I came up with the following formula and deleted that question so that I don't have two questions on the same formula.
Conjecture. Let $a, b, n \in 2\Bbb{N} + 1$ be odd natural numbers. Then the formula:
$$
H(a, b, n) = \left \lfloor \dfrac{b}{n} \right \rfloor -\left \lfloor\dfrac{a}{n}\right\rfloor + C(n)
$$
for some function $C: (\Bbb{2\Bbb{N} + 1}) \to \{-1, 0, 1\}$, counts exactly the number of holes punched into the set $P = \{(o, o+2) : o, o+2 \in [a, b]\}$, where by "punches a hole" we mean either $n \mid o$ or $n \mid (o+2)$.
Question. It seems to be true computationally, so what is a formula for computing $C(n)$ that is similar to the formula so far (can use floor, mod, +, -, /, *)?
I tried experimenting with the code to come up with the formula, but that failed, so I'm only posting the code so that you can see apparently $C(n) \in \{-1, 0, 1\}$, but not to "experiment with and find $C(n)$". What we need to do is a formal analysis based on divisibilty / modular arithmetic, possibly breaking up into modulo cases.
The code:
from math import floor, sqrt
N = 50
def actual_count(a, b, n):
count = 0
for x in range(a, b-1):
if x % 2 == 1:
if x % n == 0 or (x+2) % n == 0:
count += 1
return count
m = 0
m1 = 0
for a in range(0, N):
a = 2*a + 1
for b in range(0, N):
b = 2*b + 1
if b >= a + 4:
for n in range(1, N):
n = 2*n + 1
if n <= sqrt(b):
Hn = floor(b/n) - floor(a/n)
Cn = actual_count(a,b,n)
if Hn - Cn < m1:
m1 = Hn - Cn
if Hn - Cn > m:
m = Hn - Cn
print(a, b, n, Hn - Cn)
print (m1, m)
seems to always print a min/max of $-1, 1$ at the end of the program, regardless of how high you set $N$ to be. Hence the conjecture.
Motivation.
This is related to twin prime counting formulas, so I've tagged with that. Essentially once you have $H(n) = H(a,b,n)$ you apply the inclusion-exclusion principle so that:
$$
\pi_2(a,b) = \dfrac{b – a}{2} – \sum_{3 \leq q \leq \sqrt{b}} H(q) – \sum_{3 \leq q \lt q' \leq \sqrt{b}} H(qq') + \dots
$$
where the $i$th summation is over all distinct odd prime $i$-tuples with each prime $\leq \sqrt{b}$. That is easy to see for me since I've been messing around with primes for a while. I just wanted to mention what this applies to. So, $\pi_2(a,b)$ counts exactly the number of twin prime pairs lying in the interval $[a, b]$.
The next step would be comparing $\pi_2(a, b^2)$ with $\pi_2(a, b^2 + 2b)$ since then the summations in each formula are over the exact same ranges! So essentially all terms involving $a$ will cancel. And what you get is an expression related to the inclusion-exclusion formula for $\pi(b^2) – \pi(b) + 1$ which can be found at Algorithms for evaluating $\pi(x)$.
What should happen is that if the twin primes were finite, then there exists a natural number $N$ such that for all $b \gt a \gt N$ we have an equation involving $\pi(x)$ that will just seem absurd with respect to the prime number theorem's result.
By experimenting with the Python code. The result (for $a = 0$) seems to be:
$$
H(0, b,n) := \lfloor \dfrac{b + n – 2}{n} \rfloor + D(b,n)
$$
where $D(b,n) \in \{0, 1\}$. This is a lot closer, but still not exact.
Attempt. Assuming the solution $C(n)$ takes (and we want it to take) the form $C(n) = \lfloor\dfrac{nX + bY + aZ + W}{n} \rfloor$, we have that the computer can't find coefficients that do the job:
from math import floor, sqrt
def hole_actual_count(a, b, n):
count = 0
for x in range(a, b-2):
if x % 2 == 1:
if x % n == 0 or (x + 2) % n == 0:
count += 1
return count
def rough_hole_estimate(a, b, n):
return floor((b + n - 2)/n)
def error_in_estimate(a, b, n):
return hole_actual_count(a, b, n) - rough_hole_estimate(a, b, n)
def error_function_desired_form(X, Y, Z, W, a, b, n):
return floor((X*a + Y*b + Z*n + W) / n)
N = 20
a = 1
min_soln = (N, N, N, N)
min_solns = []
for X in range(-N, N):
for Y in range(-N, N):
for Z in range(-N, N):
for W in range(-N, N):
for b in range(0, N):
b = 2*b + 1
for n in range(1, int(floor(sqrt(b)) / 2) + 1):
n = 2*n + 1
En = error_in_estimate(a, b, n)
if error_function_desired_form(X, Y, Z, W, a, b, n) != En:
break
else:
continue
break
else:
soln = (X,Y,Z,W)
Xm,Ym,Zm,Wm = min_soln
if abs(X*Y*Z*W) < abs(Xm*Ym*Zm*Wm):
min_soln = soln
min_solns.append(soln)
print(min_solns)
prints [] since there are no apparent solutions at least with low coefficients.
Best Answer
You have that $\lfloor\frac{b}{n}\rfloor$ counts how many numbers less or equal to b are multiple of $n$. From these you have to subtract those multiple of $2n$ to have only the odd ones, getting $\lfloor\frac{b}{n}\rfloor-\lfloor\frac{b}{2n}\rfloor$. Now if you want only those in $[a,b]$ you have to subtract the count for $a−1$ to get: $\lfloor\frac{b}{n}\rfloor-\lfloor\frac{b}{2n}\rfloor-\lfloor\frac{a-1}{n}\rfloor+\lfloor\frac{a-1}{2n}\rfloor$. After this, everything must be multiplied by $2$ because most of those numbers belong to $2$ couples, but we need to subtract the contribution of $a$ and $b$ if these are multiples of $n$: this can be accomplished with $\lfloor\frac{b}{n}\rfloor-\lfloor\frac{b-1}{n}\rfloor$ and $\lfloor\frac{a}{n}\rfloor-\lfloor\frac{a-1}{n}\rfloor$. So we finally get:
$$\begin{aligned} H(a,b,n) &= 2\left(\lfloor\frac{b}{n}\rfloor-\lfloor\frac{b}{2n}\rfloor-\lfloor\frac{a-1}{n}\rfloor+\lfloor\frac{a-1}{2n}\rfloor\right)-\lfloor\frac{b}{n}\rfloor+\lfloor\frac{b-1}{n}\rfloor-\lfloor\frac{a}{n}\rfloor+\lfloor\frac{a-1}{n}\rfloor \\&= \lfloor\frac{b}{n}\rfloor-2\lfloor\frac{b}{2n}\rfloor-\lfloor\frac{a-1}{n}\rfloor+2\lfloor\frac{a-1}{2n}\rfloor+\lfloor\frac{b-1}{n}\rfloor-\lfloor\frac{a}{n}\rfloor \end{aligned}$$