Seqindignot sequence

Python 2, 92 91 89 88 bytes

a=()
i=0
exec"x=0\nwhile set(`x`)&set(`i`)or x in a:x+=1\na+=x,;i+=1;"*-~input()
print a

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Prints a list of the first n+1 numbers


Different approach, which is a lot faster:

Python 2, 96 bytes

n=input()
r=range(9*n)
i=0
exec"x=0\nwhile set(`r[x]`)&set(`i`):x+=1\nprint r.pop(x),;i+=1;"*-~n

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Haskell, 80 69 bytes

f n=[x|x<-[0..],all(`notElem`show n)$show x,all(/=x)$f<$>[0..n-1]]!!0

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Very slow for large n.

f n=
    [x|x<-[0..]     ] !!0          -- pick the first of all 'x' from [0..]
                                   -- where
      all(`notElem`show n)$show x  -- no digit of 'n' appears in 'x', and
      all(/=x)                     -- 'x' is not seen before, i.e. not in the list
               f<$>[0..n-1]        -- 'f' mapped to [0..n-1]

Edit: @Laikoni saved 10 bytes. Thanks!


Pyth, 18 bytes

u+Gf!|}[email protected]`H`T0hQY

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Note that this returns the entire sequence up to index N, but the link returns the last number only, by prepending an e (end). If you want to see the raw value returned by this program, just remove it.

How it works

u+Gf!|}[email protected]`H`T0hQY  - Full program.

u     ...      hQY  - Reduce hQ (the input incremented) from left to right, with the
                       function ...(G, H), with starting value Y (the empty list).
                       G is the current value and H is the iteration index.
   f          0     - First integer starting from 0, that satisfies the following:
      }TG             - Appears in G...
     |   @`H`T        - Or its (string) intersection with the current index (H) is
                        non-empty.
    !                 - Logical NOT (boolean negation).
 +G                 - Append the value obtained above to the current value (G).
                      This becomes the given value for the next iteration.
                    - Implicitly print all intermediate results, or add e to print 
                      the last one.