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"thinking": " Let me try different starting values and recurrence relations. With a₁=2, I get a₂=5/2 which isn't an integer, and a₁=1 gives the trivial solution. I'm wondering if the recurrence might follow a different pattern entirely—maybe something like aₙ₊₁ = aₙ² - aₙ + 1 or aₙ₊₁ = aₙ² + 1 instead.\n\nTesting aₙ₊₁ = aₙ² + 1 with a₁=2 gives 2, 5, 26, 677, but these aren't pairwise coprime. The recurrence aₙ₊₁ = aₙ² - aₙ + 1 starting from a₁=2 produces 2, 3, 7, 43, 1807—which are pairwise coprime, a classic result. But I need to reconcile this with the original formula (aₙ² + 1)/2. Maybe the starting value isn't 2, or perhaps if a₁ is odd, the formula produces integers. Let me check a₁=3: that gives a₂=5, a₃=13. I'm testing a few sequences with different starting values and recurrences to see which one produces pairwise coprime terms. The sequence with a₁ = 2 and a_{n+1} = a_n² - a_n + 1 seems like the most likely candidate—it's a classic variant related to Sylvester's sequence that should have the coprimality property.\n\nLet me verify this works: if a_{n+1} - 1 = a_n(a_n - 1), then by telescoping I get a_{n+1} - 1 = a_1 a_2 ... a_n, which means a_{n+1} ≡ 1 (mod a_i) for all earlier terms, guaranteeing they're pairwise coprime. The user's original formula might have been a transcription error or translation issue from the source problem.\n\nNow I'm checking whether the substitution b_n = 2a_n - 1 transforms this into the (b_n² + 1)/2 recurrence. The algebra checks out: b_{n+1} = (b_n² + 1)/2 with b_1 = 3 gives the sequence 3, 5, 13, 85, 3613. But here's the issue—gcd(5, 85) = 5, so the b sequence isn't pairwise coprime even though the original a sequence is. This means the transformation preserves the recurrence but not the coprimality property.\n\nI'm realizing the original problem statement might have an error. Let me test different starting values and recurrences. With a_1 = 1 and a_{n+1} = 2a_n² - 1, I just get 1, 1, 1... which is trivial. But if I try a_{n+1} = (a_n - 1)² + 1 starting with a_1 = 3, I get 3, 5, 17, 257—these are Fermat numbers, which are famously pairwise coprime. This looks like the intended recurrence. I'm going to go with the most likely interpretation: a₁ = 2 and a_{n+1} = a_n² - a_n + 1, which is the classic Sylvester sequence variant. I'll present this solution while noting the apparent typo in the original problem statement. I'm deriving the key identity: a_{n+1} - 1 equals the product of all previous terms, which means a_{n+1} ≡ 1 (mod a_i) for any i ≤ n. This immediately gives pairwise coprimality since gcd(a_m, a_n) = 1 for m > n. I should just commit to interpreting this as the Sylvester sequence with the recurrence a_{n+1} = a_n² - a_n + 1, since that's the classic problem and the given formula likely contains a typo.",
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"
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"text": "# 解答\n\n## 关于题目的说明\n\n按照题目字面 $a_1=2$ 及 $a_{n+1}=\\dfrac{a_n^2+1}{2}$ 计算会得到 $a_2=\\dfrac{5}{2}$,不是整数。结合 $a_1=2$ 且\"数列中任意两项互质\"这一经典结论,**正确的递推式应为**:\n$$a_{n+1}=a_n^2-a_n+1$$\n\n这就是著名的 **Sylvester 数列**:$2,\\ 3,\\ 7,\\ 43,\\ 1807,\\ \\ldots$ 下面按此给出证明。\n\n---\n\n## 证明思路\n\n**核心想法**:把递推式写成一个\"连乘\"的形式,使得 $a_{n+1}$ 对任何 $a_i\\,(i\\le n)$ 取余都等于 $1$。\n\n## 详细步骤\n\n### 第一步:变形递推式\n\n$$a_{n+1}-1=a_n^2-a_n=a_n(a_n-1)$$\n\n### 第二步:迭代展开\n\n将 $a_n-1=a_{n-1}(a_{n-1}-1)$ 代入上式:\n$$a_{n+1}-1=a_n\\cdot a_{n-1}(a_{n-1}-1)$$\n\n继续将 $a_{n-1}-1=a_{n-2}(a_{n-2}-1)$ 代入:\n$$a_{n+1}-1=a_n\\cdot a_{n-1}\\cdot a_{n-2}(a_{n-2}-1)$$\n\n一直迭代到 $a_1-1=2-1=1$:\n\n$$a_{n+1}-1=a_n\\cdot a_{n-1}\\cdots a_2\\cdot a_1\\cdot (a_1-1)=a_1 a_2\\cdots a_n$$\n\n### 第三步:得到关键恒等式\n\n$$\\boxed{a_{n+1}=a_1 a_2\\cdots a_n+1}$$\n\n**用数学归纳法严格验证**:\n- $n=1$:$a_2=3=2+1=a_1+1$ ✓\n- 假设 $a_{n+1}=a_1 a_2\\cdots a_n+1$,则\n$$a_{n+2}=a_{n+1}^2-a_{n+1}+1=a_{n+1}(a_{n+1}-1)+1=a_{n+1}\\cdot a_1 a_2\\cdots a_n+1=a_1 a_2\\cdots a_{n+1}+1\\ \\checkmark$$\n\n### 第四步:证明任意两项互质\n\n任取 $m>n\\ge 1$。由第三步有\n$$a_m=a_1 a_2\\cdots a_{m-1}+1$$\n\n由于 $n\\le m-1$,所以 $a_n$ 是 $a_1 a_2\\cdots a_{m-1}$ 的一个因子,即\n$$a_1 a_2\\cdots a_{m-1}\\equiv 0\\pmod{a_n}$$\n\n因此\n$$a_m\\equiv 0+1\\equiv 1\\pmod{a_n}$$\n\n设 $d=\\gcd(a_m,a_n)$,则 $d\\mid a_n$ 且 $d\\mid a_m$,从而 $d\\mid (a_m-a_1a_2\\cdots a_{m-1})=1$。\n\n故 $d=1$,即 \n$$\\gcd(a_m,a_n)=1$$\n\n### 结论\n\n数列中任意两项互质。 $\\blacksquare$\n\n---\n\n## 关键点回顾\n\n| 步骤 | 技巧 |\n|------|------|\n| 变形 | $a_{n+1}-1=a_n(a_n-1)$ —— 构造出\"连乘\"结构 |\n| 迭代 | 把每个 $a_k-1$ 继续展开,利用 $a_1-1=1$ 终止 |\n| 恒等式 | $a_{n+1}=a_1a_2\\cdots a_n+1$ —— 这是全部论证的\"发动机\" |\n| 互质 | 利用 $a_m\\equiv 1\\pmod{a_n}$ |\n\n这种\"前面各项之积加 1\"的结构正是构造两两互质的经典手法(类似欧几里得证明素数无穷多的思路,也与 Fermat 数 $F_{n+1}-2=F_0F_1\\cdots F_n$ 同源)。",
"citations": null
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Let me try different starting values and recurrence relations. With a₁=2, I get a₂=5/2 which isn't an integer, and a₁=1 gives the trivial solution. I'm wondering if the recurrence might follow a different pattern entirely—maybe something like aₙ₊₁ = aₙ² - aₙ + 1 or aₙ₊₁ = aₙ² + 1 instead.\n\nTesting aₙ₊₁ = aₙ² + 1 with a₁=2 gives 2, 5, 26, 677, but these aren't pairwise coprime. The recurrence aₙ₊₁ = aₙ² - aₙ + 1 starting from a₁=2 produces 2, 3, 7, 43, 1807—which are pairwise coprime, a classic result. But I need to reconcile this with the original formula (aₙ² + 1)/2. Maybe the starting value isn't 2, or perhaps if a₁ is odd, the formula produces integers. Let me check a₁=3: that gives a₂=5, a₃=13. I'm testing a few sequences with different starting values and recurrences to see which one produces pairwise coprime terms. The sequence with a₁ = 2 and a_{n+1} = a_n² - a_n + 1 seems like the most likely candidate—it's a classic variant related to Sylvester's sequence that should have the coprimality property.\n\nLet me verify this works: if a_{n+1} - 1 = a_n(a_n - 1), then by telescoping I get a_{n+1} - 1 = a_1 a_2 ... a_n, which means a_{n+1} ≡ 1 (mod a_i) for all earlier terms, guaranteeing they're pairwise coprime. The user's original formula might have been a transcription error or translation issue from the source problem.\n\nNow I'm checking whether the substitution b_n = 2a_n - 1 transforms this into the (b_n² + 1)/2 recurrence. The algebra checks out: b_{n+1} = (b_n² + 1)/2 with b_1 = 3 gives the sequence 3, 5, 13, 85, 3613. But here's the issue—gcd(5, 85) = 5, so the b sequence isn't pairwise coprime even though the original a sequence is. This means the transformation preserves the recurrence but not the coprimality property.\n\nI'm realizing the original problem statement might have an error. Let me test different starting values and recurrences. With a_1 = 1 and a_{n+1} = 2a_n² - 1, I just get 1, 1, 1... which is trivial. But if I try a_{n+1} = (a_n - 1)² + 1 starting with a_1 = 3, I get 3, 5, 17, 257—these are Fermat numbers, which are famously pairwise coprime. This looks like the intended recurrence. I'm going to go with the most likely interpretation: a₁ = 2 and a_{n+1} = a_n² - a_n + 1, which is the classic Sylvester sequence variant. I'll present this solution while noting the apparent typo in the original problem statement. I'm deriving the key identity: a_{n+1} - 1 equals the product of all previous terms, which means a_{n+1} ≡ 1 (mod a_i) for any i ≤ n. This immediately gives pairwise coprimality since gcd(a_m, a_n) = 1 for m > n. I should just commit to interpreting this as the Sylvester sequence with the recurrence a_{n+1} = a_n² - a_n + 1, since that's the classic problem and the given formula likely contains a typo.","type":"reasoning.text"}"
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