There is a series of numbers: l, 4, 16, 61, 232, 880, ..., where the first number is 1, the second number is 4, and from the third number onwards, each number is exactly between the previous two numbers 3 times the sum plus l, then divide the 2010th number in this column by 6, what is the remainder?
有一列数：l，4，16，61，232，880，…，其中第一个数是1，第二个数是4，从第三个数起，每个数恰好是前面两个数之和的3倍再加上l，那么这列数中的第2010个数除以6，得到的余数是多少．
<result>4</result>