cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Oct 20 2024

Keywords

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 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

Examples

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

Links

Crossrefs

Cf. A188570, A377110, A377111, A377112.

Programs

  • Mathematica
    (* 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 *)

Formula

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).

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.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Oct 24 2024

Keywords

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 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.

Examples

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

Links

Crossrefs

Cf. A377109, A377118, A377119.

Programs

  • Mathematica
    (* 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 *)

Formula

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).

A377119 a(n) = coefficient of 2^(2/3) in the expansion of (2^(1/3) + 2^(2/3))^n.

Original entry on oeis.org

0, 1, 1, 6, 12, 42, 108, 324, 900, 2592, 7344, 20952, 59616, 169776, 483408, 1376352, 3919104, 11158560, 31772736, 90465984, 257587776, 733432320, 2088322560, 5946120576, 16930529280, 48206658816, 137259899136, 390823128576, 1112799347712, 3168498166272
Offset: 0

Views

Author

Clark Kimberling, Oct 26 2024

Keywords

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 6 primes, with respective period lengths 2,2,4,5,2,5 and these periods:
p = 2: (2,1)
p = 3: (7, 1)
p = 5: (21, 5, 13, 1)
p = 7: (15, 17, 20, 43, 1)
p = 11: (9, 1)
p = 13: (102, 5, 17, 15, 1)
See A377109 for a guide to related sequences.

Examples

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

Links

Crossrefs

Cf. A377109, A377117, A377118.

Programs

  • Mathematica
    (* Program 1 generates sequences A377117-A377119. *)
tbl = Table[Expand[(2^(1/3) + 2^(2/3))^n], {n, 0, 30}];
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]];
s3   (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates (a(n)) for n>=1. *)
LinearRecurrence[{0,6,6}, {0,1,1}, 30]

Formula

a(n) = 6*a(n-2) + 6*a(n-3) for n>=1, with a(0)=0, a(1)=1, a(3)=1.
G.f.: x*(1 + x)/(1 - 6*x^2 - 6*x^3).
Showing 1-3 of 3 results.

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