# Leibniz harmonic triangle

The Leibniz harmonic triangle is a triangular arrangement of unit fractions in which the outermost diagonals consist of the reciprocals of the row numbers and each inner cell is the cell diagonally above and to the left minus the cell to the left. To put it algebraically, L(r, 1) = 1/r, where r is the number of the row, starting from 1, and c is the number, never more than r and L(r, c) = L(r – 1, c – 1) – L(r, c – 1)

Relation with pascal’s triangle
Whereas each entry in Pascal’s triangle is the sum of the two entries in the above row, each entry in the Leibniz triangle is the sum of the two entries in the row below it. For example, in the 5th row, the entry (1/30) is the sum of the two (1/60)s in the 6th row.
Just as Pascal’s triangle can be computed by using binomial coefficients, so can Leibniz’s:

Properties
If one takes the denominators of the nth row and adds them, then the result will equal n.2n-1. For example, for the 3rd row, we have 3 + 6 + 3 = 12 = 3 × 22.

Given a positive integer n. The task is to print Leibniz harmonic triangle of height n.

Examples:

Input : n = 4
Output :
1
1/2 1/2
1/3 1/6 1/3
1/4 1/12 1/12 1/4

Input : n = 3
Output :
1
1/2 1/2
1/3 1/6 1/3


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Below is the implementation of printing Leibniz harmonic triangle of height n based on above relation with Pascal triangle.

 // CPP Program to print Leibniz Harmonic Triangle  #include  using namespace std;     // Print Leibniz Harmonic Triangle  void LeibnizHarmonicTriangle(int n)  {      int C[n + 1][n + 1];         // Calculate value of Binomial Coefficient in      // bottom up manner      for (int i = 0; i <= n; i++) {          for (int j = 0; j <= min(i, n); j++) {                 // Base Cases              if (j == 0 || j == i)                  C[i][j] = 1;                 // Calculate value using previously               // stored values              else                 C[i][j] = C[i - 1][j - 1] + C[i - 1][j];          }      }         // printing Leibniz Harmonic Triangle      for (int i = 1; i <= n; i++) {          for (int j = 1; j <= i; j++)              cout << "1/" << i * C[i - 1][j - 1] << " ";             cout << endl;      }  }     // Driven Program  int main()  {      int n = 4;      LeibnizHarmonicTriangle(n);      return 0;  }

 // Java Program to print   // Leibniz Harmonic Triangle  import java.io.*;  import java.math.*;     class GFG {             // Print Leibniz Harmonic Triangle      static void LeibnizHarmonicTriangle(int n)      {          int C[][] = new int[n + 1][n + 1];                 // Calculate value of Binomial           // Coefficient in bottom up manner          for (int i = 0; i <= n; i++) {              for (int j = 0; j <= Math.min(i, n);                                             j++) {                         // Base Cases                  if (j == 0 || j == i)                      C[i][j] = 1;                         // Calculate value using                  // previously stored values                  else                     C[i][j] = C[i - 1][j - 1] +                                 C[i - 1][j];              }          }                 // printing Leibniz Harmonic Triangle          for (int i = 1; i <= n; i++) {              for (int j = 1; j <= i; j++)                  System.out.print("1/" + i * C[i - 1][j - 1]                                             + " ");                     System.out.println();          }      }             // Driven Program      public static void main(String args[])      {          int n = 4;          LeibnizHarmonicTriangle(n);      }  }     // This code is contributed by Nikita Tiwari

 # Python3 Program to print   # Leibniz Harmonic Triangle     # Print Leibniz Harmonic   # Triangle  def LeibnizHarmonicTriangle(n):      C = [[0 for x in range(n + 1)]               for y in range(n + 1)];                     # Calculate value of Binomial       # Coefficient in bottom up manner      for i in range(0, n + 1):          for j in range(0, min(i, n) + 1):                             # Base Cases              if (j == 0 or j == i):                  C[i][j] = 1;                                 # Calculate value using               # previously stored values              else:                  C[i][j] = (C[i - 1][j - 1] +                             C[i - 1][j]);                                   # printing Leibniz       # Harmonic Triangle      for i in range(1, n + 1):          for j in range(1, i + 1):              print("1/", end = "");              print(i * C[i - 1][j - 1],                              end = " ");          print();     # Driver Code  LeibnizHarmonicTriangle(4);     # This code is contributed  # by mits.

 // C# Program to print Leibniz Harmonic Triangle  using System;     class GFG {             // Print Leibniz Harmonic Triangle      static void LeibnizHarmonicTriangle(int n)      {          int [,]C = new int[n + 1,n + 1];                 // Calculate value of Binomial           // Coefficient in bottom up manner          for (int i = 0; i <= n; i++) {              for (int j = 0; j <= Math.Min(i, n);                                           j++) {                         // Base Cases                  if (j == 0 || j == i)                      C[i,j] = 1;                         // Calculate value using                  // previously stored values                  else                     C[i,j] = C[i - 1,j - 1] +                               C[i - 1,j];              }          }                 // printing Leibniz Harmonic Triangle          for (int i = 1; i <= n; i++) {              for (int j = 1; j <= i; j++)                  Console.Write("1/" + i * C[i - 1,j - 1]                                           + " ");                 Console.WriteLine();          }      }             // Driven Program      public static void Main()      {          int n = 4;                     LeibnizHarmonicTriangle(n);      }  }     // This code is contributed by vt_m.

 

Output:
1/1
1/2 1/2
1/3 1/6 1/3
1/4 1/12 1/12 1/4


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