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Program #2 Define a sequence of numbers recursively Define a sequence as, at, az, as, where ao = 1 an = (an-1+ 1) 2 an = (2* an - 2 + aa -1) ifn is even So for instance: (n = 0) ao = 1 (n = 1) a; = (ao + 1) * 2 = (1+ 1)*2= 4 (n = 2) az = (2* ao + ao) = 2*1+4 = 6 (n = 3) as = (az + 1) *2= (6+ 1) * 2 =14 The resulting sequence is 1, 4, 6, 14, - %3D if n is odd %3D You will write a function that returis the nth term of the sequence. You must do this by using a recursive function. The function will NOT print out any values. Rather, the function will return the nth term of the sequence using a recursive algorithm. In the main part of the program: 1. Input the number of terms of the sequence to output. 2. In a for loop, call the function repeatedly to get the desired number of terms. The function will take i (assuming the for index is called i) as the argument and return the ith term of the sequence. 3. In the for loop, print out each term as it is returned. Figure out the necessary prompts for the inputs and the desired outputs by looking at this example session. The number in red is a possible input and is not what you print out. Enter the number of terma> 4 Term 10> 1 Term 1> 4 Term 12> 6 Term 13> 14 HINTS: 1. The base case is when n==0 2. In the recursive case you will need to decide if n is odd or even. n is odd if there is a remainder when you divide by two. if (n % 2) !=0): #test for odd Since a nonzero number is true, the above could be shortened to: if (n % 2): #test for odd Either way, else: #must be even

Program #2 Define a sequence of numbers recursively Define a sequence as, at, az, as, where ao = 1 an = (an-1+ 1) 2 an = (2* an - 2 + aa -1) ifn is even So for instance: (n = 0) ao = 1 (n = 1) a; = (ao + 1) * 2 = (1+ 1)*2= 4 (n = 2) az = (2* ao + ao) = 2*1+4 = 6 (n = 3) as = (az + 1) *2= (6+ 1) * 2 =14 The resulting sequence is 1, 4, 6, 14, - %3D if n is odd %3D You will write a function that returis the nth term of the sequence. You must do this by using a recursive function. The function will NOT print out any values. Rather, the function will return the nth term of the sequence using a recursive algorithm. In the main part of the program: 1. Input the number of terms of the sequence to output. 2. In a for loop, call the function repeatedly to get the desired number of terms. The function will take i (assuming the for index is called i) as the argument and return the ith term of the sequence. 3. In the for loop, print out each term as it is returned. Figure out the necessary prompts for the inputs and the desired outputs by looking at this example session. The number in red is a possible input and is not what you print out. Enter the number of terma> 4 Term 10> 1 Term 1> 4 Term 12> 6 Term 13> 14 HINTS: 1. The base case is when n==0 2. In the recursive case you will need to decide if n is odd or even. n is odd if there is a remainder when you divide by two. if (n % 2) !=0): #test for odd Since a nonzero number is true, the above could be shortened to: if (n % 2): #test for odd Either way, else: #must be even

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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Program #2 Define a sequence of numbers recursively Define a sequence ao, a1, az, a3, where ao = 1 an = (an -1+ 1) * 2 an = (2* an - 2+ an-1) ifn is even So for instance: (n = 0) ao = 1 (n = 1) ai = (ao + 1) *2 = (1+ 1) *2= 4 (n = 2) az = (2* ao + ao) = 2*1+4 = 6 (n = 3) az = (az + 1) *2 = (6 + 1) *2 =14 The resulting sequence is 1, 4, 6, 14, . if n is odd !3! %3D You will write a function that returs the nth term of the sequence. You must do this by using a recursive function. The function will NOT print out any values. Rather, the function will return the nth term of the sequence using a recursive algorithm. In the main part of the program: 1. Input the number of terms of the sequence to output. 2. In a for loop, call the function repeatedly to get the desired number of terms. The function will take i (assuming the for index is called i) as the argument and return the ith term of the sequence. 3. In the for loop, print out each term as it is returned. Figure out the necessary prompts for the inputs and the desired outputs by looking at this example session. The number in red is a possible input and is not what you print out. Enter the number of terms> 4 Term #0> 1 Term #1> 4 Term #2> 6 Term #3> 14 HINTS: 1. The base case is when n==0 2. In the recursive case you will need to decide if n is odd or even. n is odd if there is a remainder when you divide by two. if (n % 2) !=0): #test for odd Since a nonzero number is true, the above could be shortened to: if (n % 2): #test for odd Either way, else: #must be even
Transcribed Image Text:Program #2 Define a sequence of numbers recursively Define a sequence ao, a1, az, a3, where ao = 1 an = (an -1+ 1) * 2 an = (2* an - 2+ an-1) ifn is even So for instance: (n = 0) ao = 1 (n = 1) ai = (ao + 1) *2 = (1+ 1) *2= 4 (n = 2) az = (2* ao + ao) = 2*1+4 = 6 (n = 3) az = (az + 1) *2 = (6 + 1) *2 =14 The resulting sequence is 1, 4, 6, 14, . if n is odd !3! %3D You will write a function that returs the nth term of the sequence. You must do this by using a recursive function. The function will NOT print out any values. Rather, the function will return the nth term of the sequence using a recursive algorithm. In the main part of the program: 1. Input the number of terms of the sequence to output. 2. In a for loop, call the function repeatedly to get the desired number of terms. The function will take i (assuming the for index is called i) as the argument and return the ith term of the sequence. 3. In the for loop, print out each term as it is returned. Figure out the necessary prompts for the inputs and the desired outputs by looking at this example session. The number in red is a possible input and is not what you print out. Enter the number of terms> 4 Term #0> 1 Term #1> 4 Term #2> 6 Term #3> 14 HINTS: 1. The base case is when n==0 2. In the recursive case you will need to decide if n is odd or even. n is odd if there is a remainder when you divide by two. if (n % 2) !=0): #test for odd Since a nonzero number is true, the above could be shortened to: if (n % 2): #test for odd Either way, else: #must be even
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