cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A275218 Numbers in 2-cycles of RATS sequences.

Original entry on oeis.org

78, 117, 156, 288, 11127, 11667, 23388, 27888, 111177, 228888, 111111777, 222888888, 1111122267, 3333337788, 111111117777, 222288888888, 111111111177777, 222228888888888, 111111111111777777, 222222888888888888
Offset: 1

Views

Author

Robert Israel, Jul 20 2016

Keywords

Comments

Numbers n such that A036839(A036839(n)) = n.
Subset of A161596.
Contains A002275(3*k) + 6*A002275(k) and 2*A002275(3*k)+6*A002275(2*k) for all k>0.
In particular, this sequence and A161596 are infinite.
Do all sufficiently large members of the sequence have the form A002275(3*k) + 6*A002275(k) or 2*A002275(3*k)+6*A002275(2*k)?

Examples

			78 is in the sequence because A036839(78) = 156 and A036839(156) = 78.
		

Crossrefs

Programs

  • Maple
    rev:= proc(n) local t,L;
       L:= convert(n,base,10);
       add(10^j*L[-1-j],j=0..nops(L)-1)
    end proc:
    sord:= proc(n) local L,t;
      L:= sort(convert(n,base,10),`>`);
      add(10^j*L[1+j],j=0..nops(L)-1)
    end proc:
    rats:= proc(n) option remember;  sord(n + rev(n)) end proc:
    Res:= NULL:
    for d from 1 to 15 do
      for x1 from 0 to d do
        for x2 from 0 to d-x1 do
          for x3 from 0 to d-x1-x2 do
             for x4 from 0 to d-x1-x2-x3 do
               for x5 from 0 to d-x1-x2-x3-x4 do
                 for x6 from 0 to d-x1-x2-x3-x4-x5 do
                   for x7 from 0 to d-x1-x2-x3-x4-x5-x6 do
                     for x8 from 0 to d-x1-x2-x3-x4-x5-x6-x7 do
                       x9:= d-x1-x2-x3-x4-x5-x6-x7-x8;
                       L:= [1$x1,2$x2,3$x3,4$x4,5$x5,6$x6,7$x7,8$x8,9$x9];
                       x:= add(L[-i]*10^(i-1),i=1..d);
                       if rats(rats(x)) = x then Res:= Res,x fi
    od od od od od od od od od:
    sort([Res]);