Factorion

In number theory, a factorion in a given number base $b$ 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. We define the sum of the factorial of the digits of $n$ for base $b>1$ $SFD_{b}:\mathbb {N} \rightarrow \mathbb {N}$ to be the following:

$SFD_{b}(n)=\sum _{i=0}^{k-1}d_{i}!$.

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 each digit of the number. A natural number $n$ is a $b$-factorion if it is a fixed point for $SFD_{b}$, which occurs if $SFD_{b}(n)=n$. $1$ and $2$ are fixed points for all $b$, 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 factorial of the digits is simply the number of digits $k$ in the base 2 representation.

A natural number $n$ is a sociable factorion if it is a periodic point for $SFD_{b}$, where $SFD_{b}^{k}(n)=n$ for a positive integer $k$, and forms a cycle of period $k$. A factorion is a sociable factorion with $k=1$, and a amicable factorion is a sociable factorion with $k=2$.

All natural numbers $n$ are preperiodic points for $SFD_{b}$, regardless of the base. This is because all natural numbers of base $b$ with $k$ digits satisfy $b^{k-1}\leq n\leq (b-1)!(k)$. However, when $k\geq b$, then $b^{k-1}>(b-1)!(k)$ for $b>2$, so any $n$ will satisfy $n>SFD_{b}(n)$ until $n<b^{b}$. There are a finite number of 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 $SFD_{b}^{i}(n)$ to reach a fixed point is the $SFD_{b}$ function's persistence of $n$, and undefined if it never reaches a fixed point.

Factorions for $SFD_{b}$

b = (k - 1)!

Let $k$ be a positive integer and the number base $b=(k-1)!$. Then:

• $n_{1}=kb+1$ is a factorion for $SFD_{b}$ for all $k$.

Proof

Let the digits of $n_{1}=d_{1}b+d_{0}$ be $d_{1}=k$, and $d_{0}=1$. Then

$SFD_{b}(n_{1})=d_{1}!+d_{0}!$

$=k!+1!$

$=k(k-1)!+1$

$=d_{1}b+d_{0}$

$=n_{1}$

Thus $n_{1}$ is a factorion for $F_{b}$ for all $k$.

• $n_{2}=kb+2$ is a factorion for $SFD_{b}$ for all $k$.

Proof

Let the digits of $n_{2}=d_{1}b+d_{0}$ be $d_{1}=k$, and $d_{0}=2$. Then

$SFD_{b}(n_{2})=d_{1}!+d_{0}!$

$=k!+2!$

$=k(k-1)!+2$

$=d_{1}b+d_{0}$

$=n_{2}$

Thus $n_{2}$ is a factorion for $F_{b}$ for all $k$.

Factorions

$k$	$b$	$n_{1}$	$n_{2}$
4	6	41	42
5	24	51	52
6	120	61	62
7	720	71	72

b = k! - k + 1

Let $k$ be a positive integer and the number base $b=k!-k+1$. Then:

• $n_{1}=b+k$ is a factorion for $SFD_{b}$ for all $k$.

Proof

Let the digits of $n_{1}=d_{1}b+d_{0}$ be $d_{1}=1$, and $d_{0}=k$. Then

$SFD_{b}(n_{1})=d_{1}!+d_{0}!$

$=1!+k!$

$=k!+1-k+k$

$=1(k!-k+1)+k$

$=d_{1}b+d_{0}$

$=n_{1}$

Thus $n_{1}$ is a factorion for $F_{b}$ for all $k$.

Factorions

$k$	$b$	$n_{1}$
3	4	13
4	21	14
5	116	15
6	715	16

Table of factorions and cycles of $SFD_{b}$

All numbers are represented in base $b$.

Base $b$	Nontrivial factorion ($n\neq 1$, $n\neq 2$)	Cycles
2	$\varnothing$	$\varnothing$
3	$\varnothing$	$\varnothing$
4	13	3 → 12 → 3
5	144	$\varnothing$
6	41, 42	$\varnothing$
7	$\varnothing$	36 → 2055 → 465 → 2343 → 53 → 240 → 36
8	$\varnothing$	3 → 6 → 1320 → 12 175 → 12051 → 175
9	62558
10	145, 40585	871 → 45361 → 871 872 → 45362 → 872

Programming example

The example below implements the sum of the factorial of the digits described in the definition above to search for factorions and cycles in Python.

def factorial(x: int) -> int:
    total = 1
    for i in range(0, x):
        total = total * (i + 1)
    return total

def sfd(x: int, b: int) -> int:
    """Sum of the factorial of the digits."""
    total = 0
    while x > 0:
        total = total + factorial(x % b)
        x = x // b
    return total

def sfd_cycle(x: int, b: int) -> List[int]:
    seen = []
    while x not in seen:
        seen.append(x)
        x = sfd(x, b)
    cycle = []
    while x not in cycle:
        cycle.append(x)
        x = sfd(x, b)
    return cycle

See also

• Arithmetic dynamics
• Dudeney number
• Happy number
• Kaprekar's constant
• Kaprekar number
• Meertens number
• Narcissistic number
• Perfect digit-to-digit invariant
• Perfect digital invariant
• Sum-product number

External links

• Factorion at Wolfram MathWorld