Question overhauled: I am attempting to add all the numbers in the 3n 1 sequence.

My expected outputs are n=20 which will return 66.n=4 returns 7 and n=31 returns 101104

Obviously the code I have written below will do some of the basic numbers but after that it just goes all over the place.

int sumSeq3nPlus1(int n) {
    if (n <= 0) {
       return 0;
    }
 
    int fib[n 1];
    fib[0] = 0; 
    fib[1] = 1;
 
    int sum = fib[0]   fib[1];

    for (int i = 2; i <= n;   i) {
        fib[i] = fib[i-1]   fib[i-2];
        sum  = fib[i];
    }
    return sum;
}

Adding the image of what was asked of us: https://imgur.com/a/o0ISNHB

CodePudding user response：

I figured out what the actual question is...

Given the 3n 1 sequence, as used in the Collatz conjecture:

next = 3*current   1 if current is odd  
next = current/2 if current is even  

Return the sum of the sequence starting at n and ending at 1.

This has nothing to do with the Fibonacci sequence.

What you need to do is compute the running sum of the sequence: start with n, then add the next term, and so on until the term you get is 1, at which point add the 1 and return the sum.

Examples:
for n=4, the sum is 4 2 1 = 7
for n=20, the sum is 20 10 5 16 8 4 2 1 = 66
for n=31, the sum is 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 = 101104

CodePudding user response：

The loop I have written add the begging number n = x and then adds it along into its sequence.

int sumSeq3nPlus1(int value) {
    
    int n = 0;
    
    while(value > 1) {
        n  = value;
        if (value % 2 != 0) {
            value = 3 * value   1;
        }
        else {
            value = value / 2;
        }
    }
    return n   1;
}