Counting Ordered Factorisations

Define $d(n,k)$ to be the number of ways to write $n$ as a product of $k$ ordered integers
\[
n = x_1\times x_2\times x_3\times \ldots\times x_k\qquad 1\le x_1\le x_2\le\ldots\le x_k
\]
Further define $D(N,K)$ to be the sum of $d(n,k)$ for $1\le n\le N$ and $1\le k\le K$.
You are given that $D(10, 10) = 153$ and $D(100, 100) = 35384$.
Find $D(10^{10},10^{10})$ giving your answer modulo $1\,000\,000\,007$.

This problem involves number theory and combinatorics to solve it.

The function $d(n, k)$ describes the number of partitions of $n$ into $k$ ordered integers. This problem can be solved by using the concept of integer partitions and prime factorization. We know that any integer can be factored uniquely into primes. Therefore, we can encode the problem of partitioning the number $n$ into $k$ ordered integers as partitioning the exponents of its prime factors.

The way to compute $d(n, k)$ is by creating a prime factorization for $n$, and by putting these prime factors into $k$ different bins. Because the prime factors are ordered (the first bin is not greater than the second bin, and so forth), this is equivalent to the problem of partitioning a number into ordered sums.

The function $D(N, K)$ is the summation of $d(n, k)$ for all $n$ from 1 to $N$ and all $k$ from 1 to $K$. This can be calculated using dynamic programming. The DP state can be $dp[i][j]$ which means the number of ways to partition the number $j$ by using the first $i$ primes.

For calculating $D(10^{10},10^{10})$ modulo $1\,000\,000\,007$, the numbers are way too large to run an algorithm or simple process to solve it. This would require an advanced solution that would involve creating a mathematical model or formula or using advanced numerical methods to handle such large numbers.

Because this is a very hard and advanced problem, the exact details of the solution would require a lot of advanced maths that goes beyond school level maths and into tertiary level mathematics.

More Answers:
Divisors of $2n^2$
Paths to Equality
Coin Loops

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