A350034 a(n) = n/gcd(n,6) if gcd(n,6)>1, 5n + 1 otherwise.

0, 6, 1, 1, 2, 26, 1, 36, 4, 3, 5, 56, 2, 66, 7, 5, 8, 86, 3, 96, 10, 7, 11, 116, 4, 126, 13, 9, 14, 146, 5, 156, 16, 11, 17, 176, 6, 186, 19, 13, 20, 206, 7, 216, 22, 15, 23, 236, 8, 246, 25, 17, 26, 266, 9, 276, 28, 19, 29, 296, 10, 306, 31, 21, 32, 326

Offset

0,2

Comments

The 5x+1 problem is as follows: start with any number n. If gcd(n,6)>1, divide it by gcd(n,6), otherwise multiply it by 5 and add 1. Do we always reach 1? This is an unsolved problem. It is conjectured that the answer is yes.

Links

Winston de Greef, Table of n, a(n) for n = 0..9999

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,2,0,0,0,0,0,-1).

Formula

a(n) = 2*a(n-6) - a(n-12). - Wesley Ivan Hurt, Oct 20 2022

Sum_{k=1..n} a(k) ~ (23/24)*n^2. - Amiram Eldar, Oct 07 2023

G.f.: x*(6 + x + x^2 + 2*x^3 + 26*x^4 + x^5 + 24*x^6 + 2*x^7 + x^8 + x^9 + 4*x^10) / (1 - x^6)^2. - Elmo R. Oliveira, May 17 2026

Mathematica

f[n_]:=If[GCD[n, 6]>1, n/GCD[n, 6], 5*n+1]; Table[f[n], {n, 0, 100}]

Prog

(PARI) a(n) = my(g = gcd(n, 6)); if (g>1, n/g, 5*n+1); \\ Michel Marcus, Dec 09 2021
(Python)
from math import gcd
def A350034(n): return n//g if (g:=gcd(n, 6)) > 1 else 5*n+1 # Chai Wah Wu, Dec 29 2021

Crossrefs

Cf. A006370, A089128, A350035.

Sequence in context: A345628 A364129 A336179 * A133419 A010134 A019713

Adjacent sequences: A350031 A350032 A350033 * A350035 A350036 A350037

Keyword

nonn,easy

Author

José María Grau Ribas, Dec 09 2021

Status

approved