Explore a Klarner-Rado sequence

Question

One of the Klarner-Rado sequences is defined as follows:

• the first term is \$1\$
• for all subsequent terms, the following rule applies: if \$x\$ is present, so are \$2x+1\$ and \$3x+1\$
• the sequence is strictly increasing

This is A002977.

The first few terms are:

1, 3, 4, 7, 9, 10, 13, 15, 19, 21, 22, 27, ...

(\$3\$ is present because it's \$2\times1+1\$, \$4\$ is present because it's \$3\times1+1\$, \$7\$ is present because it's \$2\times3+1\$, etc.)

Your task

Given \$n\$, you must return the \$n\$^th element of the sequence. You may use either 0-based or 1-based indexing. (Please specify your choice in your answer.)

0-indexed example:

input = 10
output = 22

Let's see who can get less bytes...

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Answer 1

05AB1E, 15 14 bytes

$FDx>DŠ+)˜ê}sè

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Answer 2

JavaScript (ES6),  53  50 bytes

1-indexed.

n=>(g=k=>n?g(g[k]|!k++?g[n--,k*2]=g[k*3]=k:k):k)``

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Commented

Note: On the first iteration, we have k = [''], which is zero-ish but truthy. By doing !k++, we force k to be coerced to \$0\$ right away (and not just before it's incremented), which makes the test work as expected.

n => (                    // n = requested index
  g = k =>                // h is a recursive function taking k, starting at 0
    n ?                   // if n is not equal to 0:
      g(                  //   do a recursive call:
        g[k] |            //     if g[k] is defined
        !k++ ?            //     or k = 0 (increment k after the test):
          g[n--, k * 2] = //       decrement n; set g[k * 2]
          g[k * 3] = k    //       and g[k * 3] and pass k
        :                 //     else:
          k               //       just pass k
      )                   //   end of recursive call
    :                     // else:
      k                   //   stop recursion and return k
)``                       // initial call to g with k = [''] (zero-ish)

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JavaScript (ES6), 65 bytes

I thought the '11'[k/x-2] trick was neat, but overall this initial approach is far too long.

0-indexed.

n=>(g=k=>a[n]||g(-~k,a.some(x=>'11'[k/x-2])&&a.push(k+1)))(a=[1])

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Commented

n => (                  // n = requested index
  g = k =>              // g is a recursive function taking k (starting at 1)
    a[n] ||             // if a[n] is defined, return it and stop
    g(                  // otherwise, do a recursive call:
      -~k,              //   with k + 1
      a.some(x =>       //   if there exists some x in a[]
        '11'[k / x - 2] //   such that k / x is either 2 or 3 ...
      )                 //
      && a.push(k + 1)  //   ... then push k + 1 in a[]
    )                   // end of recursive call
)(a = [1])              // initial call to g with k = a = [1]

Answer 3

R, 57 bytes

n=scan();for(i in 1:n)T=sort(unique(c(T,T%o%2:3+1)));T[n]

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1-based index.

Answer 4

APL+WIN, 54 bytes

Index origin 1

Prompts for the index of required value. Given the response that I received from Grimy and Shaggy that I can assume infinite memory and time then it is trivial to modify this function to work with a series up to the limit of your machine by increasing the value 20 in the function below. This version is limited to index position 1901981

m←1⋄(⍎∊20⍴⊂'m←(0≠1,-2-/m)/m←m[⍋m←m,∊1+mר⊂2 3]⋄')⋄m[⎕]

For some reason this function will not run on Dyalog Classic in TIO which is what I usually use.

On my machine 1 => 1, 11 => 22 and 61 => 237 (checked on http://oeis.org)

Answer 5

JavaScript (ES6), 99 bytes

x=>eval("a=[i=1],a.b=a.includes;while(i++,c=i-1,x)a.b(i)||!a.b(c/2)&&!a.b(c/3)||(a.push(i),x--);c")

Answer 6

Python 3, 73 bytes

f=lambda n,i=0,k=[1]:(i==n)*k[i]or f(n,i+1,sorted(k+[2*k[i]+1,3*k[i]+1]))

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