**The only purpose of python coding below is to prevent kids from straight forward calculations, which actually are boring rather then difficult. In doing so I intend to teach my students pretty old principle "Just think first"**

**"Problem" itself :- **

**Algorithm for calculating the value of the function F (n), where n is a natural number,**

**is given by the following relations:**

**F (n) = 1 for n = 1;**

**F (n) = n + F (n - 1), if n is even,**

**F (n) = 2 × F (n - 2) if n> 1 and n is odd.**

**What is the value of the function F (46)?**

**Just notice that number 46 could be easily replaced by 56 or 76**

**Another samples :-**

**The algorithm for calculating the function F (n) is given by the following relations:**

**F (n) = n for n≤3;**

**F (n) = n ∗ n ∗ n + F (n – 1) if n > 3 and gives remainder 0 when divided by 3**

**F (n) = 4 + F (n // 3) if n > 3 and gives remainder 1 when divided by 3**

**F (n) = n ∗ n + F (n – 2) if n > 3 and gives remainder 2 when divided by 3**

**Here // stands for integer division.**

**Output an answer as binary value of F(100).**

**Algorithm for calculating functions F (n), where n is a natural number, given;by the following ratios:**

**F (n) = n - 1 for n < 4 ,**

**F (n) = n + 2 * F (n - 1) when n > 3 and a multiple of 3 ,**

**F (n) = F (n - 2) + F (n - 3) when n > 3 and not a multiple of 3 .**

**What is the sum of the digits of the value of the function F (65)?**

**The algorithm for calculating the function F (n) is given by the following relations:**

**F (n) = n for n≤3;**

**F (n) = n ∗ n ∗ n + F (n – 1) if n > 3 and gives remainder 0 when divided by 3**

**F (n) = 4 + F (n // 3) if n > 3 and gives remainder 1 when divided by 3**

**F (n) = n ∗ n + F (n – 2) if n > 3 and gives remainder 2 when divided by 3**

**Here // stands for integer division.**

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