A377119 a(n) = coefficient of 2^(2/3) in the expansion of (2^(1/3) + 2^(2/3))^n.

0, 1, 1, 6, 12, 42, 108, 324, 900, 2592, 7344, 20952, 59616, 169776, 483408, 1376352, 3919104, 11158560, 31772736, 90465984, 257587776, 733432320, 2088322560, 5946120576, 16930529280, 48206658816, 137259899136, 390823128576, 1112799347712, 3168498166272

Offset

0,4

Comments

Conjecture: every prime divides a(n) for infinitely many n, and if K(p) = (k(1), k(2),...) is the maximal subsequence of indices n such that p divides a(n), then the difference sequence of K(p) is eventually periodic; indeed, K(p) is purely periodic for the first 6 primes, with respective period lengths 2,2,4,5,2,5 and these periods:

p = 2: (2,1)

p = 3: (7, 1)

p = 5: (21, 5, 13, 1)

p = 7: (15, 17, 20, 43, 1)

p = 11: (9, 1)

p = 13: (102, 5, 17, 15, 1)

See A377109 for a guide to related sequences.

Links

Table of n, a(n) for n=0..29.

Index entries for linear recurrences with constant coefficients, signature (0,6,6).

Formula

a(n) = 6*a(n-2) + 6*a(n-3) for n>=1, with a(0)=0, a(1)=1, a(3)=1.

G.f.: (x (1 + x))/(1 - 6 x^2 - 6 x^3).

Example

((2^(1/3) + 2^(2/3)))^3 = 4 + 2*2^(1/3) + 2^(2/3), so a(3) = 1.

Mathematica

(* Program 1 generates sequences A377117-A377119. *)
tbl = Table[Expand[(2^(1/3) + 2^(2/3))^n], {n, 0, 30}];
u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
   Map[({#1, #1 /. _^_ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
{s1, s2, s3} = Transpose[(PadRight[#1, 3] &) /@ Last /@ u][[1 ;; 3]];
s3   (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates (a(n)) for n>=1. *)
LinearRecurrence[{0, 6, 6}, {0, 1, 1}, 15].

Crossrefs

Cf. A377109, A377117, A377118.

Sequence in context: A152787 A060551 A129113 * A048025 A120471 A219522

Adjacent sequences: A377116 A377117 A377118 * A377120 A377121 A377122

Keyword

nonn,easy

Author

Clark Kimberling, Oct 26 2024

Status

approved