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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 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.)