A377315 -id:A377315 - cp's OEIS frontend

A377109 a(n) = coefficient of the term that is independent of sqrt(2), sqrt(3), and sqrt(6) in the expansion of (2 + sqrt(2) + sqrt(3))^n.

1, 2, 9, 38, 185, 922, 4689, 23998, 123217, 633458, 3258489, 16765718, 86273225, 443967370, 2284733313, 11757749038, 60508271137, 311391065570, 1602499602537, 8246883961094, 42440638964825, 218410733951098, 1123999345270833, 5784397706237854

Offset: 0

Clark Kimberling, Oct 20 2024

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 1,5,7,17,3,11,35 and these periods:
p = 2: (4)
p = 3: (8, 1, 4, 3, 8)
p = 5: (10, 20, 9, 8, 32, 21, 20)
p = 7: (2, 30, 9, 19, 6, 28, 12, 5, 16, 26, 22, 13, 2, 1, 24, 16, 57)
p = 11: (61, 29, 70)
p = 13: (9, 15, 24, 3, 21, 21, 3, 24, 15, 9, 24)
p = 17: (30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 1, 29, 29, 1, 30, 30, 30, 30, 2, 28, 30, 30, 28, 2, 17, 13, 30, 30, 30, 13, 17)
Guide to related sequences:
(1 + sqrt (2) + sqrt (3))^n, coefficients of absolute terms: A188570
(1 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(2): A188571
(1 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(3): A188572
(1 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(6): A188573
(2 + sqrt (2) + sqrt (3))^n, coefficients of independent terms: this sequence
(2 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(2): A377110
(2 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(3): A377111
(2 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(6): A377112
(3 + sqrt (2) + sqrt (3))^n, coefficients of independent terms: A377113
(3 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(2): A377114
(3 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(3): A377115
(3 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(6): A377116
(2^(1/3) + 2^(2/3))^n, coefficients of independent terms: A377117
(2^(1/3) + 2^(2/3))^n, coefficients of 2^(1/3): A377118
(2^(1/3) + 2^(2/3))^n, coefficients of 2^(2/3): A377119
(1 + 2^(1/3) + 2^(2/3))^n, coefficients of independent terms: A377314
(1 + 2^(1/3) + 2^(2/3))^n, coefficients of 2^(1/3): A377315

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

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

Cf. A188570, A377110, A377111, A377112.

(* Program 1 generates sequences A377109-A377112. *)
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]];
s1  (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates this sequence. *)
LinearRecurrence[{8, -14, -8, 23}, {1, 2, 9, 38}, 15]
(* Program 3 confirms the periodicity properties described in Comments. *)
tbl = Table[Expand[(2 + 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 *)

a(n) = 8*a(n-1) - 14*a(n-2) - 8*a(n-3) + 23*a(n-4), with a(0)=1, a(1)=2, a(3)=9, a(4)=38.

G.f.: (-1 + 6 x - 7 x^2 - 2 x^3)/(-1 + 8 x - 14 x^2 - 8 x^3 + 23 x^4).

A377314 a(n) = coefficient of the term that is independent of 2^(1/3) and 2^(2/3) in the expansion of (1 + 2^(1/3) + 2^(2/3))^n.

Original entry on oeis.org

1, 1, 5, 19, 73, 281, 1081, 4159, 16001, 61561, 236845, 911219, 3505753, 13487761, 51891761, 199644319, 768096001, 2955112721, 11369270485, 43741245619, 168286661033, 647452990441, 2490960200041, 9583526232479, 36870912288001, 141854275761481

Offset: 0

Clark Kimberling, Oct 26 2024

nonn

See A377109 for a guide to related sequences.

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

Michael De Vlieger, Table of n, a(n) for n = 0..1709
Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 4.
Index entries for linear recurrences with constant coefficients, signature (3,3,1).

Cf. A377109, A377117, A377315, A108368 (coefficients of 2^(2/3)).

(* 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]];
s1  (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates (a(n)) for n>=1. *)
LinearRecurrence[{3,3,1}, {1, 1, 5}, 15]

a(n) = 3*a(n-1) + 3*a(n-2) + a(n-3), with a(0)=1, a(1)=1, a(3)=5. [Corrected by Jianing Song, Oct 31 2024]

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

cp's OEIS Frontend

A377109 a(n) = coefficient of the term that is independent of sqrt(2), sqrt(3), and sqrt(6) in the expansion of (2 + sqrt(2) + sqrt(3))^n.

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