A377114 a(n) = coefficient of sqrt(2) in the expansion of (3 + sqrt(2) + sqrt(3))^n.

0, 1, 6, 38, 240, 1504, 9360, 57968, 357888, 2205376, 13574784, 83503232, 513469440, 3156723712, 19404782592, 119276106752, 733133340672, 4506134745088, 27696241336320, 170229576458240, 1046279833190400, 6430725296226304, 39524980495024128

Offset: 0

Author

Clark Kimberling, Oct 21 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 1,8,10,7 and these periods:

p = 2: (2)

p = 3: (4, 2, 6, 6, 1, 1, 3, 1)

p = 5: (12, 3, 9, 6, 6, 2, 7, 3, 10, 2)

p = 7: (9, 15, 3, 18, 3, 15, 9)

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) = 6.
		

Links

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

Crossrefs

Cf. A377090, A377113, A377115, A377116.

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]];
s2  (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates this sequence. *)
LinearRecurrence[{12, -44, 48, 8}, {0, 1, 6, 38}, 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)=1, a(3)=6, a(4)=38.

G.f.: x*(-1 + 6*x - 10*x^2)/(-1 + 12*x - 44*x^2 + 48*x^3 + 8*x^4).