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A390310
a(n) = if(n mod 3 = 0, n/3, 4 * n + (-1)^(n mod 3)).
0
0, 3, 9, 1, 15, 21, 2, 27, 33, 3, 39, 45, 4, 51, 57, 5, 63, 69, 6, 75, 81, 7, 87, 93, 8, 99, 105, 9, 111, 117, 10, 123, 129, 11, 135, 141, 12, 147, 153, 13, 159, 165, 14, 171, 177, 15, 183, 189, 16, 195, 201, 17, 207, 213, 18, 219, 225, 19, 231, 237, 20, 243, 249, 21, 255, 261, 22, 267, 273, 23, 279, 285, 24, 291, 297, 25
OFFSET
0,2
COMMENTS
Does iterating n -> a(n) always reach 1 (i.e., cycle 1 -> 3 -> 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
Cf. A006370.
Sequence in context: A019817 A243526 A329214 * A080322 A126179 A304249
KEYWORD
nonn,easy
AUTHOR
Werner Schulte, Nov 30 2025
STATUS
approved