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Binomial coefficients forming arithmetic progression

Source: China TST 1998, problem 1

May 22, 2005
arithmetic sequencecombinatorics unsolvedcombinatoricsarithmeticArithmetic ProgressionChinaTST

Problem Statement

Find kNk \in \mathbb{N} such that
a.) For any nNn \in \mathbb{N}, there does not exist jZj \in \mathbb{Z} which satisfies the conditions 0jnk+10 \leq j \leq n - k + 1 and (nj),(nj+1),,(nj+k1)\left( \begin{array}{c} n\\ j\end{array} \right), \left( \begin{array}{c} n\\ j + 1\end{array} \right), \ldots, \left( \begin{array}{c} n\\ j + k - 1\end{array} \right) forms an arithmetic progression.
b.) There exists nNn \in \mathbb{N} such that there exists jj which satisfies 0jnk+20 \leq j \leq n - k + 2, and (nj),(nj+1),,(nj+k2)\left( \begin{array}{c} n\\ j\end{array} \right), \left( \begin{array}{c} n\\ j + 1\end{array} \right), \ldots , \left( \begin{array}{c} n\\ j + k - 2\end{array} \right) forms an arithmetic progression.
Find all nn which satisfies part b.)