Yet Unused Pairs

Question

Let's define a sequence of positive integers. We will define the sequence on even numbers to be double the previous term. The odd indices of the sequence will be smallest positive integer not yet appearing in the sequence.

Here are the first couple terms.

1,2,3,6,4,8,5,10,7,14,9,18,11,22,12,24,13,26,15,30

You can also think of this as the list of concatenated pairs (n,2n) where n is the least unused positive integer so far.

Task

Given a number n as input calculate the nth term in this sequence.

This is code-golf so you should aim to minimize the size of your source code as measured in bytes.

OEIS A036552

Answer 1

Haskell, 40 bytes

l(a:r)=a:2*a:l[x|x<-r,x/=2*a]
(l[1..]!!)

Zero-based. l incrementally builds the sequence from a lazy list of remaining integers.

Answer 2

JavaScript (ES6), 92 82 69 67 65 bytes

n=>(F=i=>i^n?F(a[b=i&1?2*b:(g=k=>a[k]?g(k+1):k)(1)]=-~i):b)(a={})

How?

We keep track of:

• The last inserted value b.
• All previously encountered values in the lookup table a.

Internally, we're using a 0-based index i. Odd and even behaviors are therefore inverted:

• At odd positions, the next value is simply 2 * b.

• At even positions, we use the recursive function g() and the lookup table a to identify the smallest matching value:

  (g = k => a[k] ? g(k + 1) : k)(1)
  

To save a few bytes, i is initialized to {} rather than 0. This compels us to use:

• i^n to compare i with n because ({}) ^ n === n whereas ({}) - n evaluates to NaN.
• -~i to increment i, because ({}) + 1 would generate a string.

Demo

let f =

n=>(F=i=>i^n?F(a[b=i&1?2*b:(g=k=>a[k]?g(k+1):k)(1)]=-~i):b)(a={})

for(n = 1; n <= 20; n++) {
  console.log('a[' + n + '] = ' + f(n));
}

Answer 3

Python 3, 80 72 69 bytes

^{-7 bytes thanks to Mr. Xcoder!}

f=lambda n:n and[f(n-1)*2,min({*range(n+1)}-{*map(f,range(n))})][n%2]

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Answer 4

Jelly, 15 bytes

Jḟ⁸ḢðḤṭṭ
0Ç¡Ḋị@

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Answer 5

Mathics, 56 53 bytes

^{-3 bytes thanks JungHwan Min!}

(w={};Do[w~FreeQ~k&&(w=w~Join~{k,2k}),{k,#}];w[[#]])&

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Based on the Mathematica expression given in the OEIS link.

Answer 6

PHP, 64 bytes

for(;$argn-$i++;)if($i&$$e=1)for(;${$e=++$n};);else$e*=2;echo$e;

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PHP, 77 bytes

for(;$argn-$i++;$r[]=$e)if($i&1)for(;in_array($e=++$n,$r););else$e*=2;echo$e;

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PHP, 78 bytes

for(;$argn-$i++;)$e=$r[]=$i&1?end(array_diff(range($i,1),$r?:[])):2*$e;echo$e;

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Answer 7

PHP, 56 bytes

while($$n=$i++<$argn)for($n*=2;$i&$$k&&$n=++$k;);echo$n;

PHP, 75 72 bytes

for($a=range(1,$argn);$i++<$argn;)$a[~-$n=$i&1?min($a):2*$n]=INF;echo$n;

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Answer 8

05AB1E, 16 15 14 bytes

1-indexed.
Uses the fact that the binary representation of elements at odd indices in the sequence ends in an even number of zeroes: A003159.

Lʒb1¡`gÈ}€x¹<è

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Explanation

L                 # range [1 ... input]
 ʒ      }         # filter, keep only elements whose
  b               # ... binary representation
   1¡             # ... split at 1's
     `gÈ          # ... ends in an even length run
         €x       # push a doubled copy of each element in place
           ¹<è    # get the element at index (input-1)

Answer 9

Python 2, 59 51 49 bytes

f=lambda n,k=2:2/n%-3*(1-k)or f(n+~(k&-k)%-3,k+1)

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Background

Every positive n integer can be expressed uniquely as n = 2^o(n)c(n), where c(n) is odd.

Let ⟨a_n⟩_n>0 be the sequence from the challenge spec.

We claim that, for all positive integers n, o(a_2n-1) is even. Since o(a_2n) = o(2a_2n-1) = o(a_2n-1) + 1, this is equivalent to claiming that o(a_2n) is always odd.

Assume the claim is false and let 2m-1 be the first odd index of the sequence such that o(a_2m-1) is odd. Note that this makes 2m be the first even index of the sequence such that o(a_2m-1) is even.

o(a_2m-1) is odd and 0 is even, so a_2m-1 is divisible by 2. By definition, a_2m-1 is the smallest positive integer not yet appearing in the sequence, meaning that a_2m-1/2 must have appeared before. Let k be the (first) index of a_2m-1/2 in a.

Since o(a_k) = o(a_2m-1/2) = o(a_2m-1) - 1 is even, the minimality of n implies that k is odd. In turn, this means that a_k+1 = 2a_k = a_2m-1, contradicting the definition of a_2m-1.

How it works

yet to come

Answer 10

R, 70 69 65 bytes

function(n){for(i in 2*1:n)F[i-1:0]=which(!1:n%in%F)[1]*1:2
F[n]}

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An anonymous function that takes one argument. F defaults to FALSE or 0 so that the algorithm correctly assesses that no positive integers are in the sequence yet.

The algorithm generates the pairs in a for loop in the following manner (where i goes from 2 to 2n by 2):

           which(!1:n%in%l)[1]     # the missing value
                              *1:2 # keep one copy the same and double the next
l[i-1:0]=                         # store into l at the indices i-1 and i

Answer 11

Haskell, 63 bytes

g x=[2*g(x-1),[a|a<-[1..],a`notElem`map g[1..x-1]]!!0]!!mod x 2

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This one abuses Haskell's lazy evaluation to the fullest extent.

Answer 12

Perl 6, 50 bytes

{(1,{@_%2??2*@_[*-1]!!first *∉@_,1..*}...*)[$_]}

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• 1, { ... } ... * is a lazily-generated infinite sequence where each term after the first is provided by the brace-delimited code block. Since the block references the @_ array, it receives the entire current sequence in that array.
• If the current number of elements is odd (@_ % 2), we're generating an even-indexed element, so the next element is double the last element we have so far: 2 * @_[*-1].
• Otherwise, we get the first positive integer that does not yet appear in the sequence: first * ∉ @_, 1..*.
• $_ is the argument to the outer function. It indexes into the infinite sequence, giving the function's return value.

Answer 13

Mathematica, 82 bytes

(s={};a=1;f=#;While[f>0,If[s~FreeQ~a,s~AppendTo~{a,2a}];a++;f--];Flatten[s][[#]])&

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Answer 14

k, 27 bytes

*|{x,*(&^x?!y;|2*x)2!y}/!2+

1-indexed. Try it online!

Using (!y)^x instead of &^x?!y works too.

Answer 15

C# (Visual C# Interactive Compiler), 82 bytes

x=>{int y=1;for(var s="";x>2;x-=2)for(s+=2*y+":";s.Contains(++y+""););return x*y;}

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-6 bytes thanks to @ASCIIOnly!

Answer 16

Clojure, 102 bytes

#(nth(loop[l[0 1 2 3]i %](if(= i 0)l(recur(conj l(*(last l)2)(nth(remove(set l)(range))0))(dec i))))%)

Iterates n times to build up the sequence and returns the nth item, 1-indexed.

Answer 17

Ruby, 60 bytes

->n,*a{eval"v+=1while a[v];a[v]=a[2*v]=v+v*n%=2;"*(n/2+v=1)}

0-indexed. We loop n/2+1 times, generating two values each time and storing them by populating an array at their indices. v+v*n%2 gives the output, either v or v*2 depending on the parity of n.

Answer 18

Ruby, 49 bytes

->n{*s=t=0;s|[t+=1]==s||s<<t<<t*2until s[n];s[n]}

Start with [0], add pairs to the array until we have at least n+1 elements, then take the n+1th (1-based)

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Answer 19

JavaScript (ES6), 60 65 bytes

An iterative solution.

n=>eval("for(s={},i=0;n;)s[++i]?0:(s[i+i]=--n)?--n?0:i+i:i")

Less golfed

n=>{
  s = {}; //hashtable for used values
  for(i=0; n; )
  {
    if ( ! s[++i] )
    {
      s[i*2] = 1; // remember i*2 is already used
      if (--n)
        if (--n)
          0;
        else
          result = i*2;
      else
        result = i;
    }
  }
  return result;  
}

Test

F=
n=>eval("for(s={},i=0;n;)s[++i]?0:(s[i+i]=--n)?--n?0:i+i:i")

for (a=1; a < 50; a++)
  console.log(a,F(a))

Answer 20

Jelly, 13 12 10 bytes

ḤRọ2ḂĠZFị@

This uses the observation from my Python answer.

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How it works

ḤRọ2ḂĠZFị@  Main link. Argument: n

Ḥ           Unhalve; yield 2n.
 R          Range; yield [1, ... , 2n].
  ọ2        Compute the order of 2 in the factorization of each k in [1, ..., 2n].
    Ḃ       Bit; compute the parity of each order.
     G      Group the indices [1, ..., 2n] by the corresponding values.
      Z     Zip/transpose the resulting 2D array, interleaving the indices of 0
            with the indices of 1, as a list of pairs.
       F    Flatten. This yields a prefix of the sequence.
        ị@  Take the item at index n.