JavaScript Recursive Sequence

Question

I'm trying to teach myself recursion and I seem to be at a total loss on how to code this particular recursive sequence. I always end up going in some bruteforcing (even hardcoding) ways that lead to nowhere.

I have start which is my starting input (the first member will be equal to start). start = 2 for example.

The sequence is:

member1 = start;
member2 = member1 + 1;
member3 = 2 * member1 + 1;
member4 = member1 + 2;
member5 = member2 + 1;
member6 = 2 * member2 + 1;
member7 = member2 + 2;

I have to return the N-th member of this sequence. N can be any number.. for example N = 11. So then I need to return 11th member of this sequence. How do I implement this recursively?

I really appreciate your help in advance!

can you add the sequence of numbers with a start value of one? — Nina Scholz
– Nina Scholz, Commented Mar 19, 2021 at 9:29

Nina Scholz · Accepted Answer · 2021-03-19 10:39:50Z

You could take a zero based approach by returning the start value at position zero, like for sequences

a₀ = start

For all other values divide the n^th by three and return the result of wanted operation.

n    operation                     an  reference
0    member1 = start;               1

1    member2 = member1 + 1;         2      a0
2    member3 = 2 * member1 + 1;     3      a0
3    member4 = member1 + 2;         3      a0

4    member5 = member2 + 1;         3      a1
5    member6 = 2 * member2 + 1;     5      a1
6    member7 = member2 + 2;         4      a1

function getSequence(start, n) {
    if (!n--) return start;
    const value = getSequence(start, Math.floor(n / 3));
    switch (n % 3) {
        case 0: return value + 1;
        case 1: return value * 2 + 1;
        case 2: return value + 2;
    }
}

console.log(...Array.from({ length: 10 }, (_, i) => getSequence(1, i)));
console.log(...Array.from({ length: 10 }, (_, i) => getSequence(2, i)));

One approach

function getSequence(start, n) {
    if (!--n) return start;
    const value = getSequence(start, 1 + Math.floor(--n / 3));
    switch (n % 3) {
        case 0: return value + 1;
        case 1: return value * 2 + 1;
        case 2: return value + 2;
    }
}

console.log(...Array.from({ length: 10 }, (_, i) => getSequence(1, i + 1)));
console.log(...Array.from({ length: 10 }, (_, i) => getSequence(2, i + 1)));

Julien Ripet · Accepted Answer · 2021-03-19 09:42:05Z

0

As stated by @mplungjan, your question isn't very clear.
From what i'm understanding, you'd want a function that takes in any number (start in your original example) and a position (end) and gives you back its result following your pattern, which, again from what I understood, is :
x + 1
x * 2 + 1
x + 2

If all my assumptions are correct, a function like follows should work

function getNthElement(start, end) {
  let result = start
  for ( i = 0; i < end; i++) {
    if (i % 3 === 0) { // first step
      result = result + 1
    }
    if (i % 3 === 1) { // second step
      result = result * 2 + 1
    }
    if (i % 3 === 2) { //third step
      result = result + 2
    }
  }
  return result
}


console.log(getNthElement(1, 10))

answered Mar 19, 2021 at 9:42

Julien Ripet

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1 Comment

Peter B Over a year ago

If have not checked if the outcome is correct, but the asker asked specifically for help with recursion, which this solution is not using.

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