# Factorial Numbers and Their Practical Applications
## 3 Theory of Factorial Numbers  
Like any positional number system, the factorial system has a digit value, an alphabet of numbers and weighting coefficients for each digit. In terms of complexity, it is approximately the same as decimal. The difference is that the maximum value of each digit of a factorial number is equal to the number of that digit. If it is, for example, the hundredth digit, then the maximum value for it will be equal to 100, and the number of all digits in the number, considering the zero digit, will be equal to 101. The weighting coefficient in this case is 100! = 100 \(\times\) 99 \(\times\ldots\times\) 1. The preceding 99-th digit will have a maximum value of 99 and a weight of 99!. And so it goes further until the zero digit, the weight of which is 0! = 1. All these features of factorial number systems are shown by Equation (1) for the numerical factorial function. The weights of the factorial numbers in it increase according to a dependence close to exponential. A factorial number is a positional \(n\)-digit number \(F=x_{n-1}x_{n-2}\;\ldots\;x_{i}\;\ldots\;x_{1}x_{0}\), where \(x\) is the number of the \(i\)-th digit, \(0\leq x_{i}\leq i,\;\;0\leq i\leq n-1\)[3; 4], for which the numerical function is  
\[F=x_{n-1}\times(n-1)!\;+\;x_{n-2}\times(n-2)!\;+\ldots+x_{i}\times i!+\ldots+x_ {1}\times 1!\;+0\times 0!. \tag{1}\]  
In this case, the expression \(i!=i\times(i-1)\times\ldots\times 1\) is called the factorial of \(i\). Then, for example, \(4!=4\times 3\times 2\times 1=24\). In other words, the number \(F\) is written as a sequence of \(n\) digits, starting with the zero digit. It is assumed that the number \(i,i=0,1,\ldots,n-1\), corresponds to the weight \(i!\). The number of digits in the numbers of the factorial number system, as Equation (1) shows, is not related to the factorials. It increases linearly. For the zero digit there is one--\"0\", for the first digit there are two--\"0\" and \"1\", for the second three--\"0\", \"1\", \"2\" and so on. These numbers are easily converted into binary numbers, and then all processing of information presented in factorial numbers occurs in binary codes. For example, the six-digit factorial number \(N=3\;4\;0\;0\;1\;0\) is represented using a numeric function as \(F=3\times 5!+4\times 4!+0\times 3!+0\times 2!+1\times 1!+0\times 0!=360+96+0 +0\)\(+1+0=457\). For the theory and practical application of factorial numbers, what matters is the minimum \(Fmin\) and maximum \(Fmax\) value of the number \(F\), as well as the range of representation of the numbers \(P\). The minimum factorial number will be when in (1) all \(x_{i}=0\). Accordingly, \(F=Fmin=0\;\cdot(n-1)!+\ldots+0\cdot 1!+0\cdot 0!=0\). The maximum factorial number \(F=Fmax\) will be when in (1) all \(x_{i}=i\), \(i=0,1,\ldots,n-1\), since, as follows from Expression (1), any deviation of the values of the numbers \(x_{i}\) from the values of \(i\) downward will lead to a decrease in the number \(F\). **Theorem 1**. : _Maximum number in n-digit factorial number system \(Fmax=n!-1\)._  
**Proof of Theorem 1. ** Consider a function F in which \(x_{i}=i\), \(i=0\), \(1\ldots\), \(n-1\). Then, given the equality \(i\times i!\ +\ i!=i!\ (i+1)=(i+1)!\) we get \(F=(n-1)\times(n-1)!\ +\ (n-2)\times(n-2)!\ +\ \ldots\ +i\times i!+\ldots+2\times 2!\ +1\times 1!\ +\ 0\times 0!=(n-1)\times(n-1)!\ +\ (n-2)\times(n-2)!\ +\ \ldots\ +i \times i!+\ldots+3\times 3!+2\times 2!\ +2!\ -1=(n-1)\times(n-1)!\ +\ (n-2)\times(n-2)!\ +\ \ldots\ +i \times i!\ +\ldots+3\times 3!+3!\ -1=(n-1)\times(n-1)!\ +\ (n-2)\times(n-2)!\ +\ \ldots\ +i \times i!\ +i!\ -1=(n-1)\times(n-1)!\ +\ (n-1)!\ -1=n!\ -1=F_{max}\). The theorem has been proven. \(\Box\)  
**Consequence 1**.