Sum-product number

A sum-product number in a given number base $b$ is a natural number that is equal to the product of the sum of its digits and the product of its digits.

There are a finite number of sum-product numbers in any given base $b$.Proof that number of sum-product numbers in any base is finite
''PlanetMath''. by Raymond Puzio In base 10, there are exactly four sum-product numbers : 0, 1, 135, and 144.

Definition

Let $n$ be a natural number. We define the sum-product function for base $b > 1$ $F_ : \mathbb \rightarrow \mathbb$ to be the following:
: $F_(n) = \left(\sum_^k d_i\right)\left(\prod_^k d_j\right)$
where $k = \lfloor \log_ \rfloor + 1$ is the number of digits in the number in base $b$, and
: $d_i = \frac$
is the value of each digit of the number. A natural number $n$ is a sum-product number if it is a fixed point for $F_$, which occurs if $F_(n) = n$. The natural numbers 0 and 1 are trivial sum-product numbers for all $b$, and all other sum-product numbers are nontrivial sum-product numbers.

For example, the number 144 in base 10 is a sum-product number, because $1 + 4 + 4 = 9$, $(1)(4)(4) = 16$, and $(9)(16) = 144$.

A natural number $n$ is a sociable sum-product number if it is a periodic point for $F_$, where $F_^p(n) = n$ for a positive integer $p$, and forms a cycle of period $p$. A sum-product number is a sociable sum-product number with $p = 1$, and a amicable sum-product number is a sociable sum-product number with $p = 2$.

All natural numbers $n$ are preperiodic points for $F_$, regardless of the base. This is because for any given digit count $k$, the minimum possible value of $n$ is $b^$ and the maximum possible value of $n$ is $b^ - 1 = \sum_^ (b - 1)^k$. The maximum possible digit sum is therefore $k(b - 1)$ and the maximum possible digit product is $(b - 1)^k$. Thus, the sum-product function value is $F_(n) = k(b-1)^$. This suggests that $k(b - 1)^ \geq n \geq b^$, or dividing both sides by $^$, $k(b - 1)^2 \geq ^$. Since $\frac \geq 1$, this means that there will be a maximum value $k$ where $^ \leq k(b - 1)^2$, because of the exponential nature of $^$ and the linearity of $k(b - 1)^2$. Beyond this value $k$, $F_(n) \leq n$ always. Thus, there are a finite number of sum-product numbers, and any natural number is guaranteed to reach a periodic point or a fixed point less than $b^ - 1$, making it a preperiodic point.

The number of iterations $i$ needed for $F_^(n)$ to reach a fixed point is the sum-product function's persistence of $n$, and undefined if it never reaches a fixed point.

Any integer shown to be a sum-product number in a given base must, by definition, also be a Harshad number in that base.

Sum-product numbers and cycles of ''F''_''b'' for specific ''b''

All numbers are represented in base $b$.

Extension to negative integers

Sum-product numbers can be extended to the negative integers by use of a signed-digit representation to represent each integer.

Programming example

The example below implements the sum-product function described in the definition above to search for sum-product numbers and cycles in Python.

def sum_product(x: int, b: int) -> int:
"""Sum-product number."""
sum_x = 0
product = 1
while x > 0:
if x % b > 0:
sum_x = sum_x + x % b
product = product * (x % b)
x = x // b
return sum_x * product

def sum_product_cycle(x: int, b: int) -> list nt
seen = []
while x not in seen:
seen.append(x)
x = sum_product(x, b)
cycle = []
while x not in cycle:
cycle.append(x)
x = sum_product(x, b)
return cycle

See also

* Arithmetic dynamics
* Dudeney number
* Factorion
* Happy number
* Kaprekar's constant
* Kaprekar number
* Meertens number
* Narcissistic number
* Perfect digit-to-digit invariant
* Perfect digital invariant

References

{{Classes of natural numbers
Arithmetic dynamics
Base-dependent integer sequences