|
|
A192082
|
|
Let f=A038554(n)+delta(n,1), where delta is the Kronecker symbol. Then a(n) is the fixed point that arises from iterating f (a(n)=0 or 1).
|
|
5
|
|
|
0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0
|
|
COMMENTS
|
Since, for n>=2, f(n)<n, the number of iterations required to reach 0 or 1 is finite.
|
|
LINKS
|
|
|
MAPLE
|
f := proc(n) local i, b, v: v:=0: if(n=1)then return 1: fi: b:=convert(n, base, 2): for i to nops(b)-1 do v:=v+((b[i]+b[i+1]) mod 2)*2^(i-1): od: return v: end: a:= proc(n) local v: v:=n: while(v>1)do v:=f(v): od: return v: end: seq(a(n), n=0..104); # Nathaniel Johnston, Jun 30 2011
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|