Consider a series of numbers composed of only digits 4 and 7. First few numbers in the series are 4, 7, 44, 47, 74, 44744,.. etc. Given a number n, we need to find n-th number in the series.

Examples:

Input : n = 2 Output : 7 Input : n = 3 Output : 44 Input : n = 5 Output : 74 Input : n = 6 Output : 77

The idea is based on the fact that the value of last digit alternates in series. For example, if last digit of i-th number is 4, then last digit of (i-1)-th and (i+1)-th numbers must be 7.

We create an array of size (n+1) and push 4 and 7 (These two are always first two elements of series) to it. For more elements we check

1) If i is odd,

arr[i] = arr[i/2]*10 + 4;

2) If it is even,

arr[i] = arr[(i/2)-1]*10 + 7;

At last return arr[n].

## C++

`// C++ program to find n-th number in a series` `// made of digits 4 and 7` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Return n-th number in series made of 4 and 7` `int` `printNthElement(` `int` `n)` `{` ` ` `// create an array of size (n+1)` ` ` `int` `arr[n+1];` ` ` `arr[1] = 4;` ` ` `arr[2] = 7;` ` ` `for` `(` `int` `i=3; i<=n; i++)` ` ` `{` ` ` `// If i is odd` ` ` `if` `(i%2 != 0)` ` ` `arr[i] = arr[i/2]*10 + 4;` ` ` `else` ` ` `arr[i] = arr[(i/2)-1]*10 + 7;` ` ` `}` ` ` `return` `arr[n];` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 6;` ` ` `cout << printNthElement(n);` ` ` `return` `0;` `}` |

## Java

`// Java program to find n-th number in a series` `// made of digits 4 and 7` `class` `FindNth` `{` ` ` `// Return n-th number in series made of 4 and 7` ` ` `static` `int` `printNthElement(` `int` `n)` ` ` `{` ` ` `// create an array of size (n+1)` ` ` `int` `arr[] = ` `new` `int` `[n+` `1` `];` ` ` `arr[` `1` `] = ` `4` `;` ` ` `arr[` `2` `] = ` `7` `;` ` ` ` ` `for` `(` `int` `i=` `3` `; i<=n; i++)` ` ` `{` ` ` `// If i is odd` ` ` `if` `(i%` `2` `!= ` `0` `)` ` ` `arr[i] = arr[i/` `2` `]*` `10` `+ ` `4` `;` ` ` `else` ` ` `arr[i] = arr[(i/` `2` `)-` `1` `]*` `10` `+ ` `7` `;` ` ` `}` ` ` `return` `arr[n];` ` ` `} ` ` ` ` ` `// main function` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `n = ` `6` `;` ` ` `System.out.println(printNthElement(n));` ` ` `}` `}` |

## Python3

`# Python3 program to find n-th number` `# in a series made of digits 4 and 7` `# Return n-th number in series made` `# of 4 and 7` `def` `printNthElement(n) :` ` ` ` ` `# create an array of size (n + 1)` ` ` `arr ` `=` `[` `0` `] ` `*` `(n ` `+` `1` `);` ` ` `arr[` `1` `] ` `=` `4` ` ` `arr[` `2` `] ` `=` `7` ` ` `for` `i ` `in` `range` `(` `3` `, n ` `+` `1` `) :` ` ` `# If i is odd` ` ` `if` `(i ` `%` `2` `!` `=` `0` `) :` ` ` `arr[i] ` `=` `arr[i ` `/` `/` `2` `] ` `*` `10` `+` `4` ` ` `else` `:` ` ` `arr[i] ` `=` `arr[(i ` `/` `/` `2` `) ` `-` `1` `] ` `*` `10` `+` `7` ` ` ` ` `return` `arr[n]` ` ` `# Driver code` `n ` `=` `6` `print` `(printNthElement(n))` `# This code is contributed by Nikita Tiwari.` |

## C#

`// C# program to find n-th number in a series` `// made of digits 4 and 7` `using` `System;` `class` `GFG` `{` ` ` `// Return n-th number in series made of 4 and 7` ` ` `static` `int` `printNthElement(` `int` `n)` ` ` `{` ` ` `// create an array of size (n+1)` ` ` `int` `[]arr = ` `new` `int` `[n+1];` ` ` `arr[1] = 4;` ` ` `arr[2] = 7;` ` ` ` ` `for` `(` `int` `i = 3; i <= n; i++)` ` ` `{` ` ` `// If i is odd` ` ` `if` `(i % 2 != 0)` ` ` `arr[i] = arr[i / 2] * 10 + 4;` ` ` `else` ` ` `arr[i] = arr[(i / 2) - 1] * 10 + 7;` ` ` `}` ` ` `return` `arr[n];` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main ()` ` ` `{` ` ` `int` `n = 6;` ` ` `Console.Write(printNthElement(n));` ` ` `}` `}` `// This code is contributed by vt_m.` |

## PHP

`<?php` `// PHP program to find n-th` `// number in a series` `// made of digits 4 and 7` `// Return n-th number in` `// series made of 4 and 7` `function` `printNthElement(` `$n` `)` `{` ` ` ` ` `// create an array` ` ` `// of size (n+1)` ` ` `$arr` `[1] = 4;` ` ` `$arr` `[2] = 7;` ` ` `for` `(` `$i` `= 3; ` `$i` `<= ` `$n` `; ` `$i` `++)` ` ` `{` ` ` ` ` `// If i is odd` ` ` `if` `(` `$i` `% 2 != 0)` ` ` `$arr` `[` `$i` `] = ` `$arr` `[` `$i` `/ 2] *` ` ` `10 + 4;` ` ` `else` ` ` `$arr` `[` `$i` `] = ` `$arr` `[(` `$i` `/ 2) - 1] *` ` ` `10 + 7;` ` ` `}` ` ` `return` `$arr` `[` `$n` `];` `}` `// Driver code` `$n` `= 6;` `echo` `(printNthElement(` `$n` `));` `// This code is contributed by Ajit.` `?>` |

## Javascript

`<script>` `// javascript program to find n-th number in a series` `// made of digits 4 and 7` ` ` `// Return n-th number in series made of 4 and 7` ` ` `function` `printNthElement(n) {` ` ` `// create an array of size (n+1)` ` ` `var` `arr = Array(n + 1).fill(0);` ` ` `arr[1] = 4;` ` ` `arr[2] = 7;` ` ` `for` `(` `var` `i = 3; i <= n; i++) {` ` ` `// If i is odd` ` ` `if` `(i % 2 != 0)` ` ` `arr[i] = arr[i / 2] * 10 + 4;` ` ` `else` ` ` `arr[i] = arr[(i / 2) - 1] * 10 + 7;` ` ` `}` ` ` `return` `arr[n];` ` ` `}` ` ` `// main function` ` ` ` ` `var` `n = 6;` ` ` `document.write(printNthElement(n));` `// This code is contributed by Princi Singh` `</script>` |

Output:

77

Find n-th element in a series with only 2 digits (4 and 7) allowed | Set 2 (log(n) method)

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