Summing up Problem (Combinations)

Question

This is a part of a bigger problem I was solving.

Problem: $N$ is a positive integer. There are $k$ number of other positive integers ($\le N$) In how many ways can you make $N$ by summing up any number of those $k$ integers. You can use any integer, any number of times.

For example: $N = 10$, $k=1: \{ 1 \}$

then there's only $1$ way of making $10$ using integers in braces: $1+1+1+1+\cdots+1 = 10$

another example: $N = 10$, $k = 2: \{ 1, 3\}$

number of ways $= 4$:

$1,1,1,1,1,1,1,1,1,1$
$1,1,1,1,1,1,1,3$
$1,1,1,1,3,3$
$1,3,3,3$

The question is to derive a generalized logic/formula to calculate the number of ways.

Answer 1

There is a simple recursive formula for that problem.

$F(0, k) = 1$ $F(N, \emptyset) = 0$

$F(N, k) = F(N - \min(k), k) + F(N, k\backslash \{\min(k)\})$

Answer 2

You’re asking for the number $p_A(n)$ of partitions of the integer $n$ into parts that belong to a specified set $A=\{a_1,\dots,a_k\}$ of $k$ positive integers. The generating function for the sequence $\langle p_A(n):n\in\mathbb{N}\rangle$ is $$\prod_{i=1}^k\frac1{(1-x^{a_i})} = \prod_{i=1}^k(1+x^{a_i}+x^{2a_i}+x^{3a_i}+\dots)\;.\tag{1}$$ In other words, $p_A(n)$ is the coefficient of $x^n$ in the product $(1)$. For actual computation, however, a recursive approach is more efficient.

Answer 3

This sounds awfully like the Frobenius problem. There has been quite a number of threads on this, e.g. this and this. There are a number of algorithms for solving the coin problem; search around for details.

Answer 4

Using the recursion method, the problem will be solved very easily.

$F(0) = 1$; $F(n<0) = 0$;

$F(N) = F(N - I_1) + F(N - I_2) + \cdots + F(N - I_k)$

or,

$F(N) = ∑ F(N - I_i)$

But for larger values of N $( \approx 10^{18})$, the above method won't work.

Matrix Exponentiation will have to be used to solve the problem.