Smallest positive integer which is coprime to the last two predecessors and has not yet appeared; a(1)=1, a(2)=2

Question

Definition

• Two integers are coprime if they share no positive common divisors other than 1.
• a(1) = 1
• a(2) = 2
• a(n) is the smallest positive integer which is coprime to the a(n-1) and a(n-2) and has not yet appeared, for integer n >= 3.

Task

• Given positive integer n, output/print a(n).

Example

• a(11) = 6 because 6 is coprime with the last two predecessors (namely, 11 and 13) and 6 has not appeared before.

Notes

• Note that the sequence is not ascending, meaning that an element can be smaller than its predecessor.

Specs

• You must use 1-indexed.

Testcases

n      a(n)
1      1
2      2
3      3
4      5
5      4
6      7
7      9
8      8
9      11
10     13
11     6
12     17
13     19
14     10
15     21
16     23
17     16
18     15
19     29
20     14
100    139
1000   1355
10000  13387
100000 133361

Scoring

• Since coprime means that the two numbers share only one divisor (1), and 1 is a small number, your code should be as small as possible in terms of byte-count.

References

• OEIS A084937

Answer 1

Python 3.5, 160 141 126 124 121 109 bytes

This is a simple implementation of the sequence's definition. Golfing suggestions welcome.

Edit: -17 bytes thanks to Leaky Nun. -9 bytes thanks to Peter Taylor. -6 bytes thanks to Sp3000 and switching to Python 3.5.

import math;f=lambda n,r=[2,1],c=3:n<2and r[1]or(c in r)+math.gcd(c,r[0]*r[1])<2and f(n-1,[c]+r)or f(n,r,c+1)

Ungolfing:

import math
def f(n, r=[2,1], c=3):
    if n<2:
        return r[1]
    elif (c in r) + math.gcd(c,r[0]*r[1]) < 2:
        return f(n-1, [c]+r)
    else:
        return f(n, r, c+1)

Answer 2

MATL, 28 27 bytes

2:i:"`@ym1MTF_)Zdqa+}@h]]G)

The code is slow, but gives the correct result.

Try it online! Or verify the first ten cases.

A small modification of the code produces a plot of the sequence:

2:i:"`@ym1MTF_)Zdqa+}@h]]G:)XG

See it as ASCII art, or with graphical output in the offline compiler:

Explanation

2:         % Push [1 2] to initiallize the sequence
i:         % Input n. Push [1 2 ... n]
"          % For loop: repeat n times
  `        %   Do while loop
    @      %     Push iteration index, starting at 1. This is the candidate number
           %     to extend the sequence
    y      %     Duplicate vector containing the sequence so far
    m      %     Is member? Gives true if the candidate is in the sequence
    1M     %     Push candidate and vector again
    TF_)   %     Get last two elements of the vector
    Zd     %     GCD between the candidate and those two elements. Produces a
           %     two-element vector
    qa     %     True if any of the two results exceeds 1, meaning
           %     the candidate is not coprime with the latest two sequence values
    +      %     Add. This corresponds to logical "or" of the two conditions, namely
           %     whether the candidate is member of the sequence so far, and
           %     whether it is not coprime with the latest two. In either case
           %     the do...while must continue with a next iteration, to try a new
           %     candidate. Else the loop is exited, and the current candidate
           %     is the new value of the sequence
  }        %   Finally (execute when the loop is exited)
    @h     %     Push current candidate and concatenate to the sequence vector
  ]        %   End do...while
]          % End for
G)         % Get n-th value of the sequence. Implicitly display

Answer 3

C, 185 bytes

G(a,b){return a%b?G(b,a%b):b;}
i,j,k;f(n){int a[n+2];for(i=0;i++<n;){a[i]=i<3?i:0;for(j=2;!a[i];++j){for(k=i;--k;){if(a[k]==j)++j,k=i;}a[G(a[i-1],j)*G(a[i-2],j)<2?i:0]=j;}}return a[n];}

Answer 4

Actually, 38 37 35 33 31 30 bytes

This is a simple implementation of the function definition. Golfing suggestions welcome. Try it online!

Edit: -3 bytes thanks to Leaky Nun.

2R#╗,;`1";2±╜tπg@╜í+Y"£╓╖`nD╜E

Ungolfing:

2R#╗    Push [1,2] and store it in register 0
,;      Take input and duplicate
`1      Start function, push 1
  "       Start string
  ;       Duplicate i
  2±╜t    Push (list in register 0)[-2:]
  πg      gcd(i, product of list[-2:])
  @╜í     Rotate the gcd and bring up i, check for i in list (0-based, -1 if not found)
  +Y      Add the gcd and the index, negate (1 if coprime and not found in list, else 0)
  "£      End string, turn into a function
╓       Push first (1) values where f(x) is truthy, starting with f(0)
╖`      Append result to the list in register 0, end function
n       Run function (input) times
D╜E     Return (final list)[n-1]

Answer 5

Haskell, 81 73 bytes

c l@(m:n:_)=m:c([x|x<-[1..],gcd(m*n)x<2,all(/=x)l]!!0:l)
((0:1:c[2,1])!!)

Usage example: ((0:1:c[2,1])!!) 12-> 17.

Build the list of all a(n), starting with 0 to fix the 1-based index and 1 and followed by c[2,1]. c takes the head of it's argument list l followed by a recursive call with then next number that fits (co-prime, not seen before) added in front of l. Pick the nth element of this list.

Answer 6

R, 141 bytes

 f=Vectorize(function(n)ifelse(n>3,{c=3;a=f(n-1);b=f(n-2);d=f(4:n-3);while(!c%%which(!a%%1:a)[-1]||!c%%which(!b%%1:b)[-1]||c%in%d)c=c+1;c},n))

ungolfed

f=Vectorize( function(n)     #build a recursive function. Vectorize allows
    if(n>3) {                #the function to be called on vectors.
        c=3                  #Tests size. Builds some frequent variables.
        a=f(n-1)
        b=f(n-2)
        d=f(4:n-3)           #Should really golf this out, but its horribly slow.
        while(!c%%which(!a%%1:a)[-1]||!c%%which(!b%%1:b)[-1]||c%in%d)
              c=c+1          #If we are coprime and not already seen. add.
        c
     } else n)               #First three are 1,2,3.

Answer 7

Mathematica, 97 90 bytes

a@1=1;a@2=2;a@n_:=SelectFirst[Range[2n],GCD[a[n-1]a[n-2],#]<2&&!MemberQ[a/@Range[n-1],#]&]

Based on my conjecture that a(n) < 2n for all n.

To get a faster run, add a@n= after the original := so that the function doesn't need to recalculate previous values.

Saved 7 bytes thanks to Sherlock9 (if gcd(a,b)=1 then gcd(ab,m) = gcd(a,m)*gcd(b,m))

Answer 8

Pyth, 23 bytes

eu+Gf&-TGq1iT*F>2G1tQ]1

Test suite

A fairly straightforward implementation, but with some nice golfing tricks.

eu+Gf&-TGq1iT*F>2G1tQ]1
 u                 tQ]1    Apply the following function input - 1 times,
                           where G is the current state (List of values so far)
  +G                       Add to G
    f             1        The first number, counting up from 1
      -TG                  That has not been seen so far
     &                     And where
               >2G         The most recent two numbers
             *F            Multiplied together
           iT              Gcd with the current number being checked
         q1                Equals 1
e                          Output the final element of the list.