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a(n) = ceiling(n/2) if n == 2 (mod 3), a(n) = 2n otherwise.
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%I #28 Jul 26 2024 08:58:16

%S 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,

%T 13,54,56,15,60,62,16,66,68,18,72,74,19,78,80,21,84,86,22,90,92,24,96,

%U 98,25,102,104,27,108,110,28,114,116,30,120,122,31,126,128,33,132,134,34

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

%C 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.

%C The use of ceiling() is here equivalent to round().

%C 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.

%C 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.

%C 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.

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,1,0,0,1,0,0,-1).

%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>.

%F 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

%F 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

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

%e 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).

%e 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).

%t 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 *)

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

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

%K easy,nonn

%O 0,2

%A _M. F. Hasler_, Mar 28 2008