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