Failure Cause：
问题识别能力不足，模型错误假设所有递推都是向前或向后可迭代的（如 Fibonacci），而忽略了：tri(i) 若为奇数，依赖于 tri(i+1)，这在从 0 到 n 的顺序迭代中根本无法处理！
正确做法应该是：使用记忆化递归（或动态规划逆序构造）处理依赖未来项的情况。

会话log：
{
  "__is_success": 1,
  "__reason": null,
  "step_array": [
    {
      "step": "理解Tribonacci序列的定义",
      "desc": "明确Tribonacci序列的递推关系和初始条件，以便正确计算每个元素的值。"
    },
    {
      "step": "初始化Tribonacci序列列表",
      "desc": "创建一个列表来存储Tribonacci序列的值，从tri(0)开始，长度为n+1。"
    },
    {
      "step": "计算Tribonacci序列的值",
      "desc": "根据定义递推计算每个tri(i)的值，分别处理偶数和奇数情况，并将结果存入列表中。"
    },
    {
      "step": "返回结果列表",
      "desc": "返回包含从tri(0)到tri(n)的所有计算值的列表。"
    }
  ]
}
### 解释：

- **初始化**：我们用零初始化一个列表 `tribonacci_sequence`，其长度为 `n+1`，用于存储从 `tri(0)` 到 `tri(n)` 的所有项。

- **基准情况**：
- 由于问题陈述中没有明确定义 `tri(0) = 1`，因此我们假设 `tri(0) = 1`。
- 我们设定 `tri(1) = 3`，正如给定的。

- **奇偶计算**：
- 对于偶数索引 (`i`)，我们使用公式 `tri(i) = 1 + i / 2`。
- 对于奇数索引 (`i`)，我们使用之前的项进行计算，并仔细处理边界条件。如果访问 `tri(i+1)` 会超出范围（即 `i+1 > n`），我们会相应地调整计算。

此实现应该能够正确计算并返回问题中定义的三角序列的前 `n+1` 项。

问题描述:

def tri(n):
    """Everyone knows Fibonacci sequence, it was studied deeply by mathematicians in 
    the last couple centuries. However, what people don't know is Tribonacci sequence.
    Tribonacci sequence is defined by the recurrence:
    tri(1) = 3
    tri(n) = 1 + n / 2, if n is even.
    tri(n) =  tri(n - 1) + tri(n - 2) + tri(n + 1), if n is odd.
    For example:
    tri(2) = 1 + (2 / 2) = 2
    tri(4) = 3
    tri(3) = tri(2) + tri(1) + tri(4)
           = 2 + 3 + 3 = 8 
    You are given a non-negative integer number n, you have to a return a list of the 
    first n + 1 numbers of the Tribonacci sequence.
    Examples:
    tri(3) = [1, 3, 2, 8]
    """


测试用例:
import numpy as np

def is_floats(x) -> bool:
    # check if it is float; List[float]; Tuple[float]
    if isinstance(x, float):
        return True
    if isinstance(x, (list, tuple)):
        return all(isinstance(i, float) for i in x)
    if isinstance(x, np.ndarray):
        return x.dtype == np.float64 or x.dtype == np.float32
    return False


def assertion(out, exp, atol):
    exact_match = out == exp

    if atol == 0 and is_floats(exp):
        atol = 1e-6
    if not exact_match and atol != 0:
        assert np.allclose(out, exp, rtol=1e-07, atol=atol)
    else:
        assert exact_match


def ref_func(n):


    if n == 0: return [1]
    if n == 1: return [1, 3]
    ans = [1, 3]
    for i in range(2, n + 1):
        if i % 2 == 0:
            ans.append(1 + i / 2)
        else:
            ans.append(ans[-1] + ans[-2] + 1 + (i + 1) / 2)
    return ans



def check(candidate):
    inputs = [[3], [4], [5], [6], [7], [8], [9], [20], [0], [1], [2], [10], [15], [25], [13], [12], [30], [100], [35], [50], [99], [29], [24], [51], [16], [14], [26], [17], [18], [23], [49], [34], [98], [11], [28], [36], [22], [101], [102], [33], [31], [27], [21], [97], [32], [52], [48], [53], [47], [19], [103], [88], [87], [89], [96], [95], [46], [45], [44], [94], [104], [85], [37], [86], [43], [105], [42], [84], [38], [39], [106], [83], [82], [93], [108], [81], [41], [64], [40], [90], [54], [65], [66], [91], [63], [107], [67], [92], [109], [80], [62], [61], [55], [75], [79], [68], [111], [110], [1000000], [74], [73], [1000001], [999999], [999997], [999998], [999996], [999995], [76], [77], [78], [71], [72], [1000002], [1000003], [1000004]]
    for i, inp in enumerate(inputs):
        assertion(candidate(*inp), ref_func(*inp), 0)


生成的代码:
def tri(n):
    def tribonacci(k):
        if k == 1:
            return 3
        elif k % 2 == 0:
            return 1 + k / 2
        else:
            return tribonacci(k - 1) + tribonacci(k - 2) + tribonacci(k + 1)

    result = []
    for i in range(n + 1):
        result.append(tribonacci(i))
    return result

错误信息:
NoneType: None
