我现在给你提供一个思路，你重新完成代码，先用分治，然后任何反等差数列乘以2仍然是反等差数列，2n+1也会是反等差数列，之后进行递归

了解您的想法，我们可以使用分治策略来构造反等差数列，然后再将这些数列进行相应的调整以保持其性质。下面是一个基于这个思路的Python代码实现：

def is_equidistant_sequence(arr):
    if len(arr) <= 2:
        return True
    diff = arr[1] - arr[0]
    for i in range(2, len(arr)):
        if arr[i] - arr[i - 1] != diff:
            return False
    return True

def construct_equidistant_sequence(n, start=0):
    if n <= 1:
        return [start]

    mid = n // 2
    left = construct_equidistant_sequence(mid, start)
    right = construct_equidistant_sequence(n - mid, start + mid)

    # 合并左右两部分，同时考虑乘以2和翻转操作
    if n % 2 == 1:
        return left + [left[-1] + 2*diff] + list(reversed(right))  # 假设diff = left[-1] - left[-2]
    else:
        return left + list(reversed(right))

def generate_antidifference_sequences(n):
    sequences = []
    for seq in construct_equidistant_sequence(2*n + 1):
        # 对反等差数列乘以2，得到新的反等差数列
        doubled_seq = [2*x for x in seq]
        if is_equidistant_sequence(doubled_seq):
            sequences.append(' '.join(map(str, doubled_seq)))

    return sequences

def main():
    n = int(input())
    while n > 0:
        result = generate_antidifference_sequences(n)
        print(f"{n}: {result[0]}")  # 输出第一个反等差数列
        n = int(input())

if __name__ == "__main__":
    main()