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I want to find how many pairs of numbers satisfy this condition on $[1,n]$.

For given $n$ , how many pairs $(a,b)$ are there such that $gcd(a,b) = 2^t , t > 0 $ for some whole number $t$.

All pairs $(a,b)$ should be bounded by $1\leq a < b \leq n $ .

Now I know that if $2|a \land2|b \implies 2|gcd(a,b)$ and this excludes all odd numbers .

But the space is too big to count for very large $n$.

Can I make general formula for $\forall n $ ?

$a,b,n,t$ are whole numbers.

Noodle
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  • Is an asymptotic answer sufficient? It's not too hard to figure out that, if $f(n)$ is the number of such pairs, then $\lim_{n\rightarrow\infty}\frac{f(n)}{n^2}$ is positive and finite (and it's not even so hard to compute this constant). I don't see how one would get an exact answer for all $n$ though without considerable computation. – Milo Brandt Jan 02 '19 at 00:10
  • that is the main problem here, for $n = 10^{6} $ , with brute force it takes about 3-4 hours to compute them , if its not possible for a general formula to be made maybe decreasing the search space will help but this of course requires more properties for such pairs – Noodle Jan 02 '19 at 00:15

1 Answers1

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Let $\psi(n)=\#\{(x,y) : 1 \leqslant x,y \leqslant n, \gcd(x, y) = 1\}$. Then, for $1 \leqslant d \leqslant n$, $$\#\{(x,y) : 1 \leqslant x,y \leqslant n, \gcd(x, y) = d\} = \psi\big(\lfloor n/d \rfloor\big),$$ and the union of these sets is just $\{(x,y) : 1 \leqslant x,y \leqslant n\}$. This gives $$\color{blue}{\sum_{d=1}^{n}\psi\big(\lfloor n/d \rfloor\big)=n^2} \implies \psi(n)=\sum_{d=1}^{n}\mu(d)\lfloor n/d \rfloor^2$$ (by the "second" Möbius inversion formula). This leads to a solution of your problem: $$2^t \leqslant n \implies \#\{(a, b) : 1 \leqslant a < b \leqslant n, \gcd(a, b) = 2^t \} = \frac{\psi(\lfloor 2^{-t} n \rfloor) - 1}{2}.$$

There's an algorithm which computes $\psi(n)$ in $\mathcal{O}(n^{1+\epsilon})$ time and $\mathcal{O}(n^{1/2+\epsilon})$ space (without computing any values of $\mu$). The idea is to use the "blue" formula above recursively (or rather iteratively), noting that there are just $\mathcal{O}(\sqrt{n})$ different values of $\lfloor n/d \rfloor$ for $1 \leqslant d \leqslant n$, which allows to simplify the sum; namely, for any $1 \leqslant k \leqslant n$, $$n^2=\sum_{d=1}^{n}\psi\left(\left\lfloor\frac{n}{d}\right\rfloor\right)=\sum_{d=1}^{\lfloor n/(k+1) \rfloor}\psi\left(\left\lfloor\frac{n}{d}\right\rfloor\right)+\sum_{d=1}^{k}\psi(d)\left(\left\lfloor\frac{n}{d}\right\rfloor-\left\lfloor\frac{n}{d+1}\right\rfloor\right),$$ and we may put $k=\lfloor\sqrt{n}\rfloor$ here. This recurrence for $\psi(n)$ gives the algorithm.

metamorphy
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