Sum-product number explained

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

. In base 10, there are exactly four numbers : 0, 1, 135, and 144.^[1]

Definition

Let

be a natural number. We define the sum-product function for base

b>1

F_b:N → N

, to be the following:

F_b(n)=

	k
\left(\sum
	i=1

d_{i\right)\left(\prod}

	k

	j=1

d_j\right)

where

k=\lfloorlog_b{n}\rfloor+1

is the number of digits in the number in base

, and

d_i=

	n\bmod{bⁱ⁺¹
	-

n\bmodb^i}{b^i}

is the value of each digit of the number. A natural number

is a number if it is a fixed point for

F_b

, which occurs if

F_b(n)=n

. The natural numbers 0 and 1 are trivial numbers for all

, and all other numbers are nontrivial numbers.

For example, the number 144 in base 10 is a sum-product number, because

1+4+4=9

1 x 4 x 4=16

, and

9 x 16=144

A natural number

is a sociable sum-product number if it is a periodic point for

F_b

, where

	p(n)
F
	b

for a positive integer

, and forms a cycle of period

. A number is a sociable number with

p=1

, and an amicable number is a sociable number with

p=2.

All natural numbers

are preperiodic points for

F_b

, regardless of the base. This is because for any given digit count

, the minimum possible value of

b^k-1

and the maximum possible value of

b^k-1=

	k-1
\sum
	i=0

(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 function value is

F_b(n)=k(b-1)^k+1.

This suggests that

k(b-1)^k+1\geqn\geqb^k-1,

or dividing both sides by

(b-1)^k-1

k(b-1)²\geq{\left(

	b
	b-1

\right)}^k-1.

Since

	b
	b-1

\geq1,

this means that there will be a maximum value

where

{\left(	b
	b-1

\right)}^k\leqk(b-1)^2,

because of the exponential nature of

{\left(	b
	b-1

\right)}^k

and the linearity of

k(b-1)^2.

Beyond this value

F_b(n)\leqn

always. Thus, there are a finite number of numbers, and any natural number is guaranteed to reach a periodic point or a fixed point less than

b^k-1,

making it a preperiodic point.

The number of iterations

needed for

	i(n)
F
	b

to reach a fixed point is the function's persistence of

, 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

Base	Nontrivial sum-product numbers	Cycles
	(none)	(none)
	(none)	2 → 11 → 2, 22 → 121 → 22
	12	(none)
	341	22 → 31 → 22
	(none)	(none)
	22, 242, 1254, 2343, 116655, 346236, 424644
	(none)
	13, 281876, 724856, 7487248	53 → 143 → 116 → 53
	135, 144
	253, 419, 2189, 7634, 82974
	128, 173, 353
	435, A644, 268956
	328, 544, 818C
	2585
	14
	33, 3B2, 3993, 3E1E, C34D, C8A2
	175, 2D2, 4B2
	873, B1E, 24A8, EAH1, 1A78A, 6EC4B7
	1D3, 14C9C, 22DCCG
	1CC69
	24, 366C, 6L1E, 4796G
	7D2, J92, 25EH6
	33DC
	15, BD75, 1BBN8A
	81M, JN44, 2C88G, EH888
	\varnothing
	15B
	\varnothing
	976, 85MDA
	44, 13H, 1E5
	\varnothing
	1KS69, 54HSA
	25Q8, 16L6W, B6CBQ
	4U5W5
	16, 22O

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 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[int]: 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

Notes and References

A038369. Numbers n such that n = (product of digits of n) * (sum of digits of n)..