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 1-10 of 17 results. Next

A006267 Continued cotangent for the golden ratio.

Original entry on oeis.org

1, 4, 76, 439204, 84722519070079276, 608130213374088941214747405817720942127490792974404
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A000032, A000204, A001622, A001999, A002666, A002667, A002668, A006266, A002813, A045529, A271223, A268924.

Programs

  • Maple
    a := proc(n) option remember; if n = 1 then 4 else a(n-1)^3 + 3*a(n-1) end if; end: seq(a(n), n = 1..5); # Peter Bala, Nov 15 2022
  • Mathematica
    c = N[GoldenRatio, 1000]; Table[Round[c^(3^n)], {n, 1, 8}] (* Artur Jasinski, Sep 22 2008 *)
a = {}; x = 4; Do[AppendTo[a, x]; x = x^3 + 3 x, {n, 1, 10}]; a (* Artur Jasinski, Sep 24 2008 *)
  • PARI
    a(n)=fibonacci(3^n+1) + fibonacci(3^n-1) \\ Andrew Howroyd, Dec 30 2024
  • PARI
    a(n)={my(t=1); for(i=1, n, t = t^3 + 3*t); t} \\ Andrew Howroyd, Dec 30 2024

Formula

(1+sqrt(5))/2 = cot(Sum_{n>=0} (-1)^n*acot(a(n))); let b(0) = (1+sqrt(5))/2, b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)). - Benoit Cloitre, Apr 10 2003
a(n) = A000204(3^n). - Benoit Cloitre, Sep 18 2005
a(n) = round(c^(3^n)) where c = GoldenRatio = 1.6180339887498948482... = (sqrt(5)+1)/2 (A001622). - Artur Jasinski, Sep 22 2008
a(n) = a(n-1)^3 + 3*a(n-1), a(0) = 1. - Artur Jasinski, Sep 24 2008
a(n+1) = Product_{k = 0..n} A002813(k). Thus a(n) divides a(n+1). - Peter Bala, Nov 22 2012
Sum_{n>=0} a(n)^2/A045529(n+1) = 1. - Amiram Eldar, Jan 12 2022
a(n) = Product_{k=0..n-1} (Lucas(2*3^k) + 1) (Usiskin, 1973). - Amiram Eldar, Jan 29 2022
From Peter Bala, Nov 15 2022: (Start)
a(n) = Lucas(3^n) for n >= 1.
a(n) == 1 (mod 3) for n >= 1.
a(n+1) == a(n) (mod 3^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the Lucas numbers).
The smallest positive residue of a(n) mod 3^n = A268924(n).
In the ring of 3-adic integers the limit_{n -> oo} a(n) exists and is equal to A271223. Cf. A006266. (End)

Extensions

The next term is too large to include.

A006268 A continued cotangent.

Original entry on oeis.org

3, 36, 46764, 102266868132036, 1069559300034650646049671039050649693658764
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Continued cotangents: A006267, A006266, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189 (k = 1 to 15 with k=4 being A006267(n+1)).
Cf. A002813, A006273.

Programs

  • Mathematica
    a = {}; k = 3; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a (* Artur Jasinski, Oct 03 2008 *)
Table[Round[N[(3/2 + Sqrt[13]/2)^(3^(n - 1)), 1000]], {n, 1, 8}] (* Artur Jasinski, Oct 03 2008 *)
  • PARI
    a(n) = if (n==0, 3, a(n-1)^3 + 3*a(n-1)); \\ Michel Marcus, Aug 28 2020

Formula

From Artur Jasinski, Oct 03 2008: (Start)
a(n+1) = a(n)^3 + 3*a(n) and a(0)=3.
a(n) = round((3/2 + sqrt(13)/2)^(3^(n - 1))). (End)
From Peter Bala, Jan 19 2022: (Start)
a(n) = (3/2 + sqrt(13)/2)^(3^(n-1)) + (3/2 - sqrt(13)/2)^(3^(n-1))
a(n) divides a(n+1) and b(n) = a(n+1)/a(n) satisfies the recurrence b(n+1) = b(n)^3 - 3*b(n-1)^2 + 3. For remarks about this recurrence see A002813.
1 + a(n)^2 = A006273(n+1). (End)

A145180 Continued cotangent recurrence a(n+1) = a(n)^3 + 3*a(n) and a(1) = 6.

Original entry on oeis.org

6, 234, 12813606, 2103846732371087589834, 9311985549495522884757461748592522243432897275494229148348315206
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1) = a(n)3 + 3*a(n) and a(1)=k is following:
a(n) = Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))].
The next term (a(6)) has 192 digits. - Harvey P. Dale, Mar 09 2013

Links

Crossrefs

Cf. A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189 (k = 1 to 15 with k=4 being A006267(n+1)).
Cf. A002813.

Programs

  • Mathematica
    a = {}; k = 6; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((6 + Sqrt[40])/2)^(3^(n - 1))], {n, 1, 5}] (* Artur Jasinski *)
NestList[#^3+3#&,6,5] (* Harvey P. Dale, Mar 09 2013 *)

Formula

a(n+1)=a(n)^3 + 3*a(n) and a(1)=6
a(n)=Floor[((6+Sqrt[6^2+4])/2)^(3^(n-1))]
a(n) divides a(n+1) and b(n) = a(n+1)/a(n) satisfies the recurrence b(n+1) = b(n)^3 - 3*b(n-1)^2 + 3. See A002813. - Peter Bala, Nov 23 2012

A145189 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=15.

Original entry on oeis.org

15, 3420, 40001698260, 64008151994095341241755497070780, 262244184463346778261182615794616508638576477409715732397097802610370956164308073990185129764340
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189

Crossrefs

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

Programs

  • Mathematica
    a = {}; k = 15; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((15 + Sqrt[229])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
NestList[#^3+3#&,15,5] (* Harvey P. Dale, Aug 20 2017 *)

Formula

a(n+1)=a(n)3+3*a(n) and a(1)=14
a(n)=Floor[((14+Sqrt[14^2+4])/2)^(3^(n-1))]

A145181 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=7.

Original entry on oeis.org

7, 364, 48229636, 112186849649044142700364, 1411971263214164889494039458947084336929208169473485667118006013929636
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189

Crossrefs

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

Programs

  • Mathematica
    a = {}; k = 7; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((7 + Sqrt[53])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)

Formula

a(n+1)=a(n)^3 + 3*a(n) and a(1)=7
a(n)=Floor[((7+Sqrt[7^2+4])/2)^(3^(n-1))]

A145182 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=8.

Original entry on oeis.org

8, 536, 153992264, 3651713626720249047672536, 48695646535829720063008633136610768101443687873746944465180200686293744264
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189
The next term has 222 digits. - Harvey P. Dale, Mar 02 2018

Crossrefs

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

Programs

  • Mathematica
    a = {}; k = 7; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((8 + Sqrt[68])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
RecurrenceTable[{a[1]==8,a[n]==a[n-1]^3+3a[n-1]},a,{n,5}] (* Harvey P. Dale, Mar 02 2018 *)

Formula

a(n+1)=a(n)3+3*a(n) and a(1)=8
a(n)=Floor[((8+Sqrt[8^2+4])/2)^(3^(n-1))]

A145183 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=9.

Original entry on oeis.org

9, 756, 432083484, 80668317387203269343374356, 524939187888223206865848253384923777974930416670334526494423558665677185033084
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
For k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189

Crossrefs

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

Programs

  • Mathematica
    a = {}; k = 9; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((9 + Sqrt[85])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
NestList[#^3+3#&,9,5] (* Harvey P. Dale, Nov 02 2011 *)

Formula

a(n+1)=a(n)3+3*a(n) and a(1)=9
a(n)=Floor[((9+Sqrt[9^2+4])/2)^(3^(n-1))]

A145184 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=10.

Original entry on oeis.org

10, 1030, 1092730090, 1304784252725333839617919270, 2221345538213703371536935622204403026741331806706388823688859272519059871168740810
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189

Crossrefs

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

Programs

  • Mathematica
    a = {}; k = 10; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((10 + Sqrt[104])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)

Formula

a(n+1)=a(n)3+3*a(n) and a(1)=10
a(n)=Floor[((10+Sqrt[10^2+4])/2)^(3^(n-1))]

A145185 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=11.

Original entry on oeis.org

11, 1364, 2537720636, 16342986943522226847837781364, 4365101043708483494615466932242949707161871659736799144058331102381689400753867700636
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189

Crossrefs

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

Programs

  • Mathematica
    a = {}; k = 11; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((11 + Sqrt[125])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
RecurrenceTable[{a[1]==11,a[n]==a[n-1]^3+3*a[n-1]},a,{n,5}] (* Harvey P. Dale, Jul 23 2012 *)

Formula

a(n+1)=a(n)^3+3*a(n) and a(1)=11
a(n)=Floor[((11+Sqrt[11^2+4])/2)^(3^(n-1))]

A145186 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=12.

Original entry on oeis.org

12, 1764, 5489037036, 165382092777963331246695013764, 4523404750894779548516344022127873154658656755028228436816797201835023951822441803129036
Offset: 1

Views

Author

Artur Jasinski, Oct 03 2008

Keywords

Comments

General formula for continued cotangent recurrences type:
a(n+1)=a(n)^3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189
The next term (a(6)) has 263 digits. - Harvey P. Dale, May 21 2018

Crossrefs

Cf. A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189.

Programs

  • Mathematica
    a = {}; k = 12; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((12 + Sqrt[148])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
NestList[#^3+3#&,12,5] (* Harvey P. Dale, May 21 2018 *)

Formula

a(n+1)=a(n)3+3*a(n) and a(1)=12.
a(n)=Floor[((12+Sqrt[12^2+4])/2)^(3^(n-1))].
Showing 1-10 of 17 results. Next

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