HackerRank「Count Triplets」算法题部分测试用例未通过求助

我在解决HackerRank的「Count Triplets」问题时，遇到测试用例不通过的情况：当ratio不等于1时，测试用例6和10失败。其中测试用例6包含100000个整数，我的输出结果为13621903916，但预期结果是2325652489，两者差值约110亿。不过当ratio等于1时，代码逻辑是正确的。

我的C#实现代码如下：

private static long GetNumberOfTriplets(IEnumerable<long> array, long ratio)
{
    var frequencyByInteger = array
        .GroupBy(item => item)
        .ToDictionary(g => g.Key, g => g.LongCount());

    /*  If ratio is 1, then all the items in the triplet are equal.
     *  This formula can be derived from n(n+1)/2 via induction.
     *  But you don't need to understand this in order to help me,
     *  as the failing test cases are using ratio != 1.
     */
    if (ratio == 1) return frequencyByInteger.Values.Sum(v => v * (v - 1) * (v - 2) / 6);

    /*  Looping through each distinct number, then checking if the
     *  dictionary has the other 2 numbers necessary to form a triplet.
     *  The frequencies of the three numbers are then 
     *  multiplied with each other in order to handle all 
     *  combinations in the case of repeated numbers.
     *  E.g. in the array {1, 1, 3, 3, 9, 9}
     *  there are 2*2*2=8 combinations of (1, 3, 9).
     */
    return frequencyByInteger.Keys
        .Where(v => frequencyByInteger.ContainsKey(v * ratio)
                    && frequencyByInteger.ContainsKey(v * ratio * ratio))
        .Sum(v => frequencyByInteger[v] * frequencyByInteger[v * ratio] * frequencyByInteger[v * ratio * ratio]);
}

我的思路是：统计每个数字的出现频率，遍历每个不同的数字，检查是否存在v*ratio和v*ratio*ratio这两个数字，若存在则将三者的频率相乘，累加所有符合条件的结果。但显然这个逻辑在某些测试用例下有问题，恳请帮忙排查代码错误。

内容的提问来源于stack exchange，提问作者starshinata