How to generate series 4,32,128,256,...n in java

Answer 1

Martin Taylor maybe after 256 comes 128 because
256*(1/2)=128
then
128*(1/4)=32
32 * 1/8 = 4
4 * 1/16 =1/4
1/4*1/16=1/64
so on...

Answer 2

If the multiplicator is divided by 2 in each step and is stored as float, the series is 4, 32, 128, 256, 256, 128, 32, 4, 1/4, 1/128, 1/8192, ...
It quickly approaches 0 but never reaches it, i.e. limes is 0.

Recursive definition is a bit tricky, because it needs two variables:
x_i = x_i-1 * y_i-1
y_i = y_i-1 / 2
Alternative:
x_i = x_i-1 * 2 ^ z_i-1
z_i = z_i-1 - 1
Base case:
x_0 = 4, y_0 = 8, z_0 = 3

Answer 3

Martin Taylor I think there is mupltiplier is decreasing by half.
4 * 8= 32
32 *(8/2) = 128
128*((8/2)/2) = 256
so on, so forth))

Answer 4

Martin,
Same doubt here, I was waiting for the OP's response ...

Answer 5

Ashish Katta

Is this a right solution?

https://code.sololearn.com/cMLKjgqxtlg4/?ref=app

Answer 6

Thank you

Answer 7

But not working for these series

Answer 8

All the numbers are powers of 2
Is there an input taken for that series?
What numbers go beyond 256?

Answer 9

import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int num = 4,j=8;
System.out.print(num + " ");
for (int i = 1; i < n; i++) {
num=num*j;
System.out.print(num + " ");
j=j/2;
}
}
}

Answer 10

import java.util.*;
class series
{
public void main()
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter the number of terms");
int n=sc.nextInt();
for(int i=0;i<n;i++)
{
int j=i+1;
System.out.print(i*j*j+",");
}
}
}

Answer 11

Somthing like this bro
int a= 4;
for(int i=4;i<=n; i=i/2)
{
System.out.print(a + "," );
a=a*i;
}

🙃
Wish if that helps u bro