Question 1163947
Finding the recursive relations and generating functions for these sequences involves identifying the pattern of multiplication or the relationship between the coefficients of a polynomial (like the binomial expansion).

### (a) 4, 12, 24, 24, 12, -12, -48, -96, -156, -228, ...
This sequence is based on the **third differences** being constant (eventually). By calculating the differences between terms:
* **1st Differences:** 8, 12, 0, -12, -24, -36, -48, -60, -72...
* **2nd Differences:** 4, -12, -12, -12, -12, -12, -12...
* **3rd Differences:** -16, 0, 0, 0, 0, 0...

**Recursive Relation:**
After $n=3$, the second difference is constant ($-12$). The general recursive form for a sequence with constant second differences is:
$$a_n = 3a_{n-1} - 3a_{n-2} + a_{n-3}$$
*(Note: This holds true for the later part of the sequence where the pattern stabilizes.)*

**Generating Function:**
Since it eventually follows a quadratic pattern:
$$G(x) = \frac{4 + x - 8x^2 - 3x^3}{(1-x)^3}$$

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### (b) 4, 12, 24, 24, 12, 4, 1, 1/5, 1/30, 1/210, ...
This sequence is formed by dividing by an increasing integer at each step, related to **$\frac{4! \binom{n}{k}}{n!}$** or variations of **$\frac{1}{n!}$**. Specifically, starting from the 4th term:
* $24 \div 1 = 24$
* $24 \div 2 = 12$
* $12 \div 3 = 4$
* $4 \div 4 = 1$
* $1 \div 5 = 1/5$

**Recursive Relation:**
$$a_n = a_{n-1} \cdot \frac{k}{n} \text{ (where } k \text{ is a constant index shift)}$$
More simply: $a_n = \frac{4 \cdot 4!}{(n-1)!}$ for $n \ge 4$ (with manual overrides for the start).

**Generating Function:**
This is a variation of the exponential generating function $e^x$:
$$G(x) = 4 + 12x + 24x^2 + 24x^3 + 12x^4 + 4x^5 + \sum_{n=6}^{\infty} a_n x^n$$

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### (c) 1, 5, 20, 60, 120, 120, 0, 0, 0, ...
These are the coefficients of the polynomial for **$5! \binom{5}{k} / k!$** or simply the values of **$P(5, k)$** (permutations).
* $P(5,0) = 1$
* $P(5,1) = 5$
* $P(5,2) = 20$
* $P(5,3) = 60$
* $P(5,4) = 120$
* $P(5,5) = 120$
* $P(5,6) = 0$

**Recursive Relation:**
$$a_n = a_{n-1} \cdot (6-n) \text{ for } n \le 6, \text{ else } 0$$

**Generating Function:**
This is a finite polynomial:
$$G(x) = 1 + 5x + 20x^2 + 60x^3 + 120x^4 + 120x^5$$

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### (d) 1, 5, 20, 60, 120, 120, 60, 20, 5, 1, 1/6, 1/42, ...
This is a **symmetric** version of sequence (c). It represents the coefficients of the expansion of a binomial power where the terms are divided by factorials, often seen in **Hypergeometric series**.

**Recursive Relation:**
$$a_n = a_{n-1} \cdot \frac{6-n}{n}$$
*(Notice: at $n=1$, $5/1=5$; at $n=2$, $4/2=2 \to 20$; etc. Once $n > 5$, the terms become fractional.)*

**Generating Function:**
This is the generating function for the binomial coefficients $\binom{5}{n}$:
$$G(x) = (1+x)^5$$
*(However, because of the fractional tail $1/6, 1/42$, this specifically matches the coefficients of the expansion of a **generalized binomial series** or a **Hypergeometric function** $_1F_0$.)*