A377116 a(n) = coefficient of sqrt(6) in the expansion of (3 + sqrt(2) + sqrt(3))^n.

0, 0, 2, 18, 128, 840, 5328, 33264, 206080, 1271808, 7833472, 48200064, 296423424, 1822459392, 11203152896, 68863546368, 423273267200, 2601614180352, 15990421856256, 98282063536128, 604069867552768, 3712780777586688, 22819757583302656, 140256346936639488

Offset: 0

Author

Clark Kimberling, Oct 23 2024

Keywords

• nonn
• easy

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 4 primes, with respective period lengths 4,5,9,10 and these periods:

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

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

p = 5: (1, 6, 4, 1, 5, 7, 12, 12, 12)

p = 7: (1, 11, 5, 1, 18, 14, 4, 4, 2, 12)

See A377109 for a guide to related sequences.

Examples

			(3 + sqrt(2) + sqrt(3))^3 = 14 + 6*sqrt(2) + 6*sqrt(3) + 2*sqrt(6), so a(3) = 2.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (12,-44,48,8).

Crossrefs

Cf. A377109, A377113, A377114, A377115, A377117.

Programs

Mathematica

(* Program 1 generates sequences A377113-A377116. *)
tbl = Table[Expand[(3 + 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[{12, -44, 48, 8}, {0, 0, 2, 18}, 25]

Formula

a(n) = 12*a(n-1) - 44*a(n-2) + 48*a(n-3) + 8*a(n-4), with a(0)=0, a(1)=0, a(3)=2, a(4)=18.

G.f.: 2*x^2*(-1 + 3*x)/(-1 + 12*x - 44*x^2 + 48*x^3 + 8*x^4).