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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. 4
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

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

Table of n, a(n) for n=0..73.

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).

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

MATHEMATICA

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

PROG

(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

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

CROSSREFS

Cf. A093544, A093545, A014682, A047256.

Sequence in context: A026215 A026220 A138750 * A342842 A048850 A004488

Adjacent sequences:  A342366 A342367 A342368 * A342370 A342371 A342372

KEYWORD

nonn,easy

AUTHOR

Thomas Scheuerle, Mar 09 2021.

STATUS

approved

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Last modified September 19 08:05 EDT 2021. Contains 347556 sequences. (Running on oeis4.)