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In how many ways can the set  $N ={1, 2, \cdots, n}$  be partitioned in the form  $p(N) = A_{1}\cup A_{2}\cup \cdots \cup A_{k},$  where  $A_{i}$  consists of elements forming arithmetic progressions, all with the same common positive difference  $d_{p}$  and of length at least one? At least two?

<details><summary>Solution</summary>**Part 1**

Claim: There are  $2^{n}-2$  ways of performing the desired partitioning.

Let  $d(k)$  equal the number of ways  $N$  can be partitioned as above with common difference  $k.$  We are thus trying to evaluate $\sum_{i=1}^{n-1}d(i)$ **Lemma:  $d(i) = 2^{n-i}$**

We may divide  $N$  into various rows where the first term of each row denotes a residue  $\bmod{i}.$  The only exception is the last row, as no row starts with  $0$ ; the last row will start with  $i.$  We display the rows as such: $1, 1+i, 1+2i, 1+3i, \cdots$  $2, 2+i, 2+2i, 2+3i, \cdots$  $\cdots$  $i, 2i, 3i, \cdots$ Since all numbers have one lowest remainder  $\bmod{i}$  and we have covered all possible remainders, all elements of  $N$  occur exactly once in these rows.

Now, we can take  $k$  line segments and partition a given row above; all entries within two segments would belong to the same set. For example, we can have: $1| 1+i, 1+2i, 1+3i | 1+4i | 1+5i, 1+6i, 1+7i, 1+8i,$ which would result in the various subsets:  ${1},{1+i, 1+2i, 1+3i},{1+4i},{1+5i, 1+6i, 1+7i, 1+8i}.$  For any given row with  $k$  elements, we can have at most  $k-1$  segments. Thus, we can arrange any number of segments where the number lies between  $0$  and  $k-1$ , inclusive, in: $\binom{k-1}{0}+\binom{k-1}{1}+\cdots+\binom{k-1}{k-1}= 2^{k-1}$ ways. Applying the same principle to the other rows and multiplying, we see that the result always gives us  $2^{n-i},$  as desired.

We now proceed to the original proof.

Since  $d(i) = 2^{n-i}$  by the above lemma, we have: $\sum_{i=1}^{n-1}d(i) = \sum_{i=1}^{n-1}2^{n-i}= 2^{n}-2$ Thus, there are  $2^{n}-2$  ways of partitioning the set as desired.**Part 2**

Everything is the same as above, except the lemma slightly changes to  $d(i) = 2^{n-i}-i.$  Evaluating the earlier sum gives us: $\sum_{i=1}^{n-1}d(i) = \sum_{i=1}^{n-1}2^{n-i}-i = 2^{n}-\frac{n(n-1)}{2}-2$</details>

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