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Chapter 8: Problem 26

For each integer \(n>2\), determine a self-conjugate partition of \(n\) that has at least two parts.

Short Answer

Expert verified
If \(n\) is even, the self-conjugate partition that has at least two parts is \(n/2 + n/2\). If \(n\) is odd, the self-conjugate partition that has at least two parts is \((n-1)/2 + (n-1)/2 + 1\).

Step by step solution

01

Identify the simplest self-conjugate partition

The simplest non-trivial self-conjugate partition is the partition 2+2 from number 4. The Ferrers diagram for 2+2, both row-wise and column-wise, has exactly two dots.
02

Build self-conjugate partitions of larger numbers

To create a self-conjugate partition of a larger number, add an equal number of rows and columns to this base diagram. For instance, for \(n=6\), one can use the partition 3+3. The Ferrers diagram for 3+3, both row-wise and column-wise, has exactly three dots.
03

Generalize for any n

If \(n\) is even, the self-conjugate solution is \(n/2 + n/2\). If \(n\) is odd, subtract 1 before dividing by 2, add these two parts and then add 1 as the third part, so the solution is \((n-1)/2 + (n-1)/2 + 1\). So for every integer \(n>2\), it's possible to find at least one self-conjugate partition that has at least two parts.

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Most popular questions from this chapter

Use the generating function for the small Schr?der numbers to compute the first few small Schr枚der numbers.

The large Schr枚der number \(C_{n}\) counts the number of subdiagonal HVD-lattice paths from \((0,0)\) to \((n, n) .\) The small Schr枚der number counts the number of dissections of a convex polygonal region of \(n+1 .\) Since \(R_{n}=2 s_{n+1}\) for \(n \geq 1\), there are as many subdiagonal HVD-lattice paths from \((0,0)\) to \((n, n)\) as there are dissections of a convex polygonal region of \(n+1\) sides. Find a one-to-one correspondence between these lattice paths and these dissections.

Prove that the number of partitions of the positive integer \(n\) into parts each of which is at most 2 equals \(\lfloor n / 2\rfloor+1\). (Remark: There is a formula, namely the nearest integer to \(\frac{(n+3)^{2}}{12}\), for the number of partitions of \(n\) into parts each of which is at most 3 but it is much more difficult to prove. There is also one for partitions with no part more than 4, but it is even more complicated and difficult to prove.)

Prove that the following formula holds for the \(k\) th-order differences of a sequence \(h_{0}, h_{1}, \ldots, h_{n}, \ldots:\) $$\Delta^{k} h_{n}=\sum_{j=0}^{k}(-1)^{k-j}\left(\begin{array}{l}k \\\j\end{array}\right) h_{n+j}$$

Verify that \([n]_{n}=n !\), and write \(n !\) as a polynomial in \(n\) using the Stirling numbers of the first kind. Do this explicitly for \(n=6\).
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