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Chapter 8: Problem 6

a) Find a recurrence relation for the number of strictly increasing sequences of positive integers that have 1 as their first term and \(n\) as their last term, where \(n\) is a positive integer. That is, sequences \(a_{1}, a_{2}, \ldots, a_{k}\) where \(a_{1}=1, \quad a_{k}=n, \quad\) and \(\quad a_{j}

Short Answer

Expert verified
The recurrence relation is \(a(n) = \sum_{k=2}^{n-1} a(k)\) with initial conditions \(a(1) = 0\) and \(a(2) = 1\).

Step by step solution

01

- Understanding the Problem

The task is to find a recurrence relation for the number of strictly increasing sequences of positive integers starting with 1 and ending with n. Let's denote this number by \(a(n)\).
02

- Defining the Recurrence Relation

To form a sequence that starts with 1 and ends with \(n\), we must find the number of ways to choose intermediate numbers between 1 and \(n\). If \(n = 2\), the sequence is simply \(\{1, 2\}\). For larger \(n\), the sequence can be broken down into two parts: 1 to a smaller number \(k\) (where \(1 < k < n\)) and then from \(k\) to \(n\). Thus, we get the relation: \[ a(n) = \sum_{k=2}^{n-1} a(k) \]
03

- Establishing Initial Conditions

We need to determine the initial value for \(a(n)\). If \(n = 1\), there is no valid sequence, so \(a(1) = 0\). If \(n = 2\), the only sequence is \{1, 2\}, so \(a(2) = 1\).
04

- General Solution

Given the recurrence relation and the initial conditions, we can generate values of \(a(n)\) for \(n \geq 3\). Iterate the sum for each \(n\) using previous computed values until needed.
05

- Finding Number of Sequences for General \(n\)

To find the number of sequences for any \(n \geq 2\), apply the relation established and sum the possible sequences.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

strictly increasing sequences
Strictly increasing sequences are sequences where each term is greater than the previous one. For example, in the sequence \(\{a_1, a_2, a_3, ... , a_k \}\), the condition \(a_j < a_{j+1}\) must be satisfied for all \( j = 1, 2, ..., k-1\).
This means that every next element in the sequence must be larger than the previous one.

Let's break down why these sequences are significant:

  • They provide a clear structure or order, which is helpful in mathematical proofs and reasoning.
  • They reduce the complexity in arrangements, as each element has a defined range where it can fit.
These properties make them useful for establishing recurrence relations.

In our problem, we use strictly increasing sequences to count the number of ways to form sequences from 1 to \(n\), where 1 is the first term and \(n\) is the last. This specific structure helps simplify and clarify the number of possible sequences.
initial conditions
Initial conditions provide the starting point for solving recurrence relations. Without them, we cannot determine the exact values of the function we are trying to compute.

For the given problem, we establish initial conditions for small values of \(n\) to help compute larger values. Let's look at the initial conditions in our recurrence relation:

  • For \(n = 1\), there are no sequences since the sequence would just be \(\{1\}\), which doesn't satisfy our requirement of having 1 as the first term and \(n\) as the last term for \(n\ = 1\). Hence, \(a(1) = 0\).
  • For \(n = 2\), the sequence can only be \(\{1, 2\}\), providing exactly one valid sequence. Therefore, \(a(2) = 1\).
These conditions allow us to start computing values for larger \(n\) using our recurrence relation \(a(n) = \sum_{k=2}^{n-1} a(k)\). Initial conditions essentially

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Most popular questions from this chapter

a) Find the recurrence relation satisfied by \(S_{n},\) where \(S_{n}\) is the number of regions into which three-dimensional space is divided by \(n\) planes if every three of the planes meet in one point, but no four of the planes go through the same point, b) Find \(S_{n}\) using iteration.

Use generating functions to solve the recurrence rela- tion \(a_{k}=2 a_{k-1}+3 a_{k-2}+4^{k}+6\) with initial conditions \(a_{0}=20, a_{1}=60\)

Use generating functions (and a computer algebra pack- age, if available) to find the number of ways to make change for \(\$ 1\) using pennies, nickels, dimes, and quarters with a) no more than 10 pennies. b) no more than 10 pennies and no more than 10 nickels. c) no more than 10 coins.

Use generating functions to solve the recurrence relation \(a_{k}=5 a_{k-1}-6 a_{k-2}\) with initial conditions \(a_{0}=6\) and \(a_{1}=30\)

Exercises 33鈥�37 deal with a variation of the Josephus problem described by Graham, Knuth, and Patashnik in [GrKnPa94]. This problem is based on an account by the historian Flavius Josephus, who was part of a band of 41 Jewish rebels trapped in a cave by the Romans during the Jewish Roman war of the first century. The rebels preferred suicide to capture; they decided to form a circle and to repeatedly count off around the circle, killing every third rebel left alive. However, Josephus and another rebel did not want to be killed this way; they determined the positions where they should stand to be the last two rebels remaining alive. The variation we consider begins with n people, numbered 1 to n, standing around a circle. In each stage, every second person still left alive is eliminated until only one survives. We denote the number of the survivor by J(n). Show that \(J(n)\) satisfies the recurrence relation \(J(2 n)=\) \(2 J(n)-1\) and \(J(2 n+1)=2 J(n)+1,\) for \(n \geq 1,\) and \(J(1)=1\)
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