languages C++, Python

Answer 842043391019219959

Length 417 words

sequencegraphdigit_dp

Siegbert and Jo take turns playing a game with a heap of $N$ pebbles:

1.: Siegbert is the first to take some pebbles. He can take as many pebbles as he wants. (Between 1 and $N$ inclusive.)
2.: In each of the following turns the current player must take at least one pebble and at most twice the amount of pebbles taken by the previous player.
3.: The player who takes the last pebble wins.

Although Siegbert can always win by taking all the pebbles on his first turn, to make the game more interesting he chooses to take the smallest number of pebbles that guarantees he will still win (assuming both Siegbert and Jo play optimally for the rest of the game).

Let $H(N)$ be that minimal amount for a heap of $N$ pebbles.

$H(1)=1$, $H(4)=1$, $H(17)=1$, $H(8)=8$ and $H(18)=5$

Let $G(n)$ be $\displaystyle {\sum _{k=1}^n H(k)}$.

$G(13)=43$.

Find $G(23416728348467685)$.

Problem 692: Siegbert and Jo

Mathematical Foundation

Definition (Zeckendorf Representation). Every positive integer $n$ can be uniquely written as $n = \sum_{i \in I} F_i$ where $\{F_i\}$ are Fibonacci numbers (with $F_0 = 1, F_1 = 2, F_2 = 3, F_3 = 5, \ldots$ ) and no two consecutive indices appear in $I$ . This is the Zeckendorf representation.

Lemma 1 (Subset Counting). Let $A(i)$ be the number of non-empty subsets of non-consecutive elements from $\{F_0, F_1, \ldots, F_i\}$ , and let $S(i)$ be the sum of the minimum elements over all such subsets. Then:

A(0) = 1, \quad A(1) = 2, \quad A(i) = A(i-1) + 1 + A(i-2) \text{ for } i \geq 2,

S(0) = 1, \quad S(1) = 3, \quad S(i) = S(i-1) + F_i + S(i-2) \text{ for } i \geq 2.

Proof. For $A(i)$ : subsets not containing $F_i$ number $A(i-1)$ ; subsets containing $F_i$ alone contribute $1$ ; subsets containing $F_i$ together with elements from $\{F_0, \ldots, F_{i-2}\}$ (skipping $F_{i-1}$ for non-consecutivity) number $A(i-2)$ .

For $S(i)$ : subsets not containing $F_i$ contribute $S(i-1)$ to the sum of minima. The singleton $\{F_i\}$ contributes $F_i$ . For subsets containing $F_i$ plus elements from $\{F_0, \ldots, F_{i-2}\}$ , the minimum is always from the lower-index elements (since $F_i > F_j$ for all $j < i$ ), contributing $S(i-2)$ . $\square$

Theorem (Digit DP on Zeckendorf). Let the Zeckendorf representation of $n$ be $n = F_{z_0} + F_{z_1} + \cdots + F_{z_{m-1}}$ with $z_0 > z_1 > \cdots > z_{m-1}$ and $z_j \geq z_{j+1} + 2$ . Then:

G(n) = F_{z_{m-1}} + \sum_{j=0}^{m-1} C(j),

where:

C(0) = \begin{cases} S(z_0 - 1) & \text{if } z_0 > 0, \\ 0 & \text{if } z_0 = 0, \end{cases}

C(j) = F_{z_{j-1}} + \begin{cases} S(z_j - 1) & \text{if } z_j > 0, \\ 0 & \text{if } z_j = 0, \end{cases} \quad \text{for } j \geq 1.

Proof. We decompose numbers in $[1, n]$ by where their Zeckendorf representation first deviates below that of $n$ .

The term $F_{z_{m-1}}$ accounts for $H(n)$ itself (the minimum component of $n$ ‘s own representation).

For each $j \in \{0, \ldots, m-1\}$ , we consider numbers that match $n$ ‘s representation at positions $z_0, \ldots, z_{j-1}$ (all present) but have $0$ at position $z_j$ , with any valid non-consecutive subset from $\{F_0, \ldots, F_{z_j - 1}\}$ below.

Case $j = 0$ : No committed bits above. The numbers are exactly the positive integers representable by non-consecutive Fibonacci numbers from $\{F_0, \ldots, F_{z_0-1}\}$ . Their $H$ -values sum to $S(z_0 - 1)$ .

Case $j \geq 1$ : The committed bits are $\{z_0, \ldots, z_{j-1}\}$ with smallest committed Fibonacci number $F_{z_{j-1}}$ .

If the subset below is empty: the number is $\sum_{l=0}^{j-1} F_{z_l}$ , and $H = F_{z_{j-1}}$ .
If the subset below is non-empty: its minimum element is strictly less than $F_{z_{j-1}}$ (since all indices $\leq z_j - 1 < z_{j-1}$ ), so $H = \min(\text{subset})$ . The sum of these minima is $S(z_j - 1)$ .

Non-adjacency between committed and free positions is guaranteed since Zeckendorf gives $z_{j-1} \geq z_j + 2$ , so the highest free index $z_j - 1 \leq z_{j-1} - 3$ . $\square$

Verification. $G(1)$ : $n=1$ , Zeckendorf $=\{0\}$ . $G(1) = F_0 + C(0) = 1 + 0 = 1$ . Correct ( $H(1) = 1$ ).

$G(5)$ : $n=5$ , Zeckendorf $= \{3\}$ . $G(5) = F_3 + S(2) = 5 + 7 = 12$ . Indeed $H(1)+H(2)+H(3)+H(4)+H(5) = 1+2+3+1+5 = 12$ . $\checkmark$

Editorial

Fibonacci Nim variant. H(N) = smallest Fibonacci number in Zeckendorf representation of N. Find G(23416728348467685) = sum of H(k) for k=1..n. Uses digit-DP on Zeckendorf representation with precomputed: A[i] = count of non-empty non-consecutive subsets from {F[0],…,F[i]} S[i] = sum of minimums over such subsets. We precompute Fibonacci numbers up to n. We then precompute A[i], S[i]. Finally, zeckendorf representation of n (greedy, largest first).

Pseudocode

Precompute Fibonacci numbers up to n
Precompute A[i], S[i]
Zeckendorf representation of n (greedy, largest first)
Compute G(n)
else

Complexity Analysis

Time: $O(\log_\varphi n)$ for computing the Zeckendorf representation and the precomputation of $A$ , $S$ arrays, since the number of Fibonacci numbers up to $n$ is $O(\log_\varphi n)$ .
Space: $O(\log_\varphi n)$ for storing the Fibonacci sequence and auxiliary arrays.

For $n = 23416728348467685 \approx 2.3 \times 10^{16}$ , we need roughly $80$ Fibonacci numbers, so the computation is essentially instantaneous. Note that intermediate sums in $S(i)$ may exceed 64-bit integers and require 128-bit arithmetic.

Answer

\boxed{842043391019219959}

C++ project_euler/problem_692/solution.cpp

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

/*
 * Project Euler Problem 692: Siegbert and Jo
 *
 * This is a Fibonacci Nim variant. The losing positions for the first player
 * are exactly the Fibonacci numbers. The winning strategy relates to
 * Zeckendorf's representation (every positive integer can be uniquely written
 * as a sum of non-consecutive Fibonacci numbers).
 *
 * H(N) = the smallest Fibonacci number in the Zeckendorf representation of N.
 *
 * G(n) = sum of H(k) for k = 1 to n.
 *
 * We need G(23416728348467685).
 *
 * Note: 23416728348467685 = F(82)/2 - 1 where F(82) is a Fibonacci number.
 * Actually 23416728348467685 is carefully chosen.
 *
 * Key insight: We compute G(n) by processing the Zeckendorf representation.
 * For each Fibonacci number F_i, count how many numbers in [1, n] have F_i
 * as their smallest Zeckendorf component, multiply by F_i, and sum.
 *
 * Let C(n, i) = count of integers in [1, n] whose smallest Zeckendorf
 * component is F_i. Then G(n) = sum over i of F_i * C(n, i).
 *
 * Algorithm: We use a recursive/digit-DP approach on the Zeckendorf representation of n.
 */

typedef long long ll;
typedef __int128 lll;

// Fibonacci numbers (1-indexed: F[1]=1, F[2]=2, F[3]=3, F[4]=5, ...)
// Using Zeckendorf Fibonacci: 1, 2, 3, 5, 8, 13, 21, ...
vector<ll> fib;

void precompute_fib() {
    fib.push_back(1); // fib[0] = 1
    fib.push_back(2); // fib[1] = 2
    while (true) {
        ll next = fib[fib.size()-1] + fib[fib.size()-2];
        if (next > 5e16) break;
        fib.push_back(next);
    }
}

// Get Zeckendorf representation of n (greedy, largest first)
// Returns indices into fib[] array
vector<int> zeckendorf(ll n) {
    vector<int> rep;
    for (int i = fib.size() - 1; i >= 0 && n > 0; i--) {
        if (fib[i] <= n) {
            rep.push_back(i);
            n -= fib[i];
        }
    }
    return rep;
}

// Count of numbers in [1, F[k]-1] with smallest Zeckendorf component = F[j]
// This is complex, so let's use a different approach.

// G(n) = sum_{k=1}^{n} H(k)
// where H(k) = smallest Fibonacci number in Zeckendorf representation of k.
//
// Alternative: G(n) = sum over all Fibonacci numbers F_i of
//   F_i * (count of k in [1..n] where min Zeckendorf component of k is F_i)
//
// Let's think recursively. Numbers whose smallest Zeckendorf component is F[0]=1
// are: 1, 1+3, 1+5, 1+8, ..., 1+3+8, etc. - i.e., 1 + (any Zeckendorf sum
// using only F[2], F[3], ...) or just 1 alone.
//
// Actually, let's use a known result:
// G(F[n]) has a nice closed form related to Fibonacci numbers.
//
// Let me implement a digit-DP approach on Zeckendorf representation.

// f(n) = G(n) = sum of H(k) for k=1..n
// We process n's Zeckendorf representation from most significant to least significant.

// Number of integers representable using Fibonacci numbers F[0]..F[k]
// with no two consecutive = F[k+2] (including 0)
// Actually the count of subsets of {F[0], F[2], F[4], ...} etc.
// with no two consecutive indices is more nuanced.

// Let me use a cleaner approach:
// Let T(i) = number of non-negative integers representable using
// non-consecutive Fibonacci numbers from {F[0], ..., F[i]}.
// T(-1) = 1 (just 0), T(0) = 2 (0 and 1), T(1) = 3 (0, 1, 2)
// T(i) = T(i-1) + T(i-2) -- because either we don't use F[i] (T(i-1) choices)
// or we use F[i] and skip F[i-1] (T(i-2) choices).
// Note T(i) = F[i+2] in standard Fibonacci.

// For the digit DP: process Zeckendorf digits from high to low.
// When we encounter a '1' bit at position i, we can:
// 1) Set it to 0 and fill lower positions freely (count contributions)
// 2) Set it to 1 and continue to next digit

// Let cnt(i) = # of positive integers representable using non-consecutive
//              Fibonacci numbers from {F[0], ..., F[i]}
// cnt(i) = T(i) - 1 (exclude 0)

// sml(i) = sum of smallest components over all positive integers representable
//           using non-consecutive Fibonacci numbers from {F[0], ..., F[i]}

// For numbers that include F[i]: smallest component <= F[i]
// For numbers that don't include F[i]: recurse on {F[0], ..., F[i-1]}

// Let A(i) = # of non-empty subsets of non-consecutive elements from {F[0],...,F[i]}
// Let S(i) = sum of min elements over all such subsets

// A(-1) = 0, A(0) = 1, A(1) = 2
// A(i) = A(i-1) + 1 + A(i-2)
// (don't use F[i]: A(i-1) subsets; use F[i] alone: 1; use F[i] with something from {F[0],...,F[i-2]}: A(i-2))

// S(-1) = 0, S(0) = 1, S(1) = 1+2 = 3
// S(i) = S(i-1) + F[i] + S(i-2)
// (don't use F[i]: min is determined by lower; use F[i] alone: min = F[i];
//  use F[i] with something from {F[0],...,F[i-2]}: min is from lower part = S(i-2))
// Wait, that's wrong. When we use F[i] with subsets from {F[0],...,F[i-2]},
// the minimum component is the minimum of F[i] and the subset's minimum.
// Since F[i] > anything in {F[0],...,F[i-2]}, the minimum is the subset's minimum.
// So S(i) = S(i-1) + F[i]*1 + S(i-2) is correct!
// No wait: F[i] alone -> min = F[i], that's F[i]*1
// F[i] with a subset from {F[0],...,F[i-2]} -> min = min of that subset
// So sum of mins = S(i-2)
// Without F[i] -> sum of mins from {F[0],...,F[i-1]} = S(i-1)
// Total: S(i) = S(i-1) + F[i] + S(i-2) ✓

// Now for the digit DP on Zeckendorf(n) = {z[0] > z[1] > ... > z[m-1]}:
// Process from z[0] (highest) to z[m-1] (lowest).
//
// At position z[j], we know this bit is 1. We consider numbers that:
// - Match n's Zeckendorf repr at positions z[0]..z[j-1] (those are all 1)
// - Have 0 at position z[j] (i.e., they are smaller than n with this prefix)
// - Can freely choose from positions below z[j] (but respecting non-consecutive
//   and also can't be adjacent to z[j-1])
//
// Available lower positions: {F[0], ..., F[z[j]-1]} but not F[z[j-1]-1] if z[j-1] = z[j]+1
// Actually in Zeckendorf, no two consecutive indices, so z[j-1] >= z[j]+2.
// So if this bit is turned off, we can freely use {F[0], ..., F[z[j]-1]}
// but we need to check adjacency with the previous selected bit.
// Since z[j-1] >= z[j]+2, positions below z[j] are all at least 2 away from z[j-1],
// but z[j]-1 might be adjacent to z[j-1]-hmm, no. z[j-1] >= z[j]+2, so
// F[z[j]-1] has index z[j]-1 which is >= 3 away from z[j-1]. Safe.
//
// Wait, I need to reconsider. When we "turn off" bit z[j] and pick from below,
// the previous ON bit is at z[j-1] (if j>0). The highest available index below z[j]
// is z[j]-1. Since z[j-1] >= z[j]+2, we have z[j]-1 <= z[j-1]-3, so no adjacency issue.
//
// For the last bit z[m-1], if we turn it off, we just get 0 additional,
// but we already have the prefix sum. Actually, we should also account for
// the partial sum from bits we've already committed to.

// Let me restructure:
// ans = 0
// prefix = set of committed Fibonacci numbers (from high to low)
// For each bit position z[j] in the Zeckendorf representation:
//   - "Free" contribution: numbers that match prefix so far, have 0 at z[j],
//     and choose any valid subset from {F[0],...,F[z[j]-1]}.
//     But wait, we also need z[j] to be free (not adjacent to z[j-1]).
//     Since Zeckendorf guarantees z[j-1]>=z[j]+2, index z[j]-1 is fine.
//
//     For each such subset S': H(prefix + S') = min(min(prefix), min(S'))
//     But min(prefix) includes z[j]...z[m-1] which we haven't committed yet.
//     Actually, prefix committed so far = {z[0], ..., z[j-1]}.
//     So min(prefix) = F[z[j-1]] (smallest so far).
//     But we're turning off z[j] and below, so the full number is
//     sum of F[z[0]]...F[z[j-1]] + sum(S') where S' subset of {F[0],...,F[z[j]-1]}.
//
//     If S' is empty: the number is sum of F[z[0]]...F[z[j-1]], and
//       H = F[z[j-1]] (the smallest component). But only if j > 0.
//       If j = 0, we're choosing from below z[0] with nothing committed.
//       In that case, these are just numbers < F[z[0]] using {F[0],...,F[z[0]-1]}.
//
//     If S' is non-empty: H = min(F[z[j-1]], min(S')).
//       Since z[j-1] >= z[j]+2 and S' uses indices <= z[j]-1 < z[j-1],
//       min(S') < F[z[j-1]], so H = min(S').
//       Contribution = S(z[j]-1) (sum of minimums over all non-empty subsets of {F[0],...,F[z[j]-1]}).
//       But wait - we also need to respect non-adjacency with z[j-1]...
//       No, the positions available are {F[0],...,F[z[j]-1]} and z[j-1] >= z[j]+2,
//       so z[j]-1 is at most z[j-1]-3. No conflict.
//       But also need non-adjacency WITHIN the subset, which is already guaranteed by A/S.
//
//     If S' is empty and j > 0: contribution = F[z[j-1]] (one number, its H value)
//     But we should only count this once (when we first have the opportunity).
//     Actually this case is: the number equals F[z[0]]+...+F[z[j-1]] exactly,
//     and we skip all remaining bits. This will be counted when j is the first
//     bit we turn off. But we process all bits, so we might double count.
//
//     Let me think again more carefully...
//
// Actually, let's think of it differently. The Zeckendorf representation
// of n has bits at positions z[0] > z[1] > ... > z[m-1].
// Numbers in [1, n] can be enumerated by finding where they first differ from n.
//
// For k < n: find the highest position where k's Zeckendorf differs from n's.
// At that position, k has 0 while n has 1. Below that, k can be anything valid.
//
// For k = n: add H(n) = F[z[m-1]].
//
// So: G(n) = H(n) + sum over j=0..m-1 of:
//   sum of H(x) over all x that agree with n at positions z[0]..z[j-1]
//   and have 0 at position z[j], and have any valid non-consecutive subset below z[j].
//
//   But "agree at z[0]..z[j-1]" means those bits are 1. Plus position z[j] is 0.
//   The available positions below z[j] are {0, ..., z[j]-1}, but we must also
//   avoid adjacency with z[j-1] (if j>0). Since z[j-1] >= z[j]+2, the highest
//   available position z[j]-1 is at most z[j-1]-3, so no conflict.
//
//   For the numbers counted here:
//   Components = {F[z[0]], ..., F[z[j-1]]} union S' where S' is a subset of
//   non-consecutive elements from {F[0], ..., F[z[j]-1]}.
//
//   If j=0 and S'=empty: number = 0, not counted (we want [1,n]).
//   If j>0 and S'=empty: number = F[z[0]]+...+F[z[j-1]], H = F[z[j-1]].
//   If S' non-empty: H = min(F[z[j-1]], min(S')). Since j>0 => F[z[j-1]] > max(S'),
//     so H = min(S'). If j=0, then H = min(S').
//
//   Case j=0: contribution = S(z[0]-1)  (sum of mins over all non-empty subsets)
//   Case j>0: contribution = F[z[j-1]] + S(z[j]-1)
//     (F[z[j-1]] for the empty S' case, S(z[j]-1) for non-empty S')
//     But wait - when S' is non-empty and j>0, min(S') <= F[z[j]-1] < F[z[j-1]],
//     so H = min(S'). Sum of these = S(z[j]-1). ✓
//     When S' is empty and j>0, H = F[z[j-1]]. ✓
//
// But we also need to handle adjacency between z[j] and z[j]-1 correctly...
// Actually z[j] is set to 0, so no adjacency issue from z[j].
// The subset S' must be internally non-consecutive, that's handled by A/S.
// And the highest element of S' (at most z[j]-1) must not be adjacent to any
// committed element. The nearest committed element above is z[j-1] (if j>0),
// and z[j-1] >= z[j]+2, so z[j]-1 <= z[j-1]-3. No conflict. ✓

// But wait, I also need to worry about adjacency between z[j] and z[j+1]:
// Actually z[j] is the position we're turning off. Below z[j], we use S'.
// Between z[j-1] and z[j], positions are all 0 in n's Zeckendorf.
// Below z[j], we freely choose. So no issue.

// There's another subtlety: between z[j] and z[j+1] in n's representation,
// there may be intermediate positions that are 0. When we turn off z[j],
// we're allowing positions below z[j] to be freely chosen, which includes
// positions that in n are 0. That's fine.

// Actually I realize there could be adjacency issues within S' and with the
// gap between z[j-1] and positions in S'. Let me re-examine:
// Available positions for S': {0, 1, ..., z[j]-1}
// z[j-1] >= z[j] + 2 (Zeckendorf property)
// Highest available: z[j]-1. Distance to z[j-1]: z[j-1] - (z[j]-1) >= 3. Safe.
// S' must be internally non-consecutive: guaranteed by definition of A(i), S(i).
//
// But actually, I need S' to use non-consecutive indices from {0,...,z[j]-1}.
// That's exactly what A(z[j]-1) counts.

// Hmm, but z[j]-1 might be index z[j]-1. Let me reindex.
// fib[0]=1, fib[1]=2, fib[2]=3, fib[3]=5, ...
// Available: fib[0] through fib[z[j]-1], with non-consecutive indices.
// A(z[j]-1) and S(z[j]-1) handle this correctly.

// When z[j] = 0: no positions available below, S' must be empty.
// If j=0: no contribution (number would be 0).
// If j>0: contribution = F[z[j-1]].

// Final formula:
// G(n) = F[z[m-1]]  (for k=n itself)
//       + sum_{j=0}^{m-1} contribution(j)
// where contribution(j):
//   if j=0: S(z[0]-1)  [or 0 if z[0]=0]
//   if j>0: F[z[j-1]] + S(z[j]-1)  [S(z[j]-1)=0 if z[j]=0]

// Wait, when j=0 and z[0]=0, S' is empty, number is 0, not counted. Contribution = 0. ✓
// When j>0 and z[j]=0, S' must be empty, contribution = F[z[j-1]].
//   S(z[j]-1) = S(-1) = 0. So formula F[z[j-1]] + S(-1) = F[z[j-1]]. ✓

// Let me verify with small cases.
// n=1: Zeckendorf = {0} (fib[0]=1). z = [0], m=1.
// G(1) = H(1) = 1.
// Formula: F[z[0]] (=F[0]=1) + contribution(j=0) = S(z[0]-1) = S(-1) = 0.
// Total = 1 + 0 = 1. ✓

// n=2: Zeckendorf = {1} (fib[1]=2). z = [1], m=1.
// G(2) = H(1)+H(2) = 1+2 = 3.
// Formula: F[z[0]]=F[1]=2 + contribution(j=0) = S(z[0]-1) = S(0) = 1.
// Total = 2 + 1 = 3. ✓

// n=3: Zeckendorf = {2} (fib[2]=3). z = [2], m=1.
// G(3) = 1+2+3 = 6.
// Formula: F[2]=3 + S(1).
// S(0)=1 (just {fib[0]=1}, min=1).
// S(1)= S(0) + F[1] + S(-1) = 1 + 2 + 0 = 3.
// Total = 3 + 3 = 6. ✓

// n=4: Zeckendorf of 4 = {2, 0} (3+1). z = [2, 0], m=2.
// G(4) = H(1)+H(2)+H(3)+H(4) = 1+2+3+1 = 7.
// Formula: F[z[m-1]]=F[0]=1 + contribution(j=0) + contribution(j=1).
// j=0: S(z[0]-1) = S(1) = 3.
// j=1: F[z[0]]=F[2]=3 + S(z[1]-1) = S(-1) = 0. So contribution = 3.
// Total = 1 + 3 + 3 = 7. ✓

// n=5: Zeckendorf of 5 = {3} (fib[3]=5). z=[3], m=1.
// G(5) = 1+2+3+1+5 = 12.
// Formula: F[3]=5 + S(2).
// S(2) = S(1) + F[2] + S(0) = 3 + 3 + 1 = 7.
// Total = 5 + 7 = 12. ✓

// Great! The formula works. Now implement with __int128 for large numbers.

int main() {
    precompute_fib();
    int sz = fib.size();

    // Precompute A[i] and S[i]
    // A[i] = number of non-empty subsets of non-consecutive elements from {fib[0],...,fib[i]}
    // S[i] = sum of minimums over all such subsets
    // A[-1] = 0, S[-1] = 0
    // A[i] = A[i-1] + 1 + A[i-2]  (for i >= 1; A[0] = 1)
    // S[i] = S[i-1] + fib[i] + S[i-2] (for i >= 1; S[0] = fib[0] = 1)

    // These grow like Fibonacci, so they will overflow ll. Use __int128.
    vector<lll> A(sz), S(sz);
    A[0] = 1;
    S[0] = fib[0]; // = 1
    if (sz > 1) {
        A[1] = 2; // subsets: {fib[0]}, {fib[1]}
        S[1] = fib[0] + fib[1]; // = 1 + 2 = 3
    }
    for (int i = 2; i < sz; i++) {
        A[i] = A[i-1] + 1 + A[i-2];
        S[i] = S[i-1] + (lll)fib[i] + S[i-2];
    }

    ll n = 23416728348467685LL;
    vector<int> z = zeckendorf(n);
    int m = z.size();

    // G(n) = F[z[m-1]] + sum of contributions
    lll ans = fib[z[m-1]];

    for (int j = 0; j < m; j++) {
        if (j == 0) {
            if (z[0] > 0) {
                ans += S[z[0] - 1];
            }
        } else {
            ans += fib[z[j-1]]; // empty S' case
            if (z[j] > 0) {
                ans += S[z[j] - 1];
            }
        }
    }

    // Print __int128
    // ans should fit in a reasonable range. Let's check.
    // Actually for PE problems the answer is usually manageable.
    // Let me print it.

    // Convert __int128 to string
    string result;
    lll tmp = ans;
    if (tmp == 0) {
        result = "0";
    } else {
        bool neg = false;
        if (tmp < 0) { neg = true; tmp = -tmp; }
        while (tmp > 0) {
            result += ('0' + (int)(tmp % 10));
            tmp /= 10;
        }
        if (neg) result += '-';
        reverse(result.begin(), result.end());
    }
    cout << result << endl;

    return 0;
}

Python project_euler/problem_692/solution.py

"""
Project Euler Problem 692: Siegbert and Jo

Fibonacci Nim variant. H(N) = smallest Fibonacci number in Zeckendorf
representation of N. Find G(23416728348467685) = sum of H(k) for k=1..n.

Uses digit-DP on Zeckendorf representation with precomputed:
  A[i] = count of non-empty non-consecutive subsets from {F[0],...,F[i]}
  S[i] = sum of minimums over such subsets
"""

def solve():
    # Zeckendorf Fibonacci: 1, 2, 3, 5, 8, 13, ...
    fib = [1, 2]
    while fib[-1] + fib[-2] <= 5 * 10**16:
        fib.append(fib[-1] + fib[-2])

    sz = len(fib)

    # Precompute A[i] and S[i]
    A = [0] * sz
    S = [0] * sz
    A[0] = 1
    S[0] = 1
    A[1] = 2
    S[1] = 3
    for i in range(2, sz):
        A[i] = A[i-1] + 1 + A[i-2]
        S[i] = S[i-1] + fib[i] + S[i-2]

    def zeckendorf(n):
        rep = []
        for i in range(sz - 1, -1, -1):
            if fib[i] <= n:
                rep.append(i)
                n -= fib[i]
        return rep

    def G(n):
        z = zeckendorf(n)
        m = len(z)
        ans = fib[z[m-1]]
        for j in range(m):
            if j == 0:
                if z[0] > 0:
                    ans += S[z[0] - 1]
            else:
                ans += fib[z[j-1]]
                if z[j] > 0:
                    ans += S[z[j] - 1]
        return ans

    n = 23416728348467685
    print(G(n))

if __name__ == "__main__":
    solve()