你可以使用lag()来获取“前一行”和前一行的值。如果它们都等于当前行的值，那么你就得到了一个匹配。连续的大于3的相等数字会像这样匹配多次。重复这样的连续序列也会这样做。你可以使用DISTINCT来只选择它们一次。

SELECT DISTINCT
       num
       FROM (SELECT num,
                    lag(num, 1) OVER (ORDER BY id) = num
                    AND lag(num, 2) OVER (ORDER BY id) = num c3
                    FROM logs) x
       WHERE c3;

db<>fiddle

您也可以使用ROW_NUMBER()以这种方式对连续数字进行分组。

SELECT num as ConsecutiveNums 
FROM (
    SELECT *, 
           (row_number() OVER (ORDER BY id) - row_number() OVER (PARTITION BY num ORDER BY id)
            ) AS grp
    FROM logs
    ) SQ
GROUP BY num,grp
HAVING count(*) >= 3

SELECT num as ConsecutiveNums 
FROM (
    SELECT *, 
           row_number() OVER (ORDER BY id) - row_number() OVER (PARTITION BY num ORDER BY id as rw
    FROM table
    ) S
GROUP BY num,rw
HAVING count(*) >= 3