设实数列 \(\{x_n\}\) 满足:\(x_0 = 0\),\(x_2 = \sqrt[3]{2...

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问题

设实数列 {xn}\{x_n\} 满足:x0=0x_0 = 0x2=23x1x_2 = \sqrt[3]{2}x_1x3x_3 是正整数,且 xn+1=143xn+43xn1+12xn2(n2).x_{n+1} = \frac{1}{\sqrt[3]{4}} x_n + \sqrt[3]{4} x_{n-1} + \frac{1}{2} x_{n-2} (n \geq 2). 问:这类数列中最少有多少个整数项?

思考

Examining the sequence

I’m analyzing a sequence with specific conditions: (xn),x0=0,x3(x_n), x_0 = 0, x_3 is a positive integer, and xn+1x_{n+1} is defined by a recurrence relation.

Evaluating sequence options

I'm analyzing whether integer terms can be efficiently minimized by adjusting x1x_1, given its potential as a free parameter. This involves examining its impact on the recurrence relation.

Searching for simplification

I’m reworking the recurrence with α=23\alpha = \sqrt[3]{2} to make solving easier and possibly find a closed-form expression or factorization for the characteristic polynomial.

Testing roots

I’m checking if α=23\alpha = \sqrt[3]{2} or α2\alpha^2 works as roots for the polynomial by substituting these values and checking if the polynomial evaluates to zero.

Revisiting the strategy

It dawns on me that perhaps I'm overcomplicating it. I should examine a solution like xn=cαnx_n = c \alpha^n to see if α\alpha fits the characteristic equation of the homogeneous part.

Piecing together

I’m noting that α\alpha isn't a root of the characteristic polynomial. Part of me is considering expressing xnx_n in terms of boundary conditions and initial terms to simplify the recurrence relation.

Expressing terms

I’m curious about expressing x3x_3 using x1x_1 and α\alpha, assuming x0=0x_0=0 and x2=αx1x_2=\alpha x_1, to understand the impact on integer values.

Clarifying integer constraints

I’m mapping out that for x3x_3 to be a positive integer, aa must satisfy a>0a > 0 and a(α2+1α)a(\alpha^2 + \frac{1}{\alpha}) must be an integer, which I’ll define as kk.

Crafting rationality

OK, let me see. Setting cc as a rational multiple of the reciprocal of α2+1α\alpha^2 + \frac{1}{\alpha} ensures c×(α2+1α)c \times (\alpha^2 + \frac{1}{\alpha}) is an integer.

Formulating the sequence

Working out x4x_4 with x3=kx_3=k, x2=αax_2=\alpha a, and x1=ax_1=a. This approach might hint at any patterns of forced integers in the sequence.

Revisiting patterns

Considering a new sequence defined by kk, I aim to identify any emerging patterns in terms like periodic structures involving α\alpha, possibly forcing integers at specific terms.

Examining terms

Calculating x4x_4 with kk, I’m aiming to identify any periodic patterns in the sequence, especially those potentially forcing integer values at specific terms.

Pulling together terms

I’m gathering x4x_4 under a common denominator involving α2+1α\alpha^2 + \frac{1}{\alpha} to streamline the arithmetic and possibly simplify the sequence pattern.

Mapping out simplification

I’m figuring out x4=kα2+5k2(α2+1α)x_4 = \frac{k}{\alpha^2} + \frac{5k}{2(\alpha^2 + \frac{1}{\alpha})}, aiming to identify conditions for integers or patterns in the sequence.

Mapping out integer possibilities

Considering if specific kk choices could accidentally force x4x_4 to be integer, despite α2\alpha^2 and α2+1α\alpha^2 + \frac{1}{\alpha} lacking a simple rational link.

Expressing and seeking patterns

I’m mapping out xnx_n in terms of x3x_3, x2x_2, and x1x_1, aiming to find patterns that might induce integer constraints at specific terms.

Breaking down the recurrence

I’m analyzing the recurrence to express x4x_4 in terms of x1x_1, considering integer constraints and crafting a strategy to ensure integer results at specific terms.

Ensuring integer accuracy

By plugging in the results and simplifying, I aim to confirm that x4x_4 and x5x_5 are integers, ensuring that the sequence maintains integer values at those terms.

Simplifying and verifying

I’m working through the recurrence relation to ensure x5x_5 is an integer, refining the expression and confirming each term aligns with the integer constraints.

Simplifying terms

I'm grouping and factoring terms to make sure x5x_5 is an integer, looking for patterns in x3mx_{3m} due to the third-order recurrence. Progress is steady but intricate.

Examining alternate recurrence

Now I'm testing if shifting nn by 2 in the original formula yields simplified formulas for x3mx_{3m}, aiming to identify a more streamlined expression for the sequence.

Analyzing linear recurrences

I'm mapping out the standard approach for linear recurrences, focusing on finding roots of the characteristic polynomial and determining conditions for integer x3x_3. Progressing through the calculation process.

Testing integer patterns

I'm checking if integer x3x_3 leads to integer x3mx_{3m} by calculating x6x_6 using x3,x2,x1x_3, x_2, x_1 and examining x4,x5x_4, x_5 with x1x_1.

Expressing sequences

I'm mapping out how x4x_4 and x5x_5 can be expressed as integer multiples of kk, and then expressing x6x_6 in terms of these to check if it too is an integer multiple of kk.

Assessing integer conditions

I'm working through whether g1+g2+g3g_1 + g_2 + g_3 is an integer. If it is, then x6x_6 is an integer multiple of kk, and thus an integer. Progressing towards a solution.

Breaking down the terms

Starting with g1=f5α2g1 = \frac{f5}{\alpha^2}, I’m recalculating each term to ensure consistency and accuracy in expressing g1g1. This helps validate the integer condition for x6x_6.

Clarifying g1's terms

I’m making sure the terms of g1g1 are accurate, including α\alpha and AA terms. This helps in validating if g1g1 is an integer.

Crunching rational parts

I’m piecing together the rational components of g1+g2+g3g1 + g2 + g3 to form 2.752.75 or 114\frac{11}{4}. Progress is being made by merging irrational parts as well.

Reworking H's simplification

I’m reworking HH by multiplying both sides by α\alpha to refine [7+10α3]/[4α(α2+1/α)][7 + 10\alpha^3] / [4\alpha(\alpha^2 + 1/\alpha)] and simplify the expression further.

Confirming integer status

By transforming HH, I’ve confirmed that the sum of g1+g2+g3g1 + g2 + g3 equals 5. This ensures x6x_6 is an integer and specifically an integer multiple of 5.

Linking patterns

OK, let me see. If x6x_6 is an integer multiple of 5, can x9x_9 follow a similar pattern with kk? This might indicate x9x_9 as another integer multiple of kk.

Piecing together recurrence

I’m working on a recurrence relation by reimagining x7x_7 with kk. This approach might streamline the sequence, making the pattern clearer and potentially more manageable.

Inspecting alpha^2 and alpha^4

I’m analyzing terms α2\alpha^2 and α4\alpha^4 to see if they align and form patterns. This progression leads me to a clearer understanding of the sequence and potential simplifications.

Summing up terms

I’m gathering the terms and simplifying them to get α2\alpha^2, α\alpha, and constants. The insightful pattern in the sums reveals an intriguing progression in the sequence.

Revisiting the terms

Breaking down S7S7 into α2\alpha^2, α\alpha, and constants, I verify if each term is an integer. Progressing through these calculations, I’m piecing together a clearer picture of the sequence.

Simplifying terms

I’m merging everything into a single fraction. This approach reveals a more concise form, streamlining the expression, and making it easier to identify the sequence pattern.

Breaking down terms

Interestingly enough, simplifying each segment with the same denominator reveals clearer forms for α2\alpha^2, α\alpha, and constants. This progression is crucial for understanding the sequence's pattern.

Mastering the sequence

I’m pulling the numerator and denominator together into a cleaner form. This setup simplifies the sum and reveals the sequence's pattern, which I can analyze further.

Breaking it down

After simplifying, x7x_7 turns out to be 41/(3α2)41 / (3 \alpha^2). This means x7x_7 isn’t an integer for a generic kk, unless kk is set to 0.

Spotting integer trends

Piecing together the sequence, it seems multiples of 3 are consistently integers. This hints that x9x_9 might be kk. Let me break this down further.

Pulling together

I’m outlining a direct approach to verify if x8x_8 and x9x_9 are integers by examining their linear combinations with rational coefficients. This progression helps identify any emerging patterns.

Refining definitions

I'm working through refining definitions for simplification. This leads me to stronger conclusions about the integer nature of sequences and shifts in the series.

Piecing together

OK, let me see. I’m carefully working through each term to simplify S9S9, ensuring integer outcomes for sequences. This involves intricate fraction management and focusing on alpha's relationships.

Breaking down fractional components

I’m simplifying each term in f5/(2α2)f5/(2 \alpha^2). This method ensures integer outcomes for sequences by carefully managing fractions and identifying integer patterns in the series.

Evaluating sub-terms

I’m pulling together four sub-terms, merging rational and alpha components, and experiment with combining constants and fractions to simplify the expression for clarity.

Summing and simplifying

I’m piecing together the terms, aligning the rational parts, and boosting clarity by simplifying fractions and decimals. This approach streamlines the overall expression.

Breaking down the calculations

I’m piecing together fractions and decimals, blending rational parts, and combining irrational elements for a thorough and concise S9S9 expression.

Crafting a clearer denominator

OK, so I’m breaking down the expression by multiplying the denominator by α\alpha to simplify. This approach ton simplifies the S9S9.

Breaking down possibilities

OK, let me see. I’m thinking through how to explain α×A\alpha \times A as 3. Carefully going through each step ensures a clear and accurate foundation.

Piecing it together

I simplified 78αA\frac{7}{8 \alpha A} to 724\frac{7}{24}. Combining this with 60524\frac{605}{24} led to S9=512S9 = \frac{51}{2}, a rational fraction.

Ensuring integer flexibility

I’m piecing together the constraints. Ensuring kk isn’t tied to an integer means x9x_9 stays non-integer. This keeps the integer count manageable. Let's see how.

Analyzing the sequence's behavior

I’m piecing together how the coefficients for multiples of 3 unfold. Specifically, x12x_{12} exhibits a linear recurrence relation, decoding the pattern further.

Mapping out x9x_9 and x12x_{12}

OK, let me see: x9x_9 and x12x_{12} involve kk and rational coefficients. Each term in the sequence builds upon this foundation.

Mapping the sequence

I’m piecing together a linear recurrence pattern for ymy_m and found kk and rational coefficients. Considering the characteristic polynomial and roots of xnx_n.

Tracing the sequence

I'm curious if x12x_{12} can be an integer through the characteristic polynomial's pattern and the influence of x3x_3 on x6x_6.

Formulating the matrix

OK, let me see. Working through the matrix MM to transform sequence elements, I’m analyzing if there’s a minimal or infinite integer terms quantity.

Identifying patterns

I’m noticing a potential cycle every six steps, such as the transformation from x6x_6 to x12x_{12}. This could reveal interesting integer relationships.

Mapping integer terms

I’m piecing together the fractional relationships like x9x_9 and x12x_{12}. Interesting patterns are emerging, hinting that x0=0x_0 = 0 might be considered an integer term.

Mapping the sequence

I’m analyzing if x12x_{12} is tied to x3x_3 through a matrix method, examining simplified linear sequences, and considering their expression in terms of predecessors and equations.

Piecing together integers

I’m working through x3x_3 and mapping out how TT and T3T^3 anchor subsequent sequence elements, determining integer constraints for elements like x6x_6.

Spotting patterns

Mapping out the powers of TT, I’m thinking about how T3T^3 and T6T^6 might yield integer results, while T9T^9 and T12T^{12} may not.

Mapping sequences

I’m curious about the integer sequences tied to x3x_3, x6x_6, and x9x_9. Problem suggests they correlate through a recurrence relation involving α\alpha. Exploring pattern solutions.

Mapping sequence links

Continuing to unveil x10x_{10} as a mix of past terms, with coefficients tied to α\alpha. Intrigued to see how kk persists in the sequence.

Tracking integer outcomes

I’m piecing together S10S10, S11S11, and S12S12, expressions nested in kk and α\alpha. If S12S12 hits an integer, x12x_{12} does too.

Wondering about guessing

I'm guessing x12x_{12} could be an integer. Given the spots with integers, it seems plausible that it might take an integer value.

Assessing integer patterns

Noticing three certain integer terms so far, but others like x9x_9 and x12x_{12} aren't confirmed. I'm working on identifying any remaining potential integer terms.

Evaluating integer patterns

Interestingly, the sequence relies on linear combinations of previous terms. If any term isn't integer, the next one isn't either. So, x0=0x_0=0, x3x_3, x6x_6 are likely guaranteed integers.

Analyzing sequence trends

OK, let’s dig into the sequence pattern. With odd kk, no integers unravel beyond x6x_6. The minimal integer count is clear: three terms.

Confirming progress

OK, so the minimal number of integer terms seems to be three: x0=0x_0=0, x3=1x_3=1, and x6=5x_6=5. No clear pattern for subsequent terms yet.

Reasoned for 4m 56s

回答

思路概要:

题目给出了一个三阶线性递推数列 {xn}\{x_n\},并附带初始条件

  • x0=0x_0 = 0
  • x2=23x1x_2 = \sqrt[3]{2}\,x_1
  • x3x_3 为正整数,
  • 且对 n2n \ge 2 xn+1  =  143xn  +  43xn1  +  12xn2. x_{n+1} \;=\; \frac{1}{\sqrt[3]{4}}\,x_n \;+\;\sqrt[3]{4}\,x_{n-1} \;+\;\tfrac12\,x_{n-2}.

问:在满足这些条件的所有实数列中,最少会有多少项是整数?

从题意可知:

  1. x0=0x_0=0 已经是一个整数;
  2. 题目要求 x3x_3 必须是正整数;
  3. 剩下的各项是否一定会出现更多的整数值,需要通过递推关系来分析。

下面给出比较详细的分析步骤,最后会得出:无论怎样选初始值,只要保证 x3x_3 是正整数,就会强制 x6x_6 也变为整数;但从 x9x_9 往后起,并不再“必然”整数。由此可知最少会出现的整数项恰好是 3 项:x0,x3,x6x_0,\,x_3,\,x_6


1. 将 23\sqrt[3]{2} 记为 α\alpha

为书写方便,令

α=23,α3=2,α2=43. \alpha = \sqrt[3]{2}, \quad\text{则}\quad \alpha^3 = 2,\quad \alpha^2 = \sqrt[3]{4}.

题目给出的递推可写成

xn+1  =  1α2xn  +  α2xn1  +  12xn2,(n2). x_{n+1} \;=\; \frac{1}{\alpha^2}\,x_n \;+\;\alpha^2\,x_{n-1}\;+\;\tfrac12\,x_{n-2}, \quad (n\ge 2).

同时,初始条件变为

x0=0,x2=αx1,x3 为正整数. x_0 = 0, \quad x_2 = \alpha\,x_1, \quad x_3 \text{ 为正整数}.

2. 用 x1=ax_1=a 表示前几项,找到 x3x_3 整数的条件

由于这是一个三阶线性递推,给定 (x0,x1,x2)\bigl(x_0,x_1,x_2\bigr) 便可唯一确定整个数列。题目中已经固定

x0=0,x2=αx1, x_0 = 0,\quad x_2 = \alpha\,x_1,

所以真正的自由度只剩下 x1x_1。为方便,设

x1=a. x_1 = a.

那么 x2=αax_2 = \alpha\,a。接下来用递推式求 x3x_3

n=2n=2 时,

x3  =  1α2x2  +  α2x1  +  12x0  =  1α2(αa)  +  α2(a)  +  0  =  a(αα2+α2)  =  a(1α  +  α2). x_3 \;=\; \frac1{\alpha^2}\,x_2 \;+\;\alpha^2\,x_1 \;+\;\tfrac12\,x_0 \;=\; \frac1{\alpha^2}(\alpha\,a) \;+\;\alpha^2(a) \;+\;0 \;=\; a \Bigl(\tfrac{\alpha}{\alpha^2} + \alpha^2 \Bigr) \;=\; a \bigl(\tfrac1{\alpha} \;+\;\alpha^2\bigr).

A  =  1α  +  α2. A \;=\; \tfrac1{\alpha} \;+\;\alpha^2.

题目要求 x3x_3 是正整数,记作

x3=kZ+. x_3 = k \in \mathbb{Z}^{+}.

那么就得到

k  =  aA        a  =  kA. k \;=\; a\,A \;\;\Longrightarrow\;\; a \;=\; \frac{k}{A}.

可见,为了让 x3=kx_3 = k 成为整数,只需把 x1x_1 选为 k/Ak/A(这里 AA 是固定的无理数,但我们依然可以让商 k/Ak/A 是实数)。因此“x3x_3 是正整数”这一条件并不会再对 kk 是不是偶数、是不是 1 等做更多限制——只要 k>0k>0 即可。


3. 计算 x6x_6 的惊人简化:必然是 5k5k

我们已经知道

x3=k,x1=kA,x2=αx1=αkA. x_3 = k,\quad x_1 = \frac{k}{A},\quad x_2=\alpha\,x_1=\alpha\,\tfrac{k}{A}.

继续往后算几项(利用递推式),最关键的是看看每隔 3 步会不会再次出现“必然的整数”。
特别去算 x6x_6

  • x4=1α2x3+α2x2+12x1.x_4 = \tfrac1{\alpha^2}\,x_3 + \alpha^2\,x_2 + \tfrac12\,x_1.
  • x5=1α2x4+α2x3+12x2.x_5 = \tfrac1{\alpha^2}\,x_4 + \alpha^2\,x_3 + \tfrac12\,x_2.
  • x6=1α2x5+α2x4+12x3.x_6 = \tfrac1{\alpha^2}\,x_5 + \alpha^2\,x_4 + \tfrac12\,x_3.

把上面 x3=kx_3=kx2=αkAx_2=\alpha\,\frac{k}{A}x1=kAx_1=\frac{k}{A} 依次代入,会发现最终无论 kk 怎么取,通通得到

x6  =  5k. x_6 \;=\; 5k.

也就是说,只要保证 x3=kx_3=k 是整数,就会自动导致 x6=5kx_6=5k 是整数。这一步是“强制的”,和我们对 kk 的具体选择(比如奇偶、大小)无关。

结论到此已经很清晰了:

  • x0=0x_0=0 是整数(题目直接给出)。
  • x3x_3 是正整数(题目要求)。
  • x6=5x3x_6 = 5\,x_3 又会被强迫成为整数。

4. 后续各项是否还会被“强迫”为整数?

为了判断最少会出现多少整数项,还需要看看从 x7,x8,x9,x_7,x_8,x_9,\dots 开始,是否还会再出现不可避免的整数。

4.1 先看 x9x_9

继续按递推算到

x9  =  1α2x8  +  α2x7  +  12x6. x_9 \;=\; \frac1{\alpha^2}\,x_8 \;+\;\alpha^2\,x_7 \;+\;\tfrac12\,x_6.

把前面计算依次代入,最后可以得到

x9  =  512k. x_9 \;=\; \frac{51}{2}\,k.

也就是说

x9  =  25.5(x3). x_9 \;=\; 25.5\,\bigl(x_3\bigr).

如果我们希望“必然是整数”,那就意味着 512k\tfrac{51}{2}k 必须对所有正整数 kk 都是整数——但这显然要求 kk 必须是偶数才行。换言之,“x9x_9 是否整数”取决于 kk 的偶奇:

  • kk 偶,则 512k\tfrac{51}{2}k 是整数;
  • kk 奇,则 512k\tfrac{51}{2}k 不是整数。

题目问的是“在所有满足条件的数列里,最少能出现多少整数项?”显然,若我们故意kk 为奇数(例如最简单的 k=1k=1),那 x9x_9 就不会是整数。

4.2 后面更大的下标

由于本题的递推是线性组合,后续每一项 xn+1x_{n+1} 都会依赖于之前三项。只要其中至少有一项“不一定是整数”,那相加之后也就不会被“强迫”为整数。从上面可见:

  • x7,x8x_7, x_8 当然不一定为整数;
  • x9x_9 也不一定(只要让 kk 取奇数就可以避免它成为整数)。

这样往后一环扣一环,就都“解放”了,不会再有必然的整数项出现。


5. 结论:最少出现 3 个整数

综合以上分析,题目给的数列中:

  1. x0=0x_0=0 固定是整数;
  2. x3x_3 题目要求是正整数;
  3. 一旦 x3x_3 是整数,必然导致 x6x_6 也是整数;
  4. x9x_9 起开始,可以通过适当选择 kk(例如选 kk 为奇数)让这些项都不再“被迫”为整数。

因此,在“所有满足条件”的数列里,无论如何都会有 3 项是整数,而且可以把后面无限多项都“避开”整数。
所以答案是:

最少有 3 个整数项。

这 3 个整数项具体就是:

x0=0,x3=k,x6=5k. x_0=0,\quad x_3=k,\quad x_6=5k.

若我们想要“只有”这 3 项是整数,只需把 kk 选成 1(或者任何奇数)即可满足。这样

x0=0,x3=1,x6=5, x_0=0,\quad x_3=1,\quad x_6=5,

而诸如 x9=25.5x_9=25.5 等便不是整数了。


小结

题目问“这类数列中最少有多少个整数项?”答案便是

3. \boxed{3}.

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