# Grouping natural numbers into arithmetic progression

I need to find the number of ways of dividing the first 12 natural numbers into 3 equal groups (4 numbers each), so that the numbers in any particular group can be arranged in AP (Arithmetic progression).

This is what I did:

Consider the group which contains the number $1$

For this group the possibilities are:

$1- 2 -3- 4$ or,
$1- 3 -5- 7$ or,
$1- 4 -7- 10$

Clearly 12 is not included in any one of these groups, so $12$ must be part of another group.
For this group too we get three possibilities (counting backwards from $12$).
After that, I just checked every possible combination of the two groups to see if the third was in AP. (it was kind of easy, just 9 cases)

I got my final answer as 4. I’d like to know if there is more simple and elegant method to solving this, and if possible generalizing for the first n natural numbers.

#### Solutions Collecting From Web of "Grouping natural numbers into arithmetic progression"

This is an interesting question. I don’t have a real answer but I wrote a program to calculate values for different $n$ and $m$, counting the number of ways to split the first $n$ numbers into groups of arithmetic sequences of size $m$ . Here are some results with the pairs are written as $(n,m)$:

$(9,3)=5$,

$(12,3)=15$, $(12,4)=4$,

$(15,3)=55$, $(15,5)=4$,

$(16,4)=11$,

$(18,3)=232$, $(18,6)=4$, $(18,6)=4$,

$(20,4)=23$, $(20,5)=10$,

$(24,3)=6643$, $(24,4)=68$, $(24,6)=10$, $(24,8)=4$.

I skipped writing pairs that satisfy these equations:

If $m$ is not a divisor of $n$ then $(n,m)=0$. If $n$ is even then

$$(n,\frac{n}{2})=2,$$

$$(n,2)=\frac{n!}{2^{n/2}\frac{n}{2}!}.$$

EDIT

This wouldn’t fit in a comment, but here are the ways of grouping (12,3):

$$\begin{array}{ccc} \left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ 10 & 11 & 12 \end{array}\right] & \left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 9 & 11 \\ 8 & 10 & 12 \end{array}\right] & \left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 6 & 8 \\ 5 & 7 & 9 \\ 10 & 11 & 12 \end{array}\right]\\ \left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 7 & 10 \\ 5 & 8 & 11 \\ 6 & 9 & 12 \end{array}\right] & \left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 8 & 12 \\ 5 & 6 & 7 \\ 9 & 10 & 11 \end{array}\right] & \left[\begin{array}{ccc} 1 & 3 & 5 \\ 2 & 4 & 6 \\ 7 & 8 & 9 \\ 10 & 11 & 12 \end{array}\right]\\ \left[\begin{array}{ccc} 1 & 3 & 5 \\ 2 & 4 & 6 \\ 7 & 9 & 11 \\ 8 & 10 & 12 \end{array}\right] & \left[\begin{array}{ccc} 1 & 3 & 5 \\ 2 & 6 & 10 \\ 4 & 8 & 12 \\ 7 & 9 & 11 \end{array}\right] & \left[\begin{array}{ccc} 1 & 3 & 5 \\ 2 & 7 & 12 \\ 4 & 6 & 8 \\ 9 & 10 & 11 \end{array}\right]\\ \left[\begin{array}{ccc} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \\ 10 & 11 & 12 \end{array}\right] & \left[\begin{array}{ccc} 1 & 5 & 9 \\ 2 & 3 & 4 \\ 6 & 7 & 8 \\ 10 & 11 & 12 \end{array}\right] & \left[\begin{array}{ccc} 1 & 5 & 9 \\ 2 & 4 & 6 \\ 3 & 7 & 11 \\ 8 & 10 & 12 \end{array}\right]\\ \left[\begin{array}{ccc} 1 & 5 & 9 \\ 2 & 6 & 10 \\ 3 & 7 & 11 \\ 4 & 8 & 12 \end{array}\right] & \left[\begin{array}{ccc} 1 & 6 & 11 \\ 2 & 3 & 4 \\ 5 & 7 & 9 \\ 8 & 10 & 12 \end{array}\right] & \left[\begin{array}{ccc} 1 & 6 & 11 \\ 2 & 7 & 12 \\ 3 & 4 & 5 \\ 8 & 9 & 10 \end{array}\right] \end{array}$$