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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)

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  • $\begingroup$ @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? $\endgroup$ – Nilos Apr 14 at 8:55
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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.

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