A377113 a(n) = coefficient of the term that is independent of sqrt(2), sqrt(3), and sqrt(6) in the expansion of (3 + sqrt(2) + sqrt(3))^n.

1, 3, 14, 72, 400, 2328, 13904, 84192, 513472, 3143232, 19278464, 118359552, 727045120, 4467233280, 27452300288, 168714381312, 1036914921472, 6372994560000, 39169586880512, 240744913207296, 1479676193996800, 9094462273585152, 55896907276156928

Offset: 0

Author

Clark Kimberling, Oct 21 2024

Keywords

• nonn

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 1,5,9,7 and these periods:

p = 2: (2)

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

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

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

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) = 14.
		

Links

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

Crossrefs

Cf. A377109, A377114, 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]];
s1  (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates this sequence. *)
LinearRecurrence[{12, -44, 48, 8}, {1, 3, 14, 72}, 15]
(* Program 3 confirms the periodicity properties described in Comments. *)
tbl = Table[Expand[(3 + Sqrt[2] + Sqrt[3])^n], {n, 0, 1000}];
u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
   Map[({#1, #1 /. ^ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
v = {s1, s2, s3, s4} = Transpose[(PadRight[#1, 4] &) /@ Last /@ u][[1 ;; 4]];
Position[Partition[list, Length[#], 1], Flatten[{_, #, _}]] &[
  seqtofind];
period[seq_] := (If[Last[#1] == {} || Length[#1] == Length[seq] - 1,
      0, Length[#1]] &)[NestWhileList[Rest, Rest[seq], #1 != Take[seq, Length[#1]] &, 1]];
periodicityReport[seq_] := ({Take[seq, Length[seq] - Length[#1]], period[#1],
      Take[#1, period[#1]]} &)[Take[seq, -Length[
      NestWhile[Rest[#1] &, seq, period[#1] == 0 &, 1, Length[seq]]]]];
seq = s1; Take[seq, 10]
f[n_] := Flatten[Position[Mod[s1, Prime[n]], 0]];
d[n_] := Differences[f[n]];
Table[Take[f[n], 10], {n, 2, 4}]
Table[Take[d[n], 10], {n, 2, 4}]
Column[Table[{n, Prime[n], periodicityReport[d[Prime[n]]]}, {n, 1, 8}]]
(* Peter J. C. Moses, Aug 07 2014, Oct 16 2024 *)

Formula

a(n) = 12*a(n-1) - 44*a(n-2) + 48*a(n-3) + 8*a(n-4), with a(0)=1, a(1)=3, a(3)=14, a(4)=72.

G.f.: (-1 + 9 x - 22 x^2 + 12 x^3)/(-1 + 12 x - 44 x^2 + 48 x^3 + 8 x^4).