A138750 a(n) = ceiling(n/2) if n == 2 (mod 3), a(n) = 2n otherwise.

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

Offset: 0

Author

M. F. Hasler, Mar 28 2008

Keywords

• easy
• nonn

Comments

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 ceiling() 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) = ceiling(n/2) otherwise, 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.

Examples

			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 congruent to 2 (mod 3).
a(2) = ceiling(2/2) = 1, a(5) = ceiling(5/2) = 3, a(8) = ceiling(8/2) = 4, a(11) = ceiling(11/2) = 6, ... since these indices are congruent to 2 (mod 3).
		

Links

• Index entries for linear recurrences with constant coefficients, signature (0,0,1,0,0,1,0,0,-1).
• Index entries for sequences related to 3x+1 (or Collatz) problem.

Crossrefs

Cf. A001281, A124123, A138751, A138752, A138753, A008588 (trisection), A016933 (trisection), A032766 (trisection)

Programs

Mathematica

Table[If[Mod[n,3]==2,Ceiling[n/2],2n],{n,0,70}] (* or *) LinearRecurrence[{0,0,1,0,0,1,0,0,-1},{0,2,1,6,8,3,12,14,4},70] (* Harvey P. Dale, Nov 20 2013 *)

PARI

A138750(n) = if( n%3==2, ceil(n/2), 2*n )

Formula

G.f.: x*(2 + x + 6*x^2 + 6*x^3 + 2*x^4 + 6*x^5 + 4*x^6) / ( (1+x)*(x^2-x+1)*(x-1)^2*(1+x+x^2)^2 ). - R. J. Mathar, Oct 16 2013

a(n) = a(n-3) + a(n-6) - a(n-9); a(0)=0, a(1)=2, a(2)=1, a(3)=6, a(4)=8, a(5)=3, a(6)=12, a(7)=14, a(8)=4. - Harvey P. Dale, Nov 20 2013

Sum_{n>=1} (-1)^n/a(n) = log(3)/2 - log(2)/3 = log(27/4)/6. - Amiram Eldar, Jul 26 2024