Maths Olympiad Prep

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, 2013

Combinatorics Difficulty 9.1 IMO level Prove it Saudi Arabia

Determine if there exists an infinite sequence of positive integers
a1,a2,a3, a_{1}, a_{2}, a_{3}, \ldots
such that
(i) each positive integer occurs exactly once in the sequence, and
(ii) each positive integer occurs exactly once in the sequence a1a2,a2a3,,akak+1,\left|a_{1}-a_{2}\right|, \left|a_{2}-a_{3}\right|, \ldots, \left|a_{k}-a_{k+1}\right|, \ldots

Solution

We will construct such a sequence by induction:
Define a1=1a_{1}=1 and a2=2a_{2}=2. In this case we have b1=a1a2=1b_{1}=|a_{1}-a_{2}|=1.
Assume that a1,a2,,a2na_{1}, a_{2}, \ldots, a_{2n} are defined such that there is no positive integer which occurs at least twice neither in the finite sequence a1,a2,,a2na_{1}, a_{2}, \ldots, a_{2n} nor in the finite sequence b1=a1a2,b2=a2a3,,b2n1=a2n1a2nb_{1}=|a_{1}-a_{2}|, b_{2}=|a_{2}-a_{3}|, \ldots, b_{2n-1}=|a_{2n-1}-a_{2n}|.
Let MnM_{n} be the maximum element of the set of integers {a1,a2,,a2n}\{a_{1}, a_{2}, \ldots, a_{2n}\}, cnc_{n} the maximum of the integers kk such that {1,2,,k}{a1,a2,,a2n}\{1,2, \ldots, k\} \subseteq \{a_{1}, a_{2}, \ldots, a_{2n}\}, and dnd_{n} the maximum of the integers kk such that {1,2,,k}{b1,b2,,b2n1}\{1,2, \ldots, k\} \subseteq \{b_{1}, b_{2}, \ldots, b_{2n-1}\}.
Notice that for all 1i2n11 \leq i \leq 2n-1, we have
bi=aiai+1max1j2najmin1j2naj=Mn1. b_{i}=|a_{i}-a_{i+1}| \leq \max_{1 \leq j \leq 2n} a_{j} - \min_{1 \leq j \leq 2n} a_{j} = M_{n}-1.
If cn<dnc_{n}<d_{n} define a2n+1=2Mn+cn+1a_{2n+1}=2M_{n}+c_{n}+1 and a2n+2=cn+1a_{2n+2}=c_{n}+1. If cndnc_{n} \geq d_{n} define a2n+1=2Mn+1a_{2n+1}=2M_{n}+1 and a2n+2=2Mn+dn+2a_{2n+2}=2M_{n}+d_{n}+2.
It is clear that none of the two possible values for a2n+1a_{2n+1} and for a2n+2a_{2n+2} occur in the finite sequence a1,a2,,a2na_{1}, a_{2}, \ldots, a_{2n}.
In the first case, we have
b2n=2Mn+cn+1a2nMn+cn+1>Mn1bi b_{2n}=2M_{n}+c_{n}+1-a_{2n} \geq M_{n}+c_{n}+1 > M_{n}-1 \geq b_{i}
and
b2n+1=2Mn>Mn1bi b_{2n+1}=2M_{n} > M_{n}-1 \geq b_{i}
for all 1i2n11 \leq i \leq 2n-1, and b2n+1b2nb_{2n+1} \neq b_{2n} since a2ncn+1a_{2n} \neq c_{n}+1.
In the second case, we have
b2n=2Mn+1a2nMn+1>Mn1bi b_{2n}=2M_{n}+1-a_{2n} \geq M_{n}+1 > M_{n}-1 \geq b_{i}
and
b2n+1=dn+1bi b_{2n+1}=d_{n}+1 \neq b_{i}
for all 1i2n11 \leq i \leq 2n-1, and b2n+1=dn+1Mn<Mn+1b2nb_{2n+1}=d_{n}+1 \leq M_{n} < M_{n}+1 \leq b_{2n}.
Therefore, there is no positive integer which occurs at least twice in the finite sequence a1,a2,,a2n+2a_{1}, a_{2}, \ldots, a_{2n+2} or in the finite sequence b1,b2,,b2n+1b_{1}, b_{2}, \ldots, b_{2n+1}.
This proves that there is no positive integer which occurs at least twice in the infinite sequence a1,a2,,an,a_{1}, a_{2}, \ldots, a_{n}, \ldots or in the infinite sequence b1,b2,,bn,b_{1}, b_{2}, \ldots, b_{n}, \ldots.
Notice moreover, that in the first case cn+1cn+1c_{n+1} \geq c_{n}+1 and in the second case dn+1dn+1d_{n+1} \geq d_{n}+1. Therefore, if en=min{cn,dn}e_{n}=\min\{c_{n}, d_{n}\} then en+2en+1e_{n+2} \geq e_{n}+1. But c1=2c_{1}=2 and d1=1d_{1}=1. We deduce that e2nne_{2n} \geq n for all integer nn. Therefore, any positive integer nn occurs in both finite sequences a1,a2,,a4na_{1}, a_{2}, \ldots, a_{4n} and b1,b2,,b4n1b_{1}, b_{2}, \ldots, b_{4n-1}.
This proves that any positive integer nn occurs exactly once in each infinite sequence a1,a2,a_{1}, a_{2}, \ldots and b1,b2,b_{1}, b_{2}, \ldots.

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