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  "body": "\n## Proof\n\nCall a finite nonempty set $C = \\{c_1 \u003c c_2 \u003c \\dots \u003c c_r\\}$ of integers admissible if $c_1 \\ge 2$ and $c_{i+1} \\ge c_i + 2$ for $1 \\le i \u003c r$.\nWrite $\\Sigma(C) = F_{c_1} + \\dots + F_{c_r}$, where $F_k$ are the [[Fibonacci numbers|0GA9K57A39]].\nWe prove existence by [[strong induction|0KMVFSHFGZ]], and uniqueness by comparing the largest terms.\n\n1. For $m \\ge 1$ we have $F_{m+1} = F_m + F_{m-1} \\ge F_m$, by the [[Fibonacci recurrence|0D0SMQYEJA]] and $F_{m-1} \\ge 0$. So $(F_m)_{m \\ge 1}$ is nondecreasing.\n2. For $m \\ge 2$ we have $F_{m-1} \\ge F_1 = 1$ by step 1. Hence $F_{m+1} = F_m + F_{m-1} \u003e F_m$. So $(F_m)_{m \\ge 2}$ is strictly increasing, and $F_m \\ge m - 1$ for $m \\ge 2$.\n3. Claim: if $C$ is admissible with largest element $j$, then $\\Sigma(C) \u003c F_{j+1}$. We prove it by induction on the size $r$ of $C$.\n4. If $r = 1$, then $\\Sigma(C) = F_j \u003c F_{j+1}$ by step 2, since $j \\ge 2$.\n5. If $r \\ge 2$, let $C' = C \\setminus \\{j\\}$. It is admissible, and its largest element $c_{r-1}$ satisfies $c_{r-1} \\le j - 2$.\n6. By the induction hypothesis, $\\Sigma(C') \u003c F_{c_{r-1} + 1}$. Since $1 \\le c_{r-1} + 1 \\le j - 1$, step 1 gives $F_{c_{r-1}+1} \\le F_{j-1}$.\n7. Hence $\\Sigma(C) = F_j + \\Sigma(C') \u003c F_j + F_{j-1} = F_{j+1}$. This proves the claim.\n8. Existence. Let $N \\ge 1$, and assume that every positive integer smaller than $N$ is $\\Sigma(D)$ for some admissible $D$.\n9. We have $F_2 = 1 \\le N$, and $F_m \u003e N$ for $m \\ge N + 2$ by step 2. So there is a largest $j \\ge 2$ with $F_j \\le N$. Then $F_j \\le N \u003c F_{j+1}$.\n10. If $N = F_j$, then $C = \\{j\\}$ is admissible with $\\Sigma(C) = N$.\n11. Otherwise let $M = N - F_j$. Then $1 \\le M \u003c F_{j+1} - F_j = F_{j-1}$, and $M \u003c N$.\n12. By the induction hypothesis, $M = \\Sigma(D)$ for an admissible $D$ with largest element $d$. Then $F_d \\le M \u003c F_{j-1}$.\n13. If $d \\ge j - 1$, step 1 would give $F_d \\ge F_{j-1}$. So $d \\le j - 2$.\n14. Hence $C = D \\cup \\{j\\}$ is admissible, and $\\Sigma(C) = M + F_j = N$. By strong induction, every $N \\ge 1$ has an admissible representation.\n15. Uniqueness. Let $C$ and $D$ be admissible with $\\Sigma(C) = \\Sigma(D)$, and suppose $C \\ne D$.\n16. Let $C' = C \\setminus D$ and $D' = D \\setminus C$. Removing the common terms from both sums gives $\\Sigma(C') = \\Sigma(D')$, with the convention that the empty sum is $0$.\n17. Since $C \\ne D$, one of $C', D'$ is nonempty. A nonempty set gives a sum at least $F_2 = 1$, so the other set is also nonempty.\n18. A nonempty subset of an admissible set is admissible. So $C'$ and $D'$ are admissible and disjoint.\n19. Let $c$ and $d$ be the largest elements of $C'$ and $D'$. They differ, so by symmetry we may assume $d \u003c c$.\n20. By step 3 and step 1, $\\Sigma(D') \u003c F_{d+1} \\le F_c \\le \\Sigma(C')$. This contradicts $\\Sigma(C') = \\Sigma(D')$.\n21. Hence $C = D$. So each positive integer $N$ is $F_{c_1} + \\dots + F_{c_r}$ for exactly one admissible set $\\{c_1, \\dots, c_r\\}$.\n\n## Idea\n\nThe greedy choice works because, after taking the largest Fibonacci number $F_j \\le N$, the rest is smaller than $F_{j-1}$. So the next index is at most $j - 2$.\nUniqueness comes from one inequality: a sum of non-consecutive Fibonacci numbers is always smaller than the next Fibonacci number after its largest term.\n",
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