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?

[hide="Solution"]

[b]Part 1[/b]

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

$\sum_{i=1}^{n-1}d(i)$

[b]Lemma: $d(i) = 2^{n-i}$[/b]

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.

[b]Part 2[/b]

Everything is the same as above, except the lemma slightly changes to $d(i) = 2^{n-i}-i.$ Evaluating the earlier sum gives us:

$\sum_{i=1}^{n-1}d(i) = \sum_{i=1}^{n-1}2^{n-i}-i = 2^{n}-\frac{n(n-1)}{2}-2$

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1. **Understanding the Problem:**
   We need to partition the set \( N = \{1, 2, \ldots, n\} \) into subsets \( A_1, A_2, \ldots, A_k \) such that each subset \( A_i \) forms an arithmetic progression with a common positive difference \( d_p \). We need to find the number of such partitions where each subset has at least one element and then where each subset has at least two elements.

2. **Part 1: At least one element in each subset:**
   - **Claim:** There are \( 2^n - 2 \) ways to partition \( N \) as described.
   - **Proof:**
     - Let \( d(k) \) be the number of ways \( N \) can be partitioned with a common difference \( k \).
     - We need to evaluate \( \sum_{i=1}^{n-1} d(i) \).

     **Lemma:** \( d(i) = 2^{n-i} \).

     - **Explanation:**
       - Consider the set \( N \) divided into rows where the first term of each row is a residue modulo \( i \). For example, for \( i = 2 \), the rows would be:
         \[
         \begin{aligned}
         &1, 3, 5, \ldots \\
         &2, 4, 6, \ldots
         \end{aligned}
         \]
       - Each row represents an arithmetic progression with a common difference \( i \).
       - We can partition each row into segments. For a row with \( k \) elements, we can place at most \( k-1 \) segments, resulting in \( 2^{k-1} \) ways to partition the row.
       - Applying this to all rows, we get \( 2^{n-i} \) ways to partition the set \( N \) for a given \( i \).

     - **Summing up:**
       \[
       \sum_{i=1}^{n-1} d(i) = \sum_{i=1}^{n-1} 2^{n-i} = 2^n - 2
       \]
       - This follows from the geometric series sum formula.

3. **Part 2: At least two elements in each subset:**
   - **Claim:** The number of ways is \( 2^n - \frac{n(n-1)}{2} - 2 \).
   - **Proof:**
     - The lemma changes to \( d(i) = 2^{n-i} - i \).
     - **Explanation:**
       - For each \( i \), we need to exclude the cases where subsets have only one element.
       - This results in \( 2^{n-i} - i \) ways for each \( i \).

     - **Summing up:**
       \[
       \sum_{i=1}^{n-1} d(i) = \sum_{i=1}^{n-1} (2^{n-i} - i) = 2^n - \sum_{i=1}^{n-1} i - 2 = 2^n - \frac{n(n-1)}{2} - 2
       \]
       - The sum of the first \( n-1 \) integers is \( \frac{(n-1)n}{2} \).

The final answer is \( \boxed{ 2^n - 2 } \) for at least one element and \( 2^n - \frac{n(n-1)}{2} - 2 \) for at least two elements.