floor and ceiling function

Let n = 6X + a, where X is an integer and 0 <= a < 6
L.H.S. = └ n / 3 ┘ + └ (n + 2) / 6 ┘ + └ (n + 4) / 6 ┘
= └ (6X + a) / 3 ┘ + └ ((6X + a) + 2) / 6 ┘ + └ ((6X + a) + 4) / 6 ┘
= └ 2X + a/3 ┘ + └ X + (a + 2)/6 ┘ + └ X + (a + 4)/6 ┘
= 2X + X + X = 4X

2015-05-19 08:40:16 補充：
R.H.S. = └ n / 2 ┘ + └ (n + 3) / 6 ┘
= └ (6X + a) / 2 ┘ + └ ((6X + a) + 3) / 6 ┘
= └ 3X + a/2 ┘ + └ X + (a + 3)/6 ┘
= 3X + X = 4X
so, └ n / 3 ┘ + └ (n + 2) / 6 ┘ + └ (n + 4) / 6 ┘ = └ n / 2 ┘ + └ (n + 3) / 6 ┘

2015-05-19 08:41:22 補充：
第二條試過數值，好似有 d 問題，恕小弟不才

2015-05-20 16:54:50 補充：
係喎，再諗過先......

2015-05-20 17:05:44 補充：
Let n = 6X + a, where X is an integer and 0 <= a < 6

L.H.S. = └ n / 3 ┘ + └ (n + 2) / 6 ┘ + └ (n + 4) / 6 ┘
= └ (6X + a) / 3 ┘ + └ ((6X + a) + 2) / 6 ┘ + └ ((6X + a) + 4) / 6 ┘
= └ 2X + a/3 ┘ + └ X + (a + 2)/6 ┘ + └ X + (a + 4)/6 ┘
R.H.S. = └ n / 2 ┘ + └ (n + 3) / 6 ┘
= └ (6X + a) / 2 ┘ + └ ((6X + a) + 3) / 6 ┘
= └ 3X + a/2 ┘ + └ X + (a + 3)/6 ┘

case i, 0 <= a < 2
L.H.S. = 2X + X + X = 4X
R.H.S. = 3X + X = 4X

case ii, 2 <= a < 3
L.H.S. = 2X + X + X + 1 = 4X + 1
R.H.S. = 3X + 1 + X = 4X + 1

case iii, 3 <= a < 4
L.H.S. = 2X + 1 + X + X + 1 = 4X + 2
R.H.S. = 3X + 1 + X + 1 = 4X + 2

case iv, 4 <= a < 6
L.H.S. = 2X + 1 + X + 1 + X + 1 = 4X + 3
R.H.S. = 3X + 2 + X + 1 = 4X + 3

so, └ n / 3 ┘ + └ (n + 2) / 6 ┘ + └ (n + 4) / 6 ┘ = └ n / 2 ┘ + └ (n + 3) / 6 ┘

All the questions are valid and can be proved.

2015-05-22 14:07:46 補充：
要證明⌊X⌋=⌊Y⌋，其中X小於Y：
（1）Y不可能是整數；
（2）X和Y之間不可能有其他整數。

2015-05-24 08:58:20 補充：
做一題示範
Y為整數k：1/2+√(n+1/2)=k
√(n+1/2)=k-1/2
n+1/2=k^2-k+1/4
n+1/4=k^2-k (k^2-k是整數：矛盾)
(X,Y)中有整數m: 1/2+√(n+1/2)>m>1/2+√(n+1/4)
√(n+1/2)>m-1/2>√(n+1/4)
n+1/2>m^2-m+1/4>n+1/4
n+1/4>m^2-m>n(n和n+1/4 間沒有其他整數：矛盾)

Re 意見 001,

└ 2X + a/3 ┘ 未必 等於 2X 吧

Thank you ， you are too good to answer my questions. Maybe I ve jotted the questions wronly. The original questions can be found in wikipedia on the topic' floor and ceiling function'