# Would this function work as a test for prime?

Would the function

$$\sum_{a=2}^\infty H\!\left(x-a\sum_{n=0}^\infty H(x-na)\right)$$

(H(x) is the Heaviside step function) Work as a test for prime numbers

it is zero when x is prime And a positive integer when x is composite

I have added an image in case the formula didn’t work (https://i.stack.imgur.com/MDxQI.jpg)

• @Bill : Consider the integer 23. How will the Heaviside function put a value of 0 or any other positive integer against 23 without already know if 23 is a prime or not? – Nilos Apr 14 at 8:55

Your formula doesn't quite work, but a similar function will. Consider

$$f(x) = \sum_{a=2}^\infty H\!\left( \left(a\sum_{n=1}^\infty H(x-na)\right) - x\right).$$

We can write $$\sum_{n=1}^\infty H(x-na) = \sum_{1 \leq n \leq x /a} 1$$, or the number of integers $$n \geq 1$$ with $$n \leq x/a$$, or equivalently, $$n \leq \lfloor x /a \rfloor$$; thus $$\sum_{n=1}^\infty H(x-na) = \lfloor x / a \rfloor.$$ Then

$$f(x) = \sum_{a=2}^\infty H\!\left(a\lfloor x / a \rfloor -x\right).$$

Note that $$a\lfloor x / a \rfloor = x$$ when $$a | x$$, and otherwise $$a \lfloor x / a \rfloor < x$$. Therefore the summand $$H\!\left(a\lfloor x / a \rfloor -x\right) = 1$$ if $$a | x$$ and is $$0$$ otherwise. Thus when $$x$$ is prime, $$f(x) = 1$$ since $$x$$ has only one divisor $$d > 2$$, but if x is composite then $$f(x) > 1$$.

(in your original formula you get $$\sum_{a=2}^\infty H\!\left(x - a(1+\lfloor x / a \rfloor)\right)$$

which is always $$0$$ since $$a(1 + \lfloor x /a \rfloor) > x$$ always.)

Of course, evaluating this function is like performing trial division for all integers below $$x$$, so it is not very fast.