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A269590 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-4). These are the 4 mod 5 numbers (except for n=0).

%I A269590 #26 Nov 21 2022 09:29:22
%S A269590 0,4,14,114,364,989,13489,13489,169739,560364,2513489,2513489,2513489,
%T A269590 246654114,3908763489,22219310364,52736888489,52736888489,
%U A269590 3104494700989,6919191966614
%N A269590 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-4). These are the 4 mod 5 numbers (except for n=0).
%C A269590 The other approximation for the 5-adic integer sqrt(-4) with numbers 1 (mod 5) is given in A268922. See this also for more details and references.
%C A269590 For the digits of this 5-adic integer sqrt(-4), that is the scaled first differences, see A269592. See also A268922 for the 5-adic numbers u and -u written from the right to the left.
%H A269590 Seiichi Manyama, <a href="/A269590/b269590.txt">Table of n, a(n) for n = 0..1432</a>
%F A269590 Recurrence for n >= 1: a(n) = modp( a(n-1) + 3*(a(n-1)^2 + 4), 5^n), n >= 2, with a(1) = 4. Here modp(a, m) is used to pick the representative of the residue class a modulo m from the smallest nonnegative complete residue system {0, 1, ... , m-1}.
%F A269590 a(n) = 5^n - A268922(n), n >= 1.
%F A269590 a(n) == Lucas(3*(5^n)) (mod 5^n). - _Peter Bala_, Nov 14 2022
%p A269590 with(padic): D2:=op(3,op([evalp(RootOf(x^2+4),5,20)][2])):
%p A269590 0,seq(sum('D2[k]*5^(k-1)','k'=1..n),n=1..20);
%p A269590 # alternative program
%p A269590 a := proc(n) option remember; if n = 1 then 4 else irem( a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1), 5^n) end if; end: seq(a(n), n = 1..20); # _Peter Bala_, Nov 14 2022
%o A269590 (PARI) a(n) = if (n==0, 0, 5^n - truncate(sqrt(-4+O(5^(n))))); \\ _Michel Marcus_, Mar 07 2016
%Y A269590 Cf. A048899, A048898, A269591, A268922, A269592.
%K A269590 nonn,easy
%O A269590 0,2
%A A269590 _Wolfdieter Lang_, Mar 02 2016