OFFSET
0,2
COMMENTS
Does iterating n -> a(n) always reach 1 (i.e., cycle 1 -> 3 -> 1)?
LINKS
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).
FORMULA
a(3*n) = n.
a(3*n+1) = a(3*n-2) + 12 for n > 0.
a(3*n+2) = a(3*n-1) + 12 for n > 0.
a(n) = 2 * a(n-3) - a(n-6) for n > 5.
G.f.: x*(3 + 9*x + x^2 + 9*x^3 + 3*x^4) / (1 - x^3)^2.
E.g.f.: exp(-x/2)*(25*exp(3*x/2)*x + 11*x*cos(sqrt(3)*x/2) + sqrt(3)*(11*x - 6)*sin(sqrt(3)*x/2))/9. - Stefano Spezia, Nov 30 2025
MATHEMATICA
a[n_] := If[Divisible[n, 3], n/3, 4*n + (-1)^Mod[n, 3]]; Array[a, 100, 0] (* Amiram Eldar, Nov 30 2025 *)
PROG
(PARI) a(n) = if(n%3 == 0, n/3, 4*n + (-1)^(n%3))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Werner Schulte, Nov 30 2025
STATUS
approved
