Step by Step solution about How to resolve the algorithm Stern-Brocot sequence step by step in the Mathematica / Wolfram Language programming language

Wolfram Programming Language Code:

sb = {1, 1};
Do[
sb = sb~Join~{Total@sb[[i - 1 ;; i]], sb[[i]]}
,
{i, 2, 1000}
]
Take[sb, 15]
Flatten[FirstPosition[sb, #] & /@ Range[10]]
First@FirstPosition[sb, 100]
AllTrue[Partition[Take[sb, 1000], 2, 1], Apply[GCD] /* EqualTo[1]]

Explanation:

1. Fibonacci Sequence:

   • Initialize sb with the first two Fibonacci numbers, 1 and 1.
   • Use a Do loop to iteratively calculate the next Fibonacci numbers by summing up the previous two numbers and appending them to sb.

2. Printing First 15 Fibonacci Numbers:

   • Take[sb, 15] prints the first 15 Fibonacci numbers in the list sb.

3. Positions of First 10 Fibonacci Numbers:

   • Flatten[FirstPosition[sb, #] & /@ Range[10]] finds the positions of the first 10 Fibonacci numbers (from 1 to 10) in the list sb and flattens the result.

4. Position of Fibonacci Number 100:

   • First@FirstPosition[sb, 100] finds the first occurrence of the Fibonacci number 100 in the list sb.

5. GCD of Consecutive Fibonacci Numbers:

   • AllTrue[Partition[Take[sb, 1000], 2, 1], Apply[GCD] /* EqualTo[1]] checks if the greatest common divisor (GCD) of all pairs of consecutive Fibonacci numbers in the first 1000 terms is 1. If this condition is true for all pairs, it returns True, otherwise False.

sb = {1, 1};
Do[
 sb = sb~Join~{Total@sb[[i - 1 ;; i]], sb[[i]]}
 ,
 {i, 2, 1000}
 ]
Take[sb, 15]
Flatten[FirstPosition[sb, #] & /@ Range[10]]
First@FirstPosition[sb, 100]
AllTrue[Partition[Take[sb, 1000], 2, 1], Apply[GCD] /* EqualTo[1]]