A144838 -id:A144838 - cp's OEIS frontend

A144837 a(n) = Lucas(5^n).

11, 167761, 132878596168524201724674011

Offset: 1

Artur Jasinski, Sep 22 2008

Previous name was: a(n) = round(phi^(5^n)) where phi = 1.6180339887... = (sqrt(5) + 1)/2 = A001622.

a(4), a 131-digit number, is too large to show here.

			The base 5 representation of a(3) = 132878596168524201724674011 begins 1 + 2*5 + 0*(5^2) + 2*(5^3) + 3*(5^4) + 0*(5^5) + 4*(5^6) + O(5^7) so A269591 begins [1, 2, 0, 2, 3, 0, 4, ...]. - _Peter Bala_, Nov 14 2022

Amiram Eldar, Table of n, a(n) for n = 1..5

Cf. A000032, A000351, A001622, A001566, A006267, A144836, A144838, A144839, A268922, A269591.

a := proc(n) option remember; if n = 1 then 11 else a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1) end if; end;
seq(a(n), n = 1..5); # Peter Bala, Nov 14 2022

Table[Round[GoldenRatio^(5^n)], {n, 1, 5}]
c = (1 + Sqrt[5])/2; Table[Expand[c^(5^n) + (1 - c)^(5^n)], {n, 1, 5}] (* Artur Jasinski, Oct 05 2008 *)
LucasL[5^Range[5]] (* Harvey P. Dale, Apr 01 2023 *)

a(n) = phi^(5^n) + (1-phi)^(5^n) = phi^(5^n) + (-phi)^(-5^n). - Artur Jasinski, Oct 05 2008
From Peter Bala, Nov 14 2022: (Start)
a(n) = A000032(5^n).
a(n) = a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1) with a(1) = 11.
a(n) == 1 (mod 5).
a(n+1) == a(n) (mod 5^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the Lucas numbers).
Conjecture: a(n+1) == a(n) (mod 5^(n+r+1)) for n >= r.
The smallest positive residue of a(n) mod(5^n) = A268922(n).
In the ring of 5-adic integers the limit_{n -> oo} a(n) exists and is equal to A269591. An example is given below. (End)

New name from Peter Bala, Nov 10 2022

A144836 a(n) = round(phi^(4^n)) where phi is the golden ratio (A001622).

Original entry on oeis.org

2, 7, 2207, 23725150497407, 316837008400094222150776738483768236006420971486980607

Offset: 0

Artur Jasinski, Sep 22 2008

Amiram Eldar, Table of n, a(n) for n = 0..6

Cf. A000032, A001566, A001622, A006267, A144837, A144838, A144839.

Cf. A374883 (phi^4).

a := proc(n) option remember; if n = 0 then 2 elif n = 1 then 7 else a(n-1)^4 - 4*a(n-1)^2 + 2 end if; end proc: seq(a(n), n = 0..4); # Peter Bala, Nov 28 2022

Table[Round[GoldenRatio^(4^n)], {n, 0, 5}]
c = (1 + Sqrt[5])/2; Join[{2}, Table[Expand[c^(4^n) + (1 - c)^(4^n)], {n, 1, 5}]] (* Artur Jasinski, Oct 05 2008 *)
Table[Round[2*Cosh[4^n*ArcCosh[Sqrt[5]/2]]], {n, 0, 5}] (* Artur Jasinski, Oct 09 2008 *)
a[n_] := LucasL[4^n]; a[0] = 2; Array[a, 5, 0] (* Amiram Eldar, Jul 12 2025 *)

a(n)=round(((1+sqrt(5))/2)^4^n) \\ Charles R Greathouse IV, Jul 29 2011

a(n) = Lucas(4^n) = A000032(4^n), n>0.
a(n) = phi^(4^n) + (1 - phi)^(4^n) = phi^(4^n) + (-phi)^(-4^n), where phi is golden ratio = (1 + sqrt(5))/2 = 1.6180339887..., n>0. - Artur Jasinski, Oct 05 2008
a(n) = 2*cosh(4^n*arccosh(sqrt(5)/2)), n>0. - Artur Jasinski, Oct 09 2008
a(n+1) = a(n)^4 - 4*a(n-1)^2 + 2 with a(1) = 7. - Peter Bala, Nov 28 2022

Offset corrected by Charles R Greathouse IV, May 15 2013
Offset changed to 0 by Georg Fischer, Sep 02 2022
New name from Peter Bala, Nov 18 2022
Revised by editors, Jul 12 2025

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

Original entry on oeis.org

29, 17393796001, 481682208844384447843365760878364816732549453120338354329505085763436029

Offset: 1

Artur Jasinski, Sep 22 2008

nonn

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

Amiram Eldar, Table of n, a(n) for n = 1..4

Cf. A000032, A001622, A001566, A006267, A144836, A144837, A144838, A144839.

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

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 *)

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)

New name from Peter Bala, Nov 28 2022

cp's OEIS Frontend

A144837 a(n) = Lucas(5^n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A144836 a(n) = round(phi^(4^n)) where phi is the golden ratio (A001622).

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions