Prolog中series/2谓词实现求助：求解指定数列第N项

Hey there! Let's work through this Prolog exercise together—no worries, it's easier once we spot the pattern in the sequence.

First, let's break down the pattern in the given sequence: 3, 7, 15, 31...

• For N=1: 3 = 2^(1+1) - 1 = 4 - 1
• For N=2: 7 = 2^(2+1) - 1 = 8 - 1
• For N=3: 15 = 2^(3+1) - 1 = 16 - 1
• For N=4: 31 = 2^(4+1) - 1 = 32 - 1

The core rule here is Ans = 2^(N+1) - 1. We can implement this in Prolog in two common ways—let's cover both:

Option 1: Direct Mathematical Calculation #

Using Prolog's built-in arithmetic predicates, we can compute the value directly with the formula. We'll also add a check to ensure N is a positive integer (since we're referring to the Nth term of a sequence):

series(N, Ans) :-
    integer(N),
    N > 0,
    Ans is 2^(N+1) - 1.

Testing this implementation: #

• series(1, X). returns X = 3
• series(3, X). returns X = 15
• series(4, X). returns X = 31

Option 2: Recursive Implementation #

If you want to focus on how the sequence builds incrementally (a common approach in Prolog), we can use recursion with a base case and recursive step. Notice each term is the previous term multiplied by 2 plus 1:

• 3 * 2 + 1 = 7 (2nd term)
• 7 * 2 + 1 = 15 (3rd term)
• 15 * 2 + 1 = 31 (4th term)

Here's the recursive code:

% Base case: The 1st term is 3
series(1, 3).
% Recursive step: Calculate current term from the previous one
series(N, Ans) :-
    N > 1,
    PrevN is N - 1,
    series(PrevN, PrevAns),
    Ans is PrevAns * 2 + 1.

This will produce the exact same results as the formula-based version, and it's great for understanding the sequence's iterative logic.

Quick Notes #

• Both implementations validate that N is a positive integer to avoid invalid inputs (like 0 or negative numbers).
• The formula-based version is more efficient for large values of N, while the recursive version is helpful for learning how sequences can be defined step-by-step in Prolog.

内容的提问来源于stack exchange，提问作者Walker