A377112 a(n) = coefficient of sqrt(6) in the expansion of (2 + sqrt(2) + sqrt(3))^n.

0, 0, 2, 12, 68, 360, 1878, 9716, 50120, 258192, 1329322, 6842396, 35215884, 181237368, 932711806, 4800019332, 24702255760, 127124540448, 654216959826, 3366774510892, 17326314208468, 89165799266952, 458870789205926, 2361470423992852, 12152751175719000

Offset

0,3

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 7 primes, with respective period lengths 6, 5, 18, 28, 16, 14, 35:

p = 2: (1, 2, 2, 1, 3, 3),

p = 3: (1, 4, 3, 8, 8).

See A377109 for a guide to related sequences.

Links

Table of n, a(n) for n=0..24.

Index entries for linear recurrences with constant coefficients, signature (8,-14,-8,23).

Formula

a(n) = 8*a(n-1) - 14*a(n-2) - 8*a(n-3) + 23*a(n-4), with a(0)=0, a(1)=0, a(3)=2, a(4)=12.

G.f.: (2 x^2 (-1 + 2 x))/(-1 + 8 x - 14 x^2 - 8 x^3 + 23 x^4).

Example

(2 + sqrt(2) + sqrt(3))^3 = 9 + 4*sqrt(2) + 4*sqrt(3) + 2*sqrt(6), so a(3) = 2.

Mathematica

(* Program 1 generates sequences A377109-A37712. *)
tbl = Table[Expand[(2 + Sqrt[2] + Sqrt[3])^n], {n, 0, 24}];
u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
   Map[({#1, #1 /. _^_ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
{s1, s2, s3, s4}=Transpose[(PadRight[#1, 4]&)/@Last/@u][[1;; 4]];
s4  (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates this sequence. *)
LinearRecurrence[{8, -14, -8, 23}, {0, 0, 2, 12}, 12].

Crossrefs

Cf. A377109.

Sequence in context: A056636 A174327 A245997 * A128103 A359489 A329789

Adjacent sequences: A377109 A377110 A377111 * A377113 A377114 A377115

Keyword

nonn

Author

Clark Kimberling, Oct 20 2024

Status

approved