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A342369 If n is congruent to 2 (mod 3), then a(n) = (2*n - 1)/3; otherwise, a(n) = 2*n.

%I A342369 #61 Jul 09 2025 04:55:36
%S A342369 0,2,1,6,8,3,12,14,5,18,20,7,24,26,9,30,32,11,36,38,13,42,44,15,48,50,
%T A342369 17,54,56,19,60,62,21,66,68,23,72,74,25,78,80,27,84,86,29,90,92,31,96,
%U A342369 98,33,102,104,35,108,110,37,114,116,39,120,122,41,126,128,43,132,134,45,138,140,47,144,146
%N A342369 If n is congruent to 2 (mod 3), then a(n) = (2*n - 1)/3; otherwise, a(n) = 2*n.
%C A342369 This sequence is a permutation of all numbers not congruent to 4 (mod 6) (A047256).
%C A342369 This sequence has no finite cycles other than (2,1) under recursion, because of all cycles of permutation A093545(n), only one cycle (2,1) is without any number congruent to 4 (mod 6). See section "formula" first entry here. After a finite number of recursions it will reach only numbers divisible by 3 under further recursion.
%C A342369 m = a(n) is the smallest solution to A014682(m) = n.
%C A342369 If we define f(n) = 2*n and j(n) as an arbitrary recursion into a(n) and/or f(n) ( two examples: j(n) = a(a(n)) or j(n) = a(f(a(n))) ), then for all m, k and n = A014682^k(m), exists a j(n) that allows m = j(n). "^k" means recursion here.
%C A342369 Proving the Collatz conjecture could be done by proving that for all positive integers m, a function j(n) (see first comment) could be designed such that m = j(1). All numbers greater than 4 can be reached by a^k(6*p - 2) with exactly one p and k. The Collatz conjecture cannot be true if 3*p - 1 = a^k(6*p - 2) exists.
%H A342369 <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
%H A342369 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,2,0,0,-1).
%F A342369 a(n) = A093545(n) + floor((1 + A093545(n))/5).
%F A342369 a(3*n) = a(3*(n-1)) + 6.
%F A342369 a(3*n - 1) = a(3*(n-1) - 1) + 2.
%F A342369 a(3*n - 2) = a(3*(n-1) - 2) + 6.
%F A342369 a(n) = 14*n - 2*a(n-1) - 3*a(n-2) - 2*a(n-3) - a(n-4) - 29 for n >= 4.
%F A342369 A014682(a(n)) = n.
%F A342369 a(A014682(n)) = (n+2)/3 - 1 if n == 4 (mod 6).
%F A342369 a(A014682(n)) = n if n !== 4 (mod 6).
%F A342369 a^k(3*n) = (3*n)*2^k where a^2(3*n) is a(a(3*n)) = (3*n)*4.
%F A342369 G.f.: -(-x^5 - 4*x^4 - 6*x^3 - x^2 - 2*x)/(x^6 - 2*x^3 + 1).
%F A342369 a(n) = (A093544(n+1) - 1)/2. - _Hugo Pfoertner_, Mar 10 2021
%t A342369 Array[If[Mod[#, 3] == 2, (2 # - 1)/3, 2 #] &, 74, 0] (* _Michael De Vlieger_, Mar 14 2021 *)
%o A342369 (MATLAB)
%o A342369 function a = A342369( max_n )
%o A342369     a(1) = 0;
%o A342369     for n=1:max_n
%o A342369         if mod(n,3) == 2
%o A342369             a(n) = (2*n - 1)/3;
%o A342369         else
%o A342369             a(n) = 2*n;
%o A342369         end
%o A342369     end
%o A342369 end
%o A342369 (PARI) a(n) = if ((n%3)==2, (2*n - 1)/3, 2*n); \\ _Michel Marcus_, Mar 09 2021
%Y A342369 Cf. A093544, A093545, A014682, A047256.
%K A342369 nonn,easy
%O A342369 0,2
%A A342369 _Thomas Scheuerle_, Mar 09 2021