cp's OEIS Frontend

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

%I A144837 #35 Jul 13 2025 11:08:54
%S A144837 11,167761,132878596168524201724674011
%N A144837 a(n) = Lucas(5^n).
%C A144837 Previous name was: a(n) = round(phi^(5^n)) where phi = 1.6180339887... = (sqrt(5) + 1)/2 = A001622.
%C A144837 a(4), a 131-digit number, is too large to show here.
%H A144837 Amiram Eldar, <a href="/A144837/b144837.txt">Table of n, a(n) for n = 1..5</a>
%F A144837 a(n) = phi^(5^n) + (1-phi)^(5^n) = phi^(5^n) + (-phi)^(-5^n). - _Artur Jasinski_, Oct 05 2008
%F A144837 From _Peter Bala_, Nov 14 2022: (Start)
%F A144837 a(n) = A000032(5^n).
%F A144837 a(n) = a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1) with a(1) = 11.
%F A144837 a(n) == 1 (mod 5).
%F A144837 a(n+1) == a(n) (mod 5^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the Lucas numbers).
%F A144837 Conjecture: a(n+1) == a(n) (mod 5^(n+r+1)) for n >= r.
%F A144837 The smallest positive residue of a(n) mod(5^n) = A268922(n).
%F A144837 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)
%e A144837 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
%p A144837 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;
%p A144837 seq(a(n), n = 1..5); # _Peter Bala_, Nov 14 2022
%t A144837 Table[Round[GoldenRatio^(5^n)], {n, 1, 5}]
%t A144837 c = (1 + Sqrt[5])/2; Table[Expand[c^(5^n) + (1 - c)^(5^n)], {n, 1, 5}] (* _Artur Jasinski_, Oct 05 2008 *)
%t A144837 LucasL[5^Range[5]] (* _Harvey P. Dale_, Apr 01 2023 *)
%Y A144837 Cf. A000032, A000351, A001622, A001566, A006267, A144836, A144838, A144839, A268922, A269591.
%K A144837 nonn,easy
%O A144837 1,1
%A A144837 _Artur Jasinski_, Sep 22 2008
%E A144837 New name from _Peter Bala_, Nov 10 2022