设有理数数列 \(x_1, x_2, \dots\) 定义如下:\(x_1 = \frac{25}{...

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سؤال

设有理数数列 x1,x2,…x_1, x_2, \dots 定义如下:x1=2511x_1 = \frac{25}{11},且对于所有 kk 有 xk+1=13(xk+1xk−1).x_{k+1} = \frac{1}{3}\left(x_k + \frac{1}{x_k} - 1\right). 其中 x2025x_{2025} 可以表示为互质正整数 mm 和 nn 的分数 mn\frac{m}{n}。求 m+nm+n 除以 10001000 的余数。

الفكرة

Exploring recurrence

We’re given a recurrence for a sequence x1,x2,…x_1, x_2, \ldots, with the first term x1=2511x_1 = \frac{25}{11} and the recurrence formula:

xk+1=13(xk+1xk−1)x_{k+1} = \frac{1}{3} \left( x_k + \frac{1}{x_k} - 1 \right)

I’m trying to figure out how to handle this recurrence by possibly simplifying it with substitution. I wonder if it helps to introduce a new variable, like yk=1xky_k = \frac{1}{x_k}, to gradually simplify it, or maybe find a telescoping pattern.

Let’s analyze further!

Analyzing recurrence transformations

I’m trying different transformations for the recurrence: maybe xk=akbkx_k = \frac{a_k}{b_k}, or considering similarities to cosine recurrences. I think the presence of −1-1 inside the recurrence makes it not exactly like a Chebyshev recurrence, though.

Let’s try defining uk=xk+1xku_k = x_k + \frac{1}{x_k} and see if it helps simplify things, but that doesn’t seem to work either. I might consider solving for fixed points, which gives me potential limit values of x=12x = \frac{1}{2} or −1-1. Still need to figure out the exact terms for x2025x_{2025}.

None

Exploring transformations and recurrences

I’ve tried different transformations to handle the recurrence, looking for a simpler approach. Using substitutions like yk=xk−1xk+1y_k = \frac{x_k - 1}{x_k + 1} got us somewhere, as it turns the recurrence into a well-known form resembling a Chebyshev recurrence. I found y1=718y_1 = \frac{7}{18}, and with that, I computed successive terms in the sequence. But this quickly becomes messy. This recurrence likely connects with recursive structures, so let’s see if we can find a more direct approach or closed form.

Exploring recurrence patterns

I’m still trying to find a simpler way to handle the recurrence. Using the substitution yk=xk−1xk+1y_k = \frac{x_k - 1}{x_k + 1} led to a recurrence that resembles the cosine triple angle identity, but with a twist due to the coefficients. Now, I’m trying to check if the sequence is periodic by calculating the terms one after another. It looks like the sequence might oscillate after a bit, with y1≈0.3889y_1 \approx 0.3889 and y2≈−0.2731y_2 \approx -0.2731. If it’s periodic, then I’ll try using mod-based tricks to find the exact term.

Investigating recurrence structure

I’m considering if the recurrence can be represented by a projective matrix, defining the sequence as xk=akbkx_k = \frac{a_k}{b_k} with initial values a1=25a_1 = 25 and b1=11b_1 = 11. From this, I derive relationships for ak+1a_{k+1} and bk+1b_{k+1}, and I notice patterns forming in terms of sums, ak+bk=36(2k−1)a_k + b_k = 36^{(2^{k-1})}. I'll also need to track the greatest common divisor dkd_k, but the key now seems to be understanding how this recurrence impacts the power of 3 in the terms.

None

推理关于有理数数列 for 3m 59s

الإجابة

我们首先将 xkx_k 写成分数形式:设

xk=akbk(ak,bk∈Z, gcd⁡(ak,bk)=1).x_k=\frac{a_k}{b_k}\quad (a_k,b_k\in\mathbb{Z},\,\gcd(a_k,b_k)=1).

由题中递推关系

xk+1=13(xk+1xk−1),x_{k+1}=\frac{1}{3}\Bigl(x_k+\frac{1}{x_k}-1\Bigr),

写为

xk+1=13(akbk+bkak−1)=ak2+bk2−akbk3akbk.x_{k+1}=\frac{1}{3}\Bigl(\frac{a_k}{b_k}+\frac{b_k}{a_k}-1\Bigr) =\frac{a_k^2+b_k^2-a_kb_k}{3a_kb_k}.

也就是说,我们可以取

ak+1=ak2+bk2−akbk,bk+1=3akbk.a_{k+1}=a_k^2+b_k^2-a_kb_k,\qquad b_{k+1}=3a_kb_k.

求和不变量

注意到

ak+1+bk+1=(ak2+bk2−akbk)+3akbk=ak2+bk2+2akbk=(ak+bk)2.a_{k+1}+b_{k+1}=(a_k^2+b_k^2-a_kb_k)+3a_kb_k=a_k^2+b_k^2+2a_kb_k=(a_k+b_k)^2.

所以

ak+1+bk+1=(ak+bk)2.a_{k+1}+b_{k+1}=(a_k+b_k)^2.

初始时 a1=25,  b1=11a_1=25,\;b_1=11 得

a1+b1=36.a_1+b_1=36.

由递推可知

ak+bk=36 2 k−1.a_k+b_k=36^{\,2^{\,k-1}}.

因此特别地,

a2025+b2025=36 22024.a_{2025}+b_{2025}=36^{\,2^{2024}}.

整除因素(约去的因子)

由于 bk+1=3akbkb_{k+1}=3a_kb_k显然含有因子3,而 ak+1a_{k+1} 中也可能带入因子。仔细证明后可发现(也可通过前几步计算验证)当我们把 ak,bka_k,b_k 分解成

ak=DkA,bk=DkB,gcd⁡(A,B)=1,a_k=D_kA,\quad b_k=D_kB,\quad\gcd(A,B)=1,

则经过递推可以证明

Dk+1=Dk2⋅G(A,B)D_{k+1}=D_k^2\cdot G(A,B)

其中

G(A,B)=gcd⁡(A2+B2−AB, 3AB).G(A,B)=\gcd\Bigl(A^2+B^2-AB,\,3AB\Bigr).

对于初始 a1=25,  b1=11a_1=25,\;b_1=11(注意 25≡1,  11≡2(mod3)25\equiv1,\;11\equiv2\pmod3)可以证明归约后的 A,BA,B 满足 A≡1,  B≡2(mod3)A\equiv1,\;B\equiv2\pmod3,从而 G(A,B)=3G(A,B)=3;进而证明

v3(Dk)=2 k−1−1,(k≥2),v_3(D_{k})=2^{\,k-1}-1,\quad (k\ge2),

即 DkD_k 恰好吸去了所有3的幂,且

D2025=3 22024−1.D_{2025}=3^{\,2^{2024}-1}.

表示 x2025x_{2025} 的和

写出 x2025=a2025b2025x_{2025}=\frac{a_{2025}}{b_{2025}} 的既约形式,即设

a2025=m,b2025=n,gcd⁡(m,n)=1.a_{2025}=m,\quad b_{2025}=n,\quad \gcd(m,n)=1.

注意

m+n=a2025+b2025D2025=36 220243 22024−1.m+n=\frac{a_{2025}+b_{2025}}{D_{2025}} =\frac{36^{\,2^{2024}}}{3^{\,2^{2024}-1}}.

写 36=22⋅3236=2^2\cdot3^2 得

36 22024=22⋅22024⋅32⋅22024=222025⋅3 22025.36^{\,2^{2024}}=2^{2\cdot2^{2024}}\cdot3^{2\cdot2^{2024}} =2^{2^{2025}}\cdot3^{\,2^{2025}}.

因此

m+n=  222025⋅3 22025−(22024−1)=222025⋅3 22024+1.m+n=\;2^{2^{2025}}\cdot3^{\,2^{2025}-(2^{2024}-1)} =2^{2^{2025}}\cdot3^{\,2^{2024}+1}.

求 m+nm+n 模 1000 的余数

我们需要计算

N=222025⋅3 22024+1(mod1000).N=2^{2^{2025}}\cdot3^{\,2^{2024}+1}\quad\pmod{1000}.

由于 1000=8⋅1251000=8\cdot125,利用中国剩余定理分开求解。

(1)模 8 部分

注意 2220252^{2^{2025}} 的指数远大于3,因此

222025≡0(mod8).2^{2^{2025}}\equiv0\pmod{8}.

于是 N≡0(mod8)N\equiv0\pmod8.

(2)模 125 部分

利用欧拉定理,φ(125)=100\varphi(125)=100。于是我们只需要将指数模 100 化简。

计算 222025(mod125)2^{2^{2025}}\pmod{125}:

先求 E=22025(mod100)E=2^{2025}\pmod{100}。

  • 模 4:对于 n≥2n\ge2 有 2n≡0(mod4)2^n\equiv0\pmod4,故 22025≡0(mod4)2^{2025}\equiv0\pmod4。
  • 模 25:φ(25)=20\varphi(25)=20,故 22025≡22025mod  20(mod25)2^{2025}\equiv2^{2025\mod20}\pmod{25}。注意 2025÷202025\div20 得余 2025−20⋅101=52025-20\cdot101=5,故 22025≡25=32≡7(mod25)2^{2025}\equiv2^5=32\equiv7\pmod{25}.

解联立:

{x≡0(mod4),x≡7(mod25).\begin{cases} x\equiv0\pmod4,\\[1mm] x\equiv7\pmod{25}. \end{cases}

令 x=25k+7x=25k+7,要求 25k+7≡0(mod4)25k+7\equiv0\pmod4。因 25≡1(mod4)25\equiv1\pmod4,故条件为 k+7≡0(mod4)k+7\equiv0\pmod4即 k≡1(mod4)k\equiv1\pmod4。取最小 k=1k=1 得 x=25+7=32x=25+7=32. 所以

22025≡32(mod100).2^{2025}\equiv32\pmod{100}.

因此,由于 2220252^{2^{2025}} 的指数可在模125中按指数100取余,

222025≡232(mod125).2^{2^{2025}}\equiv2^{32}\pmod{125}.

计算 232(mod125)2^{32}\pmod{125}:

  • 210=1024≡24(mod125)2^{10}=1024\equiv24\pmod{125};
  • 220≡242=576≡576−4⋅125=576−500=76(mod125)2^{20}\equiv24^2=576\equiv576-4\cdot125=576-500=76\pmod{125};
  • 230≡76⋅24=1824≡1824−14⋅125=1824−1750=74(mod125)2^{30}\equiv76\cdot24=1824\equiv1824-14\cdot125=1824-1750=74\pmod{125};
  • 232=230⋅22≡74⋅4=296≡296−2⋅125=296−250=46(mod125)2^{32}=2^{30}\cdot2^2\equiv74\cdot4=296\equiv296-2\cdot125=296-250=46\pmod{125}. 记得 A=222025≡46(mod125)A=2^{2^{2025}}\equiv46\pmod{125}.
计算 3 22024+1(mod125)3^{\,2^{2024}+1}\pmod{125}:

同样先求 22024(mod100)2^{2024}\pmod{100}:

  • 模 4:22024≡0(mod4)2^{2024}\equiv0\pmod4;
  • 模 25:2024mod  20=2024−20⋅101=42024\mod20=2024-20\cdot101=4,故 22024≡24=16(mod25)2^{2024}\equiv2^4=16\pmod{25}。 解联立得 22024≡16(mod100)2^{2024}\equiv16\pmod{100}(可证明唯一解是16)。 因此 22024+1≡16+1=17(mod100)2^{2024}+1\equiv16+1=17\pmod{100}。 利用欧拉定理,322024+1≡317(mod125)3^{2^{2024}+1}\equiv3^{17}\pmod{125}。 计算 3173^{17}:
31=3,  32=9,  33=27,  34=81,  35=243≡118,  36≡118⋅3=354≡104,3^1=3,\;3^2=9,\;3^3=27,\;3^4=81,\;3^5=243\equiv118,\;3^6\equiv118\cdot3=354\equiv104, 37≡104⋅3=312≡62,  38≡62⋅3=186≡61,  39≡61⋅3=183≡58,3^7\equiv104\cdot3=312\equiv62,\;3^8\equiv62\cdot3=186\equiv61,\;3^9\equiv61\cdot3=183\equiv58, 310≡58⋅3=174≡49,  311≡49⋅3=147≡22,  312≡22⋅3=66,3^{10}\equiv58\cdot3=174\equiv49,\;3^{11}\equiv49\cdot3=147\equiv22,\;3^{12}\equiv22\cdot3=66, 313≡66⋅3=198≡73,  314≡73⋅3=219≡94,  315≡94⋅3=282≡32,3^{13}\equiv66\cdot3=198\equiv73,\;3^{14}\equiv73\cdot3=219\equiv94,\;3^{15}\equiv94\cdot3=282\equiv32, 316≡32⋅3=96,  317≡96⋅3=288≡288−2⋅125=38(mod125).3^{16}\equiv32\cdot3=96,\;3^{17}\equiv96\cdot3=288\equiv288-2\cdot125=38\pmod{125}.

记 B=322024+1≡38(mod125)B=3^{2^{2024}+1}\equiv38\pmod{125}.

于是

N≡A⋅B≡46⋅38(mod125).N\equiv A\cdot B\equiv46\cdot38\pmod{125}.

计算 46⋅38=174846\cdot38=1748,

1748−125⋅13=1748−1625=123,1748-125\cdot13=1748-1625=123,

所以 N≡123(mod125)N\equiv123\pmod{125}.

利用中国剩余定理

我们已知:

N≡0(mod8),N≡123(mod125).N\equiv0\pmod8,\qquad N\equiv123\pmod{125}.

设 N=125k+123N=125k+123 ,要求 125k+123≡0(mod8)125k+123\equiv0\pmod8。 注意 125≡5(mod8)125\equiv5\pmod8,故有

5k+123≡0(mod8).5k+123\equiv0\pmod8.

而 123≡3(mod8)123\equiv3\pmod8,所以

5k+3≡0(mod8)⟹5k≡5(mod8).5k+3\equiv0\pmod8\quad\Longrightarrow\quad5k\equiv5\pmod8.

因 55 在模8中可逆(逆元为5,因为 5⋅5=25≡1(mod8)5\cdot5=25\equiv1\pmod8),故 k≡1(mod8)k\equiv1\pmod8。 取最小正整数 k=1k=1 得

N=125⋅1+123=248.N=125\cdot1+123=248.

结论

最终,x2025=mnx_{2025}=\frac{m}{n}(两者互质)满足

m+n≡248(mod1000).m+n\equiv248\pmod{1000}.

答案是 248.

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