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 1-4 of 4 results.

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

A118242 Pierce expansion of 1/phi.

Original entry on oeis.org

1, 2, 4, 17, 19, 5777, 5779, 192900153617, 192900153619, 7177905237579946589743592924684177, 7177905237579946589743592924684179
Offset: 1

Views

Author

Eric W. Weisstein, Apr 17 2006

Keywords

Comments

Differs from A006276 in the first term only.

Links

Crossrefs

Cf. A006276. A002813, A002814.

Programs

  • Mathematica
    PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[(Sqrt[5] - 1)/2, 7!], 10] (* G. C. Greubel, Nov 14 2016 *)

Formula

A118242(n) = A006276(n-1) for n>1.
From Peter Bala, Dec 03 2012: (Start)
Odd-indexed terms give A002813; even-indexed terms give A002814.
The Pierce series expansion is the alternating series 1/phi = 1/2*(sqrt(5) - 1) = 1/1 - 1/(1*2) + 1/(1*2*4) - 1/(1*2*4*17) + 1/(1*2*4*17*19) - ....
Another series expansion is
1/phi = a(1)/a(2) + (a(1)*a(3))/(a(2)*a(4)) + (a(1)*a(3)*a(5))/(a(2)*a(4)*a(6)) + ... = 1/2 + (1*4)/(2*17) + (1*4*19)/(2*17*5777) + ....
(End)
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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