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HMMT 二月 2016 · 团队赛 · 第 6 题

HMMT February 2016 — Team Round — Problem 6

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 35 ] A nonempty set S is called well-filled if for every m ∈ S , there are fewer than m elements of Swhich are less than m . Determine the number of well-filled subsets of { 1 , 2 , . . . , 42 } .2
    1 2 n n − 1
解析

英文解析

  1. [ 35 ] A nonempty set S is called well-filled if for every m ∈ S , there are fewer than m elements of Swhich are less than m . Determine the number of well-filled subsets of { 1 , 2 , . . . , 42 } .2
    Proposed by: Casey Fu
    ( )
    Answer: − 143
    Let a be the number of well-filled subsets whose maximum element is n (setting a = 1). Then it’s 21
    n 0
    easy to see thata = a + a + · · · + a
    2 k +1 2 k 2 k − 1 0
    a = ( a − C ) + a + · · · + a .
    2 k +2 2 k +1 k 2 k 0
    where C is the number of well-filled subsets of size k + 1 with maximal element 2 k + 1.
    We proceed to compute C . One can think of such a subset as a sequence of numbers 1 ≤ s < · · · <kk 1
    s ≤ 2 k + 1 such that s ≥ 2 i − 1 for every 1 ≤ i ≤ k + 1. Equivalently, letting s = i + 1 + t it’sk +1 i i ithe number of sequences 0 ≤ t ≤ · · · ≤ t ≤ k + 1 such that t ≥ i for every i . This gives the list of
    1 k +1 ix -coordinates of steps up in a Catalan path from (0 , 0) to ( k + 1 , k + 1), so
    ( )
    1 2( k + 1)
    C =
    k + 2 ( k + 1)kis equal to the ( k + 1)th Catalan number.
    From this we can solve the above recursion to derive that
    ( )
    a = .nnb ( n − 1) / 2 c
    Consequently, for even n ,
    ( )
    n + 1
    a + · · · + a = a = .
    0 n n +1
    b n/ 2 c
    Putting n = 42 gives the answer, after subtracting off the empty set (counted in a ).
    1 2 n n − 10