# Binomial Number System
## 3 Quantitative Characteristics of BNS
### BNS Classes  
The BNS comprises two classes. The first contains numbers with a constant amount of \(k\) units and variables of \(i=0\), \(1\),..., \(n-k-1\) number of zeros. The second class of the BNS consists of the constant number \(n-k\) number of zeros and a variable amount of units \(q=0\), \(1\),..., \(k-1\). **Theorem 7. **_Conditions (2)-(5) divide binomial codes into two disjoint classes, the first of which contains \(k\) ones and \(0\leq l\leq n-k-1\) zeros, and the second includes \(l=n-k\) zeros and \(0\leq q\leq k-1\) units._  
**Proof of Theorem 7. ** From Equation (3), it follows that conditions (2) and (3) generate binomial codes containing \(k\) ones, and from Theorem 2, those conditions (4) and (5) generatenumbers that constantly have \(n-k\) zeros. It also follows from Theorem 1 that the number of zeros in binomial codes specified by conditions (2) and (3), \(l=0\), \(1\),..., \(n-k-1\), and the number of ones in numbers specified by conditions (4) and (5), \(q=0\), \(1\),..., \(k-1\). Moreover, binomial codes corresponding to conditions (2) and (3) do not correspond to requirements (4) and (5) and vice versa. By definition, there are no other binomial codes, except for the numbers generated by conditions (2)-(5). Therefore, each binomial code can belong only to one of two classes of binomial numbers, corresponding to conditions (2) and (3) and consisting of \(k\) ones and \(0\leq l\leq n-k-1\) zeros or resembling conditions (4) and (5) and consisting of \(n-k\) zeros and \(0\leq q\leq k-1\) units. \(\square\)  
**Corollary 3**. : _Binomial codes of the first-class contain \(1\) in the LSB, and the second class include \(0\) in the LSB since when the \(k\)-th unit or (\(n-k\))-th zero appears, is the critical sign of the formation of the binomial codes._  
**Lemma 1**. : _The sum of binomial codes of the first-class contains \(1\) in the LSB, with a fixed number of zeros \(l\), \(0\leq l\leq n-k-1\):_  
\[N_{l}=C_{k+l-1}^{l}. \tag{9}\]  
**Proof of Lemma 1. ** Since in the LSB of each binomial code combinations of the first class, by Corollary 1 of Theorem 7, there is \(1\), then the total number of binomial code of code length \(r\) for a given value \(l\)\(N_{l}=C_{r-1}^{l}\). Since the code lengths of binomial code from this class are \(r=k+l\), then \(r-1=k+l-1\), and accordingly \(N_{l}=C_{k+l-1}^{l}\). \(\square\)  
**Corollary 4**. : _For \(l=l_{max}=n-k-1\):_  
\[N_{l}=N_{lmax}=C_{k+l_{\max}-1}^{l_{\max}}=C_{n-2}^{n-k-1}=C_{n-2}^{k-1}. \tag{10}\]  
**Corollary 5**. : _For \(l=l_{min}=0\):_  
\[N_{l}=N_{l_{\min}}=C_{k-1}^{0}=1. \tag{11}\]  
For parameters \(n=6\), \(k=4\), the number of binomial codewords, which contain \(1\) in the LSB and do not contain zeros, by Corollary 2, equal to \(1\), and with one \(0\)--\(N_{l=1}=C_{r-1}^{l}=C_{5-1}^{1}=4\). These are the codewords: \(1111\)\(u01111\), \(10111\), \(11011\), \(11101\). No other codewords contain more than one zero and have \(1\) in the LSB in this class of binomial codes, since with two zeros, the binomial codes, by Theorem 7, transform to the type of numbers containing \(0\) in the LSB. **Theorem 8**. : _Sum of all binomial code combinations of the first-class contain \(1\) in the LSB:_  
\[N_{k}=C_{n-1}^{k}. \tag{12}\]  
**Proof of Theorem 8. ** Summing the \(N_{l}\) values obtained in Lemma 1 over all possible \(l=0\),..., \(n-k-1\), provides the sum of all binomial code combinations containing \(k\) ones:  
\[N_{k}=\sum_{l=0}^{n-k-1}N_{l}=\sum_{l=0}^{n-k-1}C_{k+l-1}^{l}=C_{n-2}^{n-k+1}+ \ldots+C_{k}^{1}+C_{k-1}^{0}=\sum_{j=0}^{n-k-1}C_{n-2-j}^{n-k-1-j}=C_{n-1}^{n -k-1}=C_{n-1}^{k}.\]  
Theorem 8 that the sum of binomial code combinations for \(n=6\), \(k=4\) containing \(1\) in the LSB is equal to \(C_{n-1}^{k}=C_{6-1}^{4}=5\). The sum of the numbers given in the previous example, the LSB of which includes \(1\), also using Table 1, their total number is \(5\), which confirms this theorem. \(\square\)  
**Lemma 2**.