We will construct such a sequence by induction:
Define a1=1 and a2=2. In this case we have b1=∣a1−a2∣=1.
Assume that a1,a2,…,a2n are defined such that there is no positive integer which occurs at least twice neither in the finite sequence a1,a2,…,a2n nor in the finite sequence b1=∣a1−a2∣,b2=∣a2−a3∣,…,b2n−1=∣a2n−1−a2n∣.
Let Mn be the maximum element of the set of integers {a1,a2,…,a2n}, cn the maximum of the integers k such that {1,2,…,k}⊆{a1,a2,…,a2n}, and dn the maximum of the integers k such that {1,2,…,k}⊆{b1,b2,…,b2n−1}.
Notice that for all 1≤i≤2n−1, we have
bi=∣ai−ai+1∣≤1≤j≤2nmaxaj−1≤j≤2nminaj=Mn−1.
If cn<dn define a2n+1=2Mn+cn+1 and a2n+2=cn+1. If cn≥dn define a2n+1=2Mn+1 and a2n+2=2Mn+dn+2.
It is clear that none of the two possible values for a2n+1 and for a2n+2 occur in the finite sequence a1,a2,…,a2n.
In the first case, we have
b2n=2Mn+cn+1−a2n≥Mn+cn+1>Mn−1≥bi
and
b2n+1=2Mn>Mn−1≥bi
for all 1≤i≤2n−1, and b2n+1=b2n since a2n=cn+1.
In the second case, we have
b2n=2Mn+1−a2n≥Mn+1>Mn−1≥bi
and
b2n+1=dn+1=bi
for all 1≤i≤2n−1, and b2n+1=dn+1≤Mn<Mn+1≤b2n.
Therefore, there is no positive integer which occurs at least twice in the finite sequence a1,a2,…,a2n+2 or in the finite sequence b1,b2,…,b2n+1.
This proves that there is no positive integer which occurs at least twice in the infinite sequence a1,a2,…,an,… or in the infinite sequence b1,b2,…,bn,….
Notice moreover, that in the first case cn+1≥cn+1 and in the second case dn+1≥dn+1. Therefore, if en=min{cn,dn} then en+2≥en+1. But c1=2 and d1=1. We deduce that e2n≥n for all integer n. Therefore, any positive integer n occurs in both finite sequences a1,a2,…,a4n and b1,b2,…,b4n−1.
This proves that any positive integer n occurs exactly once in each infinite sequence a1,a2,… and b1,b2,….