A213890 For any n >= 0, write all permutations of {0,1,...,n} in reverse lexicographic order. The last elements of the permutations will be the initial terms of this sequence.

0, 1, 0, 2, 1, 2, 0, 1, 0, 3, 1, 3, 0, 2, 0, 3, 2, 3, 1, 2, 1, 3, 2, 3, 0, 1, 0, 2, 1, 2, 0, 1, 0, 4, 1, 4, 0, 2, 0, 4, 2, 4, 1, 2, 1, 4, 2, 4, 0, 1, 0, 3, 1, 3, 0, 1, 0, 4, 1, 4, 0, 3, 0, 4, 3, 4, 1, 3, 1, 4, 3, 4, 0, 2, 0, 3, 2, 3, 0, 2, 0, 4, 2, 4, 0, 3, 0, 4, 3, 4, 2, 3, 2, 4, 3, 4, 1, 2, 1, 3, 2, 3, 1, 2, 1, 4, 2, 4, 1, 3, 1, 4, 3, 4, 2, 3, 2, 4, 3, 4, 0, 1, 0, 2, 1, 2, 0, 1, 0, 3

Offset

0,4

Comments

a(A007489(n)) = n is the first occurrence of n in the sequence.

Links

David W. Wilson, Table of n, a(n) for n = 0..5039

David W. Wilson, C++ function to efficiently compute a(n).

Example

For n = 2, the permutations of {0,1,2} in reverse lexical order are: (2,1,0),(2,0,1),(1,2,0),(1,0,2),(0,2,1),(0,1,2). The final elements of these permutations are 0,1,0,2,1,2 which are the initial terms of this sequence. The same holds for any n >= 0.

Maple

a:= proc(n) local h, k, m;
      h, k:= n, 0;
      for m from 2 while h>0 do
        k:= k+ `if`(irem(h, m, 'h')+k+1 < m, 0, 1)
      od; k
    end:
seq(a(n), n=0..130); # Alois P. Heinz, Jun 23 2012

Mathematica

a[n_] := Module[{h, k, m, r}, h = n; k = 0; For[m = 2, h>0, m++, k = k + If[{h, r} = QuotientRemainder[h, m ]; r+k+1 < m, 0, 1]]; k]; Table[a[n], {n, 0, 130}] (* Jean-François Alcover, Jun 11 2015, after Alois P. Heinz *)

Crossrefs

Cf. A159880, A191246, A191247.

Sequence in context: A125925 A340795 A137853 * A226460 A358360 A094114

Adjacent sequences: A213887 A213888 A213889 * A213891 A213892 A213893

Keyword

nonn

Author

David W. Wilson, Jun 23 2012

Status

approved