A188570 -id:A188570 - 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).

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

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

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

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

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

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