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Factorion

In number theory, a factorion in a given number base is a natural number that equals the sum of the factorials of its digits. The name factorion was coined by the author Clifford A. Pickover.

Definition
Let n be a natural number. For a base b > 1, we define the sum of the factorials of the digits of n, \operatorname{SFD}_b : \mathbb{N} \rightarrow \mathbb{N}, to be the following: :\operatorname{SFD}_b(n) = \sum_{i=0}^{k - 1} d_i!. where k = \lfloor \log_b n \rfloor + 1 is the number of digits in the number in base b, n! is the factorial of n and :d_i = \frac{n \bmod{b^{i+1}} - n \bmod{b^{i}}}{b^{i}} is the value of the ith digit of the number. A natural number n is a b-factorion if it is a fixed point for \operatorname{SFD}_b, i.e. if \operatorname{SFD}_b(n) = n. 1 and 2 are fixed points for all bases b, and thus are trivial factorions for all b, and all other factorions are nontrivial factorions. For example, the number 145 in base b = 10 is a factorion because 145 = 1! + 4! + 5!. For b = 2, the sum of the factorials of the digits is simply the number of digits k in the base 2 representation since 0! = 1! = 1. A natural number n is a sociable factorion if it is a periodic point for \operatorname{SFD}_b, where \operatorname{SFD}_b^c(n) = n for a positive integer c, and forms a cycle of period c. A factorion is a sociable factorion with c = 1, and a amicable factorion is a sociable factorion with c = 2. All natural numbers n are preperiodic points for \operatorname{SFD}_b, regardless of the base. This is because all natural numbers of base b with k digits satisfy b^{k-1} \leq n . Given that each of the k digits is at most b-1, \operatorname{SFD}_b \leq (b-1)!k. However, when k \geq b, then b^{k-1} > (b-1)!(k) for b > 2, so any n will satisfy n > \operatorname{SFD}_b(n) until n . There are finitely many natural numbers less than b^b, so the number is guaranteed to reach a periodic point or a fixed point less than b^b, making it a preperiodic point. For b = 2, the number of digits k \leq n for any number, once again, making it a preperiodic point. This means also that there are a finite number of factorions and cycles for any given base b. The number of iterations i needed for \operatorname{SFD}_b^i(n) to reach a fixed point is the \operatorname{SFD}_b function's persistence of n, and undefined if it never reaches a fixed point. ==Factorions for ==
Factorions for {{Math|{{math|SFDb}}}}
===b = (m − 1)!=== Let m be a positive integer and the number base b = (m - 1)!. Then: • n_1 = mb + 1 is a factorion for \operatorname{SFD}_b for all m\geq 4. {{Math proof|title=Proof|drop=hidden|proof= Let the digits of n_1 = d_1 b + d_0 be d_1 = m, and d_0 = 1. Then : \operatorname{SFD}_b(n_1) = d_1! + d_0! :: = m! + 1! :: = m(m - 1)! + 1 :: = d_1 b + d_0 :: = n_1 Thus n_1 is a factorion for F_b for all k. }} • n_2 = mb + 2 is a factorion for \operatorname{SFD}_b for all m\geq 4. {{Math proof|title=Proof|drop=hidden|proof= Let the digits of n_2 = d_1 b + d_0 be d_1 = m, and d_0 = 2. Then : \operatorname{SFD}_b(n_2) = d_1! + d_0! :: = m! + 2! :: = m(m - 1)! + 2 :: = d_1 b + d_0 :: = n_2 Thus n_2 is a factorion for F_b for all k. }} ===b = m! − m + 1=== Let k be a positive integer and the number base b = m! - m + 1. Then: • n_1 = b + m is a factorion for \operatorname{SFD}_b for all m\geq 3. {{Math proof|title=Proof|drop=hidden|proof= Let the digits of n_1 = d_1 b + d_0 be d_1 = 1, and d_0 = m. Then : \operatorname{SFD}_b(n_1) = d_1! + d_0! :: = 1! + m! :: = m! + 1 - m + m :: = 1(m! - m + 1) + m :: = d_1 b + d_0 :: = n_1 Thus n_1 is a factorion for F_b for all m. }} Table of factorions and cycles of All numbers are represented in base b. ==See also==
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