Least Common Multiple Count

Let $f(n)$ be the number of couples $(x, y)$ with $x$ and $y$ positive integers, $x \le y$ and the least common multiple of $x$ and $y$ equal to $n$.

Let $g$ be the summatory function of $f$, i.e.:
$g(n) = \sum f(i)$ for $1 \le i \le n$.

You are given that $g(10^6) = 37429395$.

Find $g(10^{12})$.

This problem involves a mixture of combinatorics and number theory in mathematics.

In simple terms, the function f(n) is counting the number of pairs of positive integers that have the least common multiple of n. This means that the problem is essentially asking for a summation of all the different divisors that a number has, since each divisor pair represents one way that two numbers can have an LCM of n.

The deterministic ways to compute this might involve counting the number of divisors that each number has, and then summing these all up. However, this is computationally infeasible when g(n) and f(n) are big. Instead, we could use the properties of divisor functions and prime numbers to find an efficient way to compute g(10^{12}).

Define T(p, a) to denote the number of ordered pairs (x, y) such that lcm(x, y) = p^a. It’s easy to see that T(p, a) = 2a+1 for a > 0 and T(p, 0) = 0. This can be explained as follows: Given p^a, where p is a prime, we have a+1 ways to choose the power of p for our first number (x). After that, we have a+1 ways to assign the remaining power of p to our second number (y), given that we have x <= y. In fact, it doesn't matter what p is or how large a is; as long as they form a prime power, the number of pairs will be T(p, a). Therefore, it is clear that for any arbitrary n = p1^{a1} * p2^{a2} *... *pk^{ak}, where p1, p2, ..., pk are distinct primes and a1, a2, ..., ak are their respective powers, f(n) = (2*a1 + 1) * (2*a2 + 1) * ... * (2*ak + 1). The function g(n) is then a summation of f(i) for 1 <= i <= n. Given g(10^6) = 37429395, we can use the formula f(n) = (2*a1 + 1) * (2*a2 + 1) * ... * (2*ak + 1) and the sieve algorithm (to factorize the integers and sum up the contributions for all n up to a certain bound) to compute g(10^{12}). However, this computation is beyond the capabilities of a human without a computer. It requires a good understanding of combinatorics, number theory and algorithm design to write a computer program to calculate it.

More Answers:
Nontransitive Sets of Dice
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