PUMaC 2014 · 代数（A 组） · 第 4 题

PUMaC 2014 — Algebra (Division A) — Problem 4

[ 4 ]There is a sequence with a (2) = 0 , a (3) = 1 and a ( n ) = a + a for n ≥ 4. Find 2 2 ⌊ ⌋ ⌈ ⌉ n n n a (2014). [Note that and denote the floor function (largest integer ≤ ) and the ceiling 2 2 2 n function (smallest integer ≥ ), respectively.] 2

解析

[ 4 ]There is a sequence with a (2) = 0 , a (3) = 1 and a ( n ) = a + a for n ≥ 4. 2 2 ⌊ ⌋ ⌈ ⌉ n n n Find a (2014). [Note that and denote the floor function (largest integer ≤ ) and 2 2 2 n the ceiling function (smallest integer ≥ ), respectively.] 2 Solution: We can see that if n is a power of 2, then a = 0. Then playing with some values, we see that n a a − 1 the sequence has the property that a = a + 1 if 2 < n ≤ 3 ∗ 2 and a = a − 1 n n − 1 n n − 1 a − 1 a +1 if 3 ∗ 2 < n ≤ 2 . So, the sequence goes: 0 , 1 , 0 , 1 , 2 , 1 , 0 , 1 , 2 , 3 , 4 , 3 , 2 , 1 , 0 , ... . So since a = 0, we have that a = 1 , a = 2 , ..., a = 34 . 2048 2047 2046 2014