A377315 a(n) = coefficient of 2^(1/3) in the expansion of (1 + 2^(1/3) + 2^(2/3))^n.

0, 1, 4, 15, 58, 223, 858, 3301, 12700, 48861, 187984, 723235, 2782518, 10705243, 41186518, 158457801, 609638200, 2345474521, 9023795964, 34717449655, 133569211378, 513883779063, 1977076420978, 7606449811501, 29264462476500, 112589813284981, 433169277095944

Offset

0,3

Comments

See A377109 for a guide to related sequences.

Links

Table of n, a(n) for n=0..26.

Index entries for linear recurrences with constant coefficients, signature (3,3,1).

Formula

a(n) = 3*a(n-1) + 3*a(n-2) + a(n-3), with a(0)=0, a(1)=1, a(3)=4.

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

Example

(1 + 2^(1/3) + 2^(2/3))^3 = 19 + 15 2^(1/3) + 12 2^(2/3), so a(3) = 15.

Mathematica

(* Program 1 generates sequences A377314-A377315 and A108368. *)
tbl = Table[Expand[(1 + 2^(1/3) + 2^(2/3))^n], {n, 0, 24}];
Take[tbl, 6]
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]];
s2  (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates (a(n)) for n>=1. *)
LinearRecurrence[{3, 3, 1}, {0, 1, 4}, 15]

Crossrefs

Cf. A377109, A377117, A377314, A108368 (coefficients of 2^(2/3)).

Sequence in context: A109642 A307570 A164589 * A017950 A003126 A160156

Adjacent sequences: A377312 A377313 A377314 * A377316 A377317 A377318

Keyword

nonn,easy

Author

Clark Kimberling, Oct 26 2024

Status

approved