Given the properties of the euclidean summation, for the moment we can prove only:
- that the sequences in L) and R), due to their boundedness, both contain convergent subsequences.
We can of course prove the above for discrete cases:
[; x = \frac{1}{p}, \; where \; p \in \mathbf{N}, p \ge 2 ;]
For proving L, if p divides n we have:
[; n = k \cdot p \Rightarrow \lceil n \cdot x \rfloor = k - 1 \Rightarrow n = p \cdot (k-1) + p ;]
which becomes a valid euclidean division (with quotient p and remainder p), if k-1>p. Since we are dealing with a limit, this assumption is safe.
Now on the other hand we get:
[; \tau_1(n,k-1) = p + \tau_1(k-1, p) \Rightarrow ;]
[; 0 < \tau_1(n, k-1) < p + \max\{r + \tau_1(p,r) | r \in \mathbf{N}, r < p \} \Rightarrow;]
[; 0< \tau_1(n, k-1) < 3 \cdot p \Rightarrow \lim_{n \rightarrow \infty, \; p|n}} \frac{\tau_1(n, \lceil n \cdot \frac{1}{p} \rfloor)}{n} = 0 ;]
If p does not divide n then we have:
[; n = k \cdot p + r, \; where \; r \in \mathbf{N}, \; 0<r<p \Rightarrow ;]
[; \lceil n \cdot x \rfloor = k \Rightarrow n = p \cdot k + r ;]
which becomes a valid euclidean divison if k>p.
In a similar fashion as above we get:
[; 0 < \tau_1(n,k) < 3 \cdot r \Rightarrow 0 < \tau_1(n,k) < 3 \cdot p \Rightarrow ;]
[; \lim_{n \rightarrow \infty, p \nmid n} \frac{\tau_1(n, \lceil n \cdot \frac{1}{p} \rfloor)}{n} = 0 ;]
For proving R we can use the general case:
[; n = k \cdot p + r, \; where \; r \in \mathbf{N}, \; 0 \le r < p \Rightarrow ;]
[; \lfloor n \cdot x \rceil = k+1 \Rightarrow n = (p-1) \cdot (k+1) + k + 1 - p + r ;]
which becomes a valid euclidean division if k>p.
Therefore:
[; \tau_1(n, k + 1) = k + 1 - p + r + \tau_1(k + 1, k + 1 - p + r) ;]
If we raise the boundary of k we get:
[; k > 2 \cdot p \Rightarrow ;]
[; \tau_1(n, k + 1) = k + 1 + \tau_1(k + 1 - p + r, p - r) \Rightarrow ;]
[; k + 1 \le \tau_1(n, k + 1) < k + 1 + 2 \cdot (p - r) \Rightarrow ;]
[; k + 1 \le \tau_1(n, k + 1) < k + 1 + 2 \cdot p \Rightarrow ;]
[; \lim_{n \rightarrow \infty} \frac{\tau_1(n, \lfloor n \cdot \frac{1}{p} \rceil )}{n} = \frac{1}{p} ;]
Thus we have proven not only that L and R are well defined for x=1/p but that the limits obtained make some of the conjectured properties of the left and right mappings true.
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