# Binomial Number System
## 3 Quantitative Characteristics of BNS  
### BNS Property  
Each PNS, including BNS, must meet several requirements for such number systems. First of all, this is the requirement to represent the digits of the number system in bits. After this, the implementation of transitions from these codes to the corresponding codes of the natural number system. The transition for PNS with redundant information leads to its elimination and, accordingly, information compression. Moreover, the requirement for PNS is the ability to implement the reverse transition from the obtained natural number to the digit of the original redundant number system. At the same time, in the transition from one natural number system to another, there is no compression of information since natural number systems do not contain redundant information. As shown above, binomial numbers effectively represent using the system conditions (2) and (3), (4) and (5), and from the numerical function for the BNS (1), it follows that it is calculating entirely and, therefore, is capable of transforming binomial numbers to their natural digits. The inverse problem of converting the digits of the natural number system into binomial numbers can also be solved efficiently using this numerical function [21]. Thus, the requirement for efficient representation of binomial numbers using BNS is performers. Algorithms for constructing binomial numbers obtain if the conditions meet the specified restrictions. Further theoretical constructions aim to prove the theoretical foundations for creating all binomial numbers from a given range. **Theorem 1**. : _The sum of zeros contained in the binomial code combinations formed by conditions (2) and (3):_  
\[l=0,1,\ldots,n-k-1. \tag{7}\]  
Proof of Theorem 1.: The sum of zeros \(l\) in binomial code combinations is equal to \(l=r-q\). Condition (3) shows that \(q=k\). Therefore, \(l=r-k\). Since it follows from condition (2) of the system of conditions (2) and (3) that the minimum value is \(r=r_{min}=k\), and the maximum value is \(r=r_{max}=n-1\). Then, substituting the minimum value \(r_{min}\) and maximum value into equation \(l=r-k\), as a result \(l=l_{min}=0\) and \(l=l_{max}=n-k-1\). Other possible values are between \(l_{min}\) and \(l_{max}\), respectively \(l=0\), \(1\), \(\ldots\), \(n-k-1\). \(\square\)  
**Theorem 2**. : _The sum of zeros in binomial codes, which comply with the conditions (4) and (5):_  
\[l=n-k. \tag{8}\]  
Proof of Theorem 2.: The sum of zeros \(l\) in binomial codes, including specified by the conditions (4) and (5), equals \(l=r-q\). It follows from Equation (4) that the number of ones in the corresponding binomial code combinations is \(q=r-(n-k)\). This is substituting the last equality in the previous: \(l=n-k\). \(\square\)  
**Theorem 3**. : _The number of units \(q\), which correspond to conditions (2) and (3), equals \(k\), and conditions (4) and (5)--0, \(1\), \(\ldots\), \(k-1\)._  
Proof of Theorem 3.: The first condition of the theorem follows from Equation (3) of the system of conditions (2) and (3). From Equation (5) of the system of conditions (4) and (5) follows that \(q=q_{min}=0\)\(uq=q_{max}=k-1\). Other possible values of \(q\) are in the range between \(q_{min}\) and \(q_{max}\). Therefore, the number of ones in binomial codes with zero in the LSB is \(q=0\), \(1\), \(\ldots\), \(k-1\). \(\square\)  
**Theorem 4**. : _The maximum code length of binomial codes specified by conditions (4) and (5) is \(r_{max}=n-1\), the minimum is \(r_{min}=n-k\)._  
**Proof of Theorem 4. ** From Equation (4) of the system of conditions (4) and (5), it follows that \(r=n-k+q\), and from Equation (5) of the same system, it follows that the maximum value \(q=q_{max}=k-1\). Then, \(r=r_{max}=n-k+q_{max}=n-k+k-1=n-1\). It also follows from (5) that the minimum value \(q=q_{min}=0\) and, accordingly, \(r=r_{min}=n-k+q_{min}=n-k\). \(\square\)  
**Theorem 5.