|
Search: id:A138750
|
|
|
| A138750 |
|
a(n)=ceil(n/2) if n=2 (mod 3), a(n)=2n else. |
|
+0 8
|
|
| 0, 2, 1, 6, 8, 3, 12, 14, 4, 18, 20, 6, 24, 26, 7, 30, 32, 9, 36, 38, 10, 42, 44, 12, 48, 50, 13, 54, 56, 15, 60, 62, 16, 66, 68, 18, 72, 74, 19, 78, 80, 21, 84, 86, 22, 90, 92, 24, 96, 98, 25, 102, 104, 27, 108, 110, 28, 114, 116, 30, 120, 122, 31, 126, 128, 33, 132, 134, 34
(list; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
This map is inspired by A124123, which hides in
fact a variation of the Collatz problem,
defined on the set of primes, and working mod 3 instead of mod 2. See
A138751 for more information.
The use of ceil() is here equivalent to round().
The main reason for defining this function is to write A124123 as
complement of A007918( A138750( A000040 )),
and to express the recursion function occurring there in terms of this map.
It might have been more natural to define this map as a(n)=2n if n=1
(mod 3), a(n)=ceil(n/2) else,
which is equivalent for all primes > 3 (which are either =1 or =2
(mod 3)) and would have "better"
properties regarding the analysis of orbits of all integers under this map.
However, for the prime n=3 it does make a difference,
and in order to reproduce the map occurring in A124123, we had to
adopt the present convention.
|
|
LINKS
|
Georges Brougnard, Definition of GB-sequences.
Index entries for sequences related to 3x+1 (or Collatz) problem
|
|
EXAMPLE
|
a(0)=2*0=0, a(1)=2*1=2, a(3)=2*3=6, a(4)=2*4=8,... since
these indices are not equal to 2 (mod 3).
a(2)=ceil(2/2)=1, a(5)=ceil(5/2)=3, a(8)=ceil(8/2)=4,
a(11)=ceil(11/2)=6,... since these indices equal 2 (mod 3).
|
|
PROGRAM
|
(PARI) A138750(n) = if( n%3==2, ceil(n/2), 2*n )
|
|
CROSSREFS
|
Cf. A001281, A124123, A138751-A138753.
Adjacent sequences: A138747 A138748 A138749 this_sequence A138751 A138752 A138753
Sequence in context: A138510 A026215 A026220 this_sequence A048850 A004488 A011419
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
M. F. Hasler (www.univ-ag.fr/~mhasler), Mar 28 2008
|
|
|
Search completed in 0.002 seconds
|