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

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, 17, 54, 56, 19, 60, 62, 21, 66, 68, 23, 72, 74, 25, 78, 80, 27, 84, 86, 29, 90, 92, 31, 96, 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

Offset: 0

Author

Thomas Scheuerle, Mar 09 2021

Keywords

• nonn
• easy

Comments

This sequence is a permutation of all numbers not congruent to 4 (mod 6) (A047256).

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.

m = a(n) is the smallest solution to A014682(m) = n.

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.

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.

Links

• Index entries for sequences related to 3x+1 (or Collatz) problem
• Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Crossrefs

Cf. A093544, A093545, A014682, A047256.

Programs

MATLAB

function a = A342369( max_n )
    a(1) = 0;
    for n=1:max_n
        if mod(n,3) == 2
            a(n) = (2*n - 1)/3;
        else
            a(n) = 2*n;
        end
    end
end

Mathematica

Array[If[Mod[#, 3] == 2, (2 # - 1)/3, 2 #] &, 74, 0] (* Michael De Vlieger, Mar 14 2021 *)

PARI

a(n) = if ((n%3)==2, (2*n - 1)/3, 2*n); \\ Michel Marcus, Mar 09 2021

Formula

a(n) = A093545(n) + floor((1 + A093545(n))/5).

a(3*n) = a(3*(n-1)) + 6.

a(3*n - 1) = a(3*(n-1) - 1) + 2.

a(3*n - 2) = a(3*(n-1) - 2) + 6.

a(n) = 14*n - 2*a(n-1) - 3*a(n-2) - 2*a(n-3) - a(n-4) - 29 for n >= 4.

A014682(a(n)) = n.

a(A014682(n)) = (n+2)/3 - 1 if n == 4 (mod 6).

a(A014682(n)) = n if n !== 4 (mod 6).

a^k(3*n) = (3*n)*2^k where a^2(3*n) is a(a(3*n)) = (3*n)*4.

G.f.: -(-x^5 - 4*x^4 - 6*x^3 - x^2 - 2*x)/(x^6 - 2*x^3 + 1).

a(n) = (A093544(n+1) - 1)/2. - Hugo Pfoertner, Mar 10 2021