### Proof

When \(D \geq 3\) is odd, the cycles are obtained from the cycles of \(D-1\) by putting a \(0\) or a \(9\) in the middle of each element of the cycle. And the number of cycles of \(N_{D}\) is equal to the number of cycles of \(N_{D-1}\).

Let us begin with an example. Consider \(D=5\) and \(n=21978\). Subtracting \(n\) to \(i_{5}(n)\) we get \(65\mathbf{9}34\). The natural number \(n\) belongs to a cycle of period \(2\), namely \(\{21\mathbf{9}78, 65\mathbf{9}34\}\). Notice that, if \(n'=2178\), then subtracting \(n'\) to \(i_{4}(n')\) we get \(6534\), with \(\{2178, 6534\}\) being a cycle of period \(2\) of \(N_{4}\). Similarly, if \(n=089108910\), then subtracting \(n\) to \(i_{9}(n)\) we obtain \(069306930\), and the difference from \(i_{8}(n')\) to \(n'=08918910\) is \(06936930\). The natural \(n\) belongs to a cycle of period \(5\), namely \[\{009900990, 089108910, 069306930, 029702970, 049504950\},\] and \[\{00990990, 08918910, 06936930, 02972970, 04954950\}\] is a cycle of \(N_{8}\) with the same period.

In general, if \(D>1\) is odd and \[n = a_{D-1} a_{D-2} ... a_{\frac{D+1}{2}} a_{\frac{D-1}{2}} a_{\frac{D-3}{2}}... a_{0}\] belongs to \(N_{D}\), we have \[f_{D}(n) = \left|\left[10^{D-1}-1\right]\left[a_{D-1}-a_{0}\right]+\left[10^{D-2}-10\right]\left[a_{D-2}-a_{1}\right]+...+ \\ \;\;\;\;+\left[10^{\frac{D+1}{2}}-10^{\frac{D-3}{2}}\right]\left[a_{\frac{D+1}{2}}-a_{\frac{D-3}{2}}\right]+\left[10^{\frac{D-1}{2}}-10^{\frac{D-1}{2}}\right]\left[a_{\frac{D-1}{2}} - a_{\frac{D+1}{2}}\right]\right|=\\ =\left|\left[10^{D-1}-1\right]\left[a_{D-1}-a_{0}\right]+\left[10^{D-2}-10\right]\left[a_{D-2}-a_{1}\right]+...+ \\ \;\;\;\;+ \left[10^{\frac{D+1}{2}}-10^{\frac{D-3}{2}}\right]\left[a_{\frac{D+1}{2}}-a_{\frac{D-3}{2}}\right]\right|=\\= f_{D-1}(n'), \mbox{ modulo a 9 or a 0 at the central position,}\] where \(n'\) is the natural number which is obtained from \(n\) by pulling out the central digit. Notice that the central digit of \(f_{D}(n)\) is \(9\) if the subtraction \(|a_{\frac{D+1}{2}} - a_{\frac{D-3}{2}}\)| leaves a reminder, and in that case such a reminder also exists in the computation of the value of \(f_{D-1}(n')\); and it is \(0\) in both computations otherwise. That is, the orbit of \(n\) by \(f_{D}(n)\) must end in a cycle \(\gamma\) which comes from the cycle \(\alpha\) of \(N_{D-1}\) where the orbit of \(f_{D-1}(n')\) ends, putting a \(0\) or a \(9\) in the middle of each element of \(\alpha\).

Since the orbit of each element of \(N_{D}\) ends in a cycle, it follows that each cycle \(\alpha\) of \(N_{D-1}\) must produce some cycle \(\gamma\) of \(N_{D}\) when putting a \(0\) or a \(9\) at the cental position of some of the elements of \(\alpha\). And which digit should one place there? Notice that, in the tables for \(D\) odd between \(3\) and \(9\), all elements of such a cycle \(\gamma\) have digit \(0\) at the central position, or all have digit \(9\) in that position; for \(D=11\), however, in the cycle \[\{01143966780, 07622967330, 04246044660, 02398019580, 06193069740,\\ 01397030580, 07106048730, 03321988560, 03266923770, 04466042460,\\ 01958023980, 06974061930, 03058013970, 04873071060\}\] some elements of the cycle have a central \(0\) and others a \(9\). In general, if we fix an element \(n'\) of the cycle \(\alpha\) of \(N_{D-1}\), with period \(p\), the digit that is added to the central position of \(\alpha\) in order to obtain \(n\) from a cycle of \(N_{D}\) is that one which is consistent with the equality \((f_{D})^{p}(n)=n\). These remarks guarantee that the number of cycles of \(N_{D}\) is greater or equal than the number of cycles of \(N_{D-1}\).

But, in fact, the number of cycles in \(N_{D}\) is equal to the number of cycles in \(N_{D-1}\). This is a consequence of the following property: if we obtain a cycle of \(N_{D}\) by adding a \(9\) in the central position of a cycle of \(N_{D-1}\), then if, instead of \(9\), we add a \(0\) in that position, we do not obtain a cycle; and conversely. Let us see why.

Given \(D\) odd, suppose we obtain a cycle \(\gamma\) of \(N_{D}\) \[\gamma = a_{D-2} a_{D-3} ... a_{\frac{D-1}{2}} \,c\, a_{\frac{D-3}{2}}...a_{1} a_{0}\] by adding a digit \(c\), which we already know to be either \(0\) or \(9\), in the central position of a cycle \(\alpha\) of \(N_{D-1}\), \[\alpha = a_{D-2} a_{D-3} ... a_{\frac{D-1}{2}} a_{\frac{D-3}{2}}... a_{1} a_{0}.\]

Then the central digit of \(f_{D}(\gamma)\) depends on the existence of a reminder \(1\) coming from the difference \(\left|a_{\frac{D-1}{2}} - a_{\frac{D-3}{2}}\right|\), naturally influenced by subtractions on digits placed at the right side of these ones: if the difference has a reminder, then the central digit of \(f_{D}(\gamma)\) is \(9\) independently of the value of \(c\); if there is no reminder, then it is \(0\), whatever the value of \(c\). This indicates that, if \(c=9\) and, instead of \(9\), we had placed a \(0\) at the central position of \(\alpha\), we would obtain, applying \(f_{D}\), the same value \(f_{D}(\gamma)\). That is, that other element of \(N_{D}\) is in the preimage of \(\gamma\), it does not give a new cycle. A similar argument is used if \(c=0\), thus showing that each cycle of \(N_{D-1}\) generates one and only one cycle of \(N_{D}\).