A377114 -id:A377114 - 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).

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.

Original entry on oeis.org

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

Clark Kimberling, Oct 21 2024

nonn

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.

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

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

Cf. A377109, A377114, A377115, A377116.

(* 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 *)

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).

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

Original entry on oeis.org

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

Clark Kimberling, Oct 23 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 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.

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

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

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

(* 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]

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).

A377115 a(n) = coefficient of sqrt(3) in the expansion of (3 + sqrt(2) + sqrt(3))^n.

Original entry on oeis.org

0, 1, 6, 36, 216, 1304, 7920, 48320, 295680, 1812672, 11124864, 68320000, 419719680, 2579051008, 15849305088, 97406521344, 598661038080, 3679444570112, 22614556631040, 138994100486144, 854291341737984, 5250689954316288, 32272093691707392, 198352703517884416

Offset: 0

Clark Kimberling, Oct 21 2024

nonn

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 4,8,10,10 and these periods:
 p = 2: (2, 1, 1, 2)
 p = 3: (6, 1, 1, 3, 1, 4, 2, 6)
 p = 5: (6, 2, 7, 3, 10, 2, 12, 3, 9, 6)
 p = 7: (14, 4, 4, 2, 12, 1, 11, 5, 1, 18)
See A377109 for a guide to related sequences.

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

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

Cf. A377090, A377113, A377114, A377116.

(* 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]];
s3  (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates this sequence. *)
LinearRecurrence[{12, -44, 48, 8}, {0, 1, 6, 36}, 25]

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)=36.

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

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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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.

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

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A377115 a(n) = coefficient of sqrt(3) in the expansion of (3 + sqrt(2) + sqrt(3))^n.

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