A377109 -id:A377109 - cp's OEIS frontend

A377117 a(n) = coefficient of the term that is independent of 2^(1/3) and 2^(2/3) in the expansion of (2^(1/3) + 2^(2/3))^n.

1, 1, 4, 6, 24, 60, 180, 504, 1440, 4104, 11664, 33264, 94608, 269568, 767232, 2185056, 6220800, 17713728, 50435136, 143607168, 408893184, 1164253824, 3315002112, 9438882048, 26875535616, 76523304960, 217886505984, 620393043456, 1766458865664, 5029677296640

Offset: 0

Clark Kimberling, Oct 24 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 6 primes, with respective period lengths 1,5,3,17,2,13 and these periods:
p = 2: (1)
p = 3: (1, 2, 8, 9, 4)
p = 5: (21, 9, 10)
p = 7: (8, 42, 7, 19, 1, 2, 10, 31, 4, 11, 6, 34, 60, 23, 5, 9, 16)
p = 11: (9, 21)
p = 13: (15, 51, 45, 1, 2, 17, 41, 28, 54, 119, 13, 9, 25)
See A377109 for a guide to related sequences.

			((2^(1/3) + 2^(2/3)))^3 = 4 + 2 2^(1/3) + 2^(2/3), so a(3) = 4.

Index entries for linear recurrences with constant coefficients, signature (0,6,6).

Cf. A377109, A377118, A377119.

(* Program 1 generates sequences A377117-A377119. *)
tbl = Table[Expand[(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[{0,6,6}, {1, 1, 4, 6, 24}, 15]
(* Program 3 confirms the periodicity properties described in Comments. *)
tbl = Table[Expand[(2^(1/3) + 2^(2/3))^n], {n, 0, 1000}];
u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
   Map[({#1, #1 /. ^ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
v = {s1, s2, s3} = Transpose[(PadRight[#1, 3] &) /@ Last /@ u][[1 ;; 3]];
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) = 6*a(n-2) + 6*a(n-3) for n >= 5. [Corrected by Georg Fischer, Oct 29 2024]

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

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).

A188570 a(n) = coefficient of the term that is independent of sqrt(2) and sqrt(3) in the expansion of (1 + sqrt(2) + sqrt(3))^n.

Original entry on oeis.org

1, 1, 6, 16, 80, 296, 1296, 5216, 21952, 90304, 375936, 1555456, 6456320, 26754560, 110963712, 460015616, 1907494912, 7908659200, 32792076288, 135963148288, 563742310400, 2337417887744, 9691567030272, 40183767891968, 166612591968256, 690819710058496

Offset: 0

Mateusz Szymański, Dec 28 2012

nonn

From Clark Kimberling, Oct 23 2024: (Start)
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 6 primes, with respective period lengths 1,5,5,3,3,5 and these periods:
p = 2: (4)
p = 3: (8, 1, 4, 3, 8)
p = 5: (9, 10, 1, 20, 20)
p = 7: (9, 9, 18)
p = 11: (10, 11, 43)
p = 13: (7, 21, 21, 7, 28)
See A377109 for a guide to related sequences. (End)

			a(3) = 16 because (1+sqrt(2)+sqrt(3))^3 = 16 + 14*sqrt(2) + 12*sqrt(3) + 6*sqrt(6).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (4,4,-16,8).

Cf. A188571, A188572, A188573, A377109.

a[n_] := Sum[Sum[2^(Floor[n/2] - k - j) 3^j Multinomial[2 k + n - 2 Floor[n/2], 2 j, 2 Floor[n/2] - 2 k - 2 j], {j, 0, Floor[n/2] - k}], {k, 0, Floor[n/2]}]; Table[a[n], {n, 0, 25}]
a[n_] := Expand[(1 + Sqrt[2] + Sqrt[3])^n] /. Sqrt[] -> 0; Table[a[n], {n, 0, 25}] (* _Jean-François Alcover, Jan 08 2013 *)
LinearRecurrence[{4,4,-16,8},{1,1,6,16},30] (* Harvey P. Dale, Jan 25 2019 *)

Recurrence: a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3) + 8*a(n-4). - Vaclav Kotesovec, Aug 13 2013

a(n) ~ (1+sqrt(3)+sqrt(2))^n/4. - Vaclav Kotesovec, Aug 13 2013

Edited by Clark Kimberling, Oct 20 2024

A188571 a(n) = coefficient of sqrt(2) in the expansion of (1 + sqrt(2) + sqrt(3))^n.

Original entry on oeis.org

0, 1, 2, 14, 48, 224, 880, 3760, 15360, 64192, 265088, 1101440, 4561920, 18925568, 78447616, 325313536, 1348730880, 5592420352, 23187169280, 96141172736, 398624489472, 1652807303168, 6852965761024, 28414229807104, 117812861337600, 488483370827776

Offset: 0

Mateusz Szymański, Dec 28 2012

From Clark Kimberling, Oct 23 2024: (Start)
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 6 primes, with respective period lengths 4,8,14,3,3,5 and these periods:
p = 2: (4)
p = 3: (6, 1, 1, 3, 1, 4, 2, 6)
p = 5: (6, 6, 6, 2, 4, 6, 1, 5, 4, 1, 1, 6, 6, 6)
p = 7: (9, 18, 9)
p = 11: (43, 10, 11)
p = 13: (21, 7, 28, 7, 21)
See A377109 for a guide to related sequences. (End)

			a(3) = 14 because (1+sqrt(2)+sqrt(3))^3 = 16 + 14*sqrt(2) + 12*sqrt(3) + 6*sqrt(6).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Sela Fried, On the coefficients of (r + sqrt(p) + sqrt(q))^n
Index entries for linear recurrences with constant coefficients, signature (4,4,-16,8).

Cf. A188570, A377109.

a[n_] := Sum[Sum[2^(Floor[(n - 1)/2] - k - j) 3^j Multinomial[2 Floor[(n - 1)/2] + 1 - 2 j - 2 k, 2 j, 2 k + 1 - n + 2 Floor[n/2]], {j, 0, Floor[(n - 1)/2] - k + 1}], {k, 0, Floor[(n - 1)/2]}]; Table[a[n], {n, 0, 25}]
a[n_] := Coefficient[ Expand[(1 + Sqrt[2] + Sqrt[3])^n], Sqrt[2]] /. Sqrt[3] -> 0; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jan 08 2013 *)

Conjectures from Colin Barker, Jan 08 2013: (Start)
a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3) + 8*a(n-4).
G.f.: -x*(2*x^2-2*x+1) / (8*x^4-16*x^3+4*x^2+4*x-1). (End)
The conjectures by Barker are true. See link. - Sela Fried, Jan 01 2025

Edited by Clark Kimberling, Oct 20 2024

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).

A188572 a(n) = coefficient of sqrt(3) in the expansion of (1 + sqrt(2) + sqrt(3))^n sequence.

Original entry on oeis.org

0, 1, 2, 12, 40, 184, 720, 3072, 12544, 52416, 216448, 899328, 3724800, 15452672, 64052224, 265617408, 1101234176, 4566192128, 18932244480, 78498938880, 325475532800, 1349511512064, 5595423113216, 23200121487360, 96193798471680, 398845002121216

Offset: 0

Mateusz Szymański, Dec 28 2012

nonn

From Clark Kimberling, Oct 23 2024: (Start)
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 6 primes, with respective period lengths 6,8,5,4,4,8 and these periods:
p = 2: (3, 3, 1, 2, 2, 1)
p = 3: (4, 2, 6, 6, 1, 1, 3, 1)
p = 5: (20, 20, 9, 10, 1)
p = 7: (18, 1, 16, 1)
p = 11: (32, 1, 30, 1)
p = 13: (28, 14, 1, 10, 3, 17, 10, 1)
See A377109 for a guide to related sequences. (End)
Cf. A377109.

			a(3) = 12 because (1+sqrt(2)+sqrt(3))^3 = 16 + 14*sqrt(2) + 12*sqrt(3) + 6*sqrt(6).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)
Sela Fried, On the coefficients of (r + sqrt(p) + sqrt(q))^n
Index entries for linear recurrences with constant coefficients, signature (4,4,-16,8).

Cf. A188570, A188571, A188573, A377109.

a[n_] := Sum[Sum[3^(Floor[(n - 1)/2] - k - j) 2^j Multinomial[2 Floor[(n - 1)/2] + 1 - 2 j - 2 k, 2 j, 2 k + 1 - n + 2 Floor[n/2]], {j, 0, Floor[(n - 1)/2] - k + 1}], {k, 0, Floor[(n - 1)/2]}]; Table[a[n], {n, 0, 25}]
a[n_] := Coefficient[ Expand[(1 + Sqrt[2] + Sqrt[3])^n], Sqrt[3]] /. Sqrt[2] -> 0; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jan 08 2013 *)

Conjectures from R. J. Mathar, Jan 09 2013: (Start)
a(n) = +4*a(n-1) +4*a(n-2) -16*a(n-3) +8*a(n-4).
G.f.: x*(-1+2*x)/( -1+4*x+4*x^2-16*x^3+8*x^4 ). (End)
The conjectures by Mathar are true. See link. - Sela Fried, Jan 01 2025

Edited by Clark Kimberling, Oct 20 2024

A188573 a(n) = coefficient of the sqrt(6) term in (1 + sqrt(2) + sqrt(3))^n.

Original entry on oeis.org

0, 0, 2, 6, 32, 120, 528, 2128, 8960, 36864, 153472, 635008, 2635776, 10922496, 45300736, 187800576, 778731520, 3228696576, 13387309056, 55506722816, 230146834432, 954246856704, 3956565671936, 16404954546176, 68019305840640, 282025965649920

Offset: 0

Mateusz Szymański, Dec 28 2012

nonn

From Clark Kimberling, Oct 23 2024: (Start)
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 6 primes, with respective period lengths 6,5,14,4,4,8 and these periods:
p = 2: (1, 2, 2, 1, 3, 3)
p = 3: (1, 4, 3, 8, 8)
p = 5: (1, 5, 4, 1, 1, 6, 6, 6, 6, 6, 6, 2, 4, 6)
p = 7: (1, 16, 1, 18)
p = 11: (1, 30, 1, 32)
p = 13: (1, 10, 3, 17, 10, 1, 28, 14)
See A377109 for a guide to related sequences. (End)

			a(3) = 6, because (1+sqrt(2)+sqrt(3))^3 = 16 + 14 sqrt(2) + 12 sqrt(3) + 6 sqrt(6).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)
Sela Fried, On the coefficients of (r + sqrt(p) + sqrt(q))^n
Index entries for linear recurrences with constant coefficients, signature (4,4,-16,8).

Cf. A188570, A188571, A188572, A377109.

a[n_] := Sum[Sum[2^(Floor[n/2] - j - 1 - k) 3^j Multinomial[2 k + n - 2 Floor[n/2], 2 j + 1, 2 Floor[n/2] - 2 k - 1 - 2 j], {j, 0, Floor[n/2] - k - 1}], {k, 0, Floor[n/2] - 1}]; Table[a[n], {n, 0, 25}]
a[n_] := Coefficient[ Expand[(1 + Sqrt[2] + Sqrt[3])^n], Sqrt[6]]; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jan 08 2013 *)

From G. C. Greubel, Apr 10 2018: (Start)
Empirical: a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3) + 8*a(n-4).
Empirical: G.f.: 2*x^2*(1-x)/(1 - 4*x - 4*x^2 + 16*x^3 - 8*x^4). (End)
The conjectures by Greubel are true. See link. - Sela Fried, Jan 01 2025

Keyword tabl removed by Michel Marcus, Apr 11 2018

Edited by Clark Kimberling, Oct 23 2024

A377112 a(n) = coefficient of sqrt(6) in the expansion of (2 + sqrt(2) + sqrt(3))^n.

Original entry on oeis.org

0, 0, 2, 12, 68, 360, 1878, 9716, 50120, 258192, 1329322, 6842396, 35215884, 181237368, 932711806, 4800019332, 24702255760, 127124540448, 654216959826, 3366774510892, 17326314208468, 89165799266952, 458870789205926, 2361470423992852, 12152751175719000

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 6, 5, 18, 28, 16, 14, 35:
   p = 2: (1, 2, 2, 1, 3, 3),
   p = 3: (1, 4, 3, 8, 8).
See A377109 for a guide to related sequences.

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

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

Cf. A377109.

(* 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]];
s4  (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates this sequence. *)
LinearRecurrence[{8, -14, -8, 23}, {0, 0, 2, 12}, 25]

a(n) = 8*a(n-1) - 14*a(n-2) - 8*a(n-3) + 23*a(n-4), with a(0)=0, a(1)=0, a(3)=2, a(4)=12.

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

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

Original entry on oeis.org

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

Clark Kimberling, Oct 21 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 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.

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

G.f.: x*(-1 + 6*x - 10*x^2)/(-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

A377117 a(n) = coefficient of the term that is independent of 2^(1/3) and 2^(2/3) in the expansion of (2^(1/3) + 2^(2/3))^n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A188570 a(n) = coefficient of the term that is independent of sqrt(2) and sqrt(3) in the expansion of (1 + sqrt(2) + sqrt(3))^n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A188571 a(n) = coefficient of sqrt(2) in the expansion of (1 + sqrt(2) + sqrt(3))^n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A188572 a(n) = coefficient of sqrt(3) in the expansion of (1 + sqrt(2) + sqrt(3))^n sequence.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A188573 a(n) = coefficient of the sqrt(6) term in (1 + sqrt(2) + sqrt(3))^n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica