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.

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A144838 a(n) = Lucas(6^n).

Original entry on oeis.org

18, 33385282, 1384619022984618483717737087933569992335566082
Offset: 1

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Author

Artur Jasinski, Sep 22 2008

Keywords

Comments

Previous name was: a(n) = round(phi^(6^n)) where phi = 1.6180339887498948482... = (sqrt(5)+1)/2.
General (hyperbolic) trigonometric formula for a(n) = round(phi^((2*k)^n)) = 2*cosh((2*k)^n*arccosh(sqrt(5)/2)) where phi = 1.6180339887498948482... = (sqrt(5)+1)/2. - Artur Jasinski, Oct 09 2008

Crossrefs

Programs

  • Maple
    a := proc(n) option remember; if n = 1 then 18 else a(n-1)^6 - 6*a(n-1)^4 + 9*a(n-1)^2 - 2 end if; end;
    seq(a(n), n = 1..5); # Peter Bala, Nov 28 2022
  • Mathematica
    Table[Round[GoldenRatio^(6^n)], {n, 1, 5}]
    c = (1 + Sqrt[5])/2; Table[Expand[c^(6^n) + (1 - c)^(6^n)], {n, 1, 5}] (* Artur Jasinski, Oct 05 2008 *)
    Table[Round[2*Cosh[6^n*ArcCosh[Sqrt[5]/2]]], {n, 1, 4}] (* Artur Jasinski, Oct 09 2008 *)
    Table[LucasL[6^n], {n, 1, 4}] (* Amiram Eldar, Jul 13 2025 *)

Formula

a(n) = G^(6^n) + (1 - G)^(6^n) = G^(6^n) + (-G)^(-6^n) where G is the golden ratio A001622. - Artur Jasinski, Oct 05 2008
a(n) = 2*cosh(6^n*arccosh(sqrt(5)/2)). - Artur Jasinski, Oct 09 2008
From Peter Bala, Nov 28 2022: (Start)
a(n) = Lucas(6^n).
a(n+1) = a(n)^6 - 6*a(n)^4 + 9*a(n)^2 - 2 with a(1) = 18. (End)

Extensions

New name from Peter Bala, Nov 28 2022

A144839 a(n) = Lucas(7^n).

Original entry on oeis.org

29, 17393796001, 481682208844384447843365760878364816732549453120338354329505085763436029
Offset: 1

Views

Author

Artur Jasinski, Sep 22 2008

Keywords

Comments

Previous name was: a(n) = round(phi^(7^n)) where phi = 1.6180339887498948482... = (sqrt(5)+1)/2.

Crossrefs

Programs

  • Maple
    a := proc(n) option remember; if n = 0 then 1 else a(n-1)^7 + 7*a(n-1)^5 + 14*a(n-1)^3 +7*a(n-1) end if; end;
    seq(a(n), n = 1..5); # Peter Bala, Nov 28 2022
  • Mathematica
    Table[Round[GoldenRatio^(7^n)], {n, 1, 5}]
    c = (1 + Sqrt[5])/2; Table[Expand[c^(7^n) + (1 - c)^(7^n)], {n, 1, 5}] (* Artur Jasinski, Oct 05 2008 *)
    Table[LucasL[7^n], {n, 1, 4}] (* Amiram Eldar, Jul 13 2025 *)

Formula

a(n) = G^(7^n) + (1 - G)^(7^n) = G^(7^n) + (-G)^(-7^n) where G is the golden ratio A001622. [Artur Jasinski, Oct 05 2008]
From Peter Bala, Nov 28 2022: (Start)
a(n) = Lucas(7^n).
a(n+1) = a(n)^7 + 7*a(n)^5 + 14*a(n)^3 + 7*a(n) with a(0) = 1.
a(n) == 1 (mod 7).
a(n+1) == a(n) (mod 7^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the Lucas numbers).
In the ring of 7-adic integers, the limit_{n -> oo} a(n) exists and is a root of the quartic equation x^4 + 4*x^2 + 2 = 0. (End)

Extensions

New name from Peter Bala, Nov 28 2022

A002000 a(n+1) = a(n)*(a(n)^2 - 3) with a(0) = 7.

Original entry on oeis.org

7, 322, 33385282, 37210469265847998489922, 51522323599677629496737990329528638956583548304378053615581043535682
Offset: 0

Views

Author

Keywords

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    [n eq 1 select 7 else Self(n-1)^3 - 3*Self(n-1): n in [1..6]]; // Vincenzo Librandi, Feb 09 2017
  • Maple
    a := proc(n) option remember; if n = 0 then 7 else a(n-1)^3 - 3*a(n-1) end if; end;
    seq(a(n), n = 0..4); # Peter Bala, Nov 15 2022
  • Mathematica
    RecurrenceTable[{a[0] == 7, a[n] == a[n - 1]^3 - 3 a[n - 1]}, a, {n, 0, 8}]
    (* Vincenzo Librandi, Feb 09 2017 *)
    NestList[#(#^2-3)&,7,4] (* Harvey P. Dale, Aug 11 2021 *)

Formula

From Peter Bala, Feb 01 2017: (Start)
a(n) = ((7 + sqrt(45))/2)^(3^n) + ((7 - sqrt(45))/2)^(3^n).
a(n) = 2*T(3^n,7/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
Product_{n >= 0} (1 + 2/(a(n) - 1)) = 3*sqrt(5)/5.
Cf. A001999 and A219161. (End)
From Peter Bala, Nov 15 2022: (Start)
a(n) = Lucas(4*(3^n)).
a(n+1) == a(n) (mod 3^(n+1)) (a particular case of the Gauss congruences for the Lucas numbers).
Conjecture: a(n+1) == a(n) (mod 3^(n+r+2)) for n >= r.
The least positive residue of a(n) mod(3^n) = 3^n - 2 = A058481(n). In the ring of 3-adic integers the limit_{n -> oo} a(n) exists and is equal to -2.
Product_{k = 0..n} (a(k) - 1) = (1/3)*Lucas(6*(3^n)). (End)
Previous Showing 21-23 of 23 results.