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A210850 Digits of one of the two 5-adic integers sqrt(-1).

%I A210850 #42 Jan 08 2025 10:43:55
%S A210850 2,1,2,1,3,4,2,3,0,3,2,2,0,4,1,3,2,4,0,4,3,4,0,4,1,2,4,1,4,1,1,3,1,4,
%T A210850 1,4,2,0,1,1,3,3,2,2,4,0,4,2,4,0,3,1,2,4,0,3,3,0,3,0,0,0,3,1,3,1,1,0,
%U A210850 3,0,0,3,4,1,3,3,3,4,0,2,2,0,2,0,1,0,4,1,1,4,4,2,1,0,2,0,0,3,0,4
%N A210850 Digits of one of the two 5-adic integers sqrt(-1).
%C A210850 See A048898 for the successive approximations to this 5-adic integer, called there u.
%C A210850 The digits of -u, the other 5-adic integer sqrt(-1), are given in A210851.
%C A210850 a(n) is the unique solution of the linear congruence 2*A048898(n)*a(n) + A210848(n) == 0 (mod 5), n>=1. Therefore only the values 0, 1, 2, 3 and 4 appear. See the Nagell reference given in A210848, eq. (6) on p. 86 adapted to this case. a(0)=2 follows from the formula given below.
%C A210850 If n>0, a(n) == A210848(n) (mod 5), since A048898(n) == 2 (mod 5). - _Álvar Ibeas_, Feb 21 2017
%C A210850 If a(n)=0 then A048899(n+1) and A048899(n) coincide.
%C A210850 a(n) + A210851(n) = 4 for n >= 1. - _Robert Israel_, Mar 04 2016
%C A210850 From _Jianing Song_, Sep 06 2022: (Start)
%C A210850 With a(0) = 1, this is the digits of one of the four 4th root of -4 in the ring of 5-adic integers, the one that is congruent to 1 modulo 5.
%C A210850 With a(0) = 3, this is the digits of one of the four 4th root of -4 in the ring of 5-adic integers, the one that is congruent to 3 modulo 5. (End)
%C A210850 This square root of -1 in the 5-adic integers is equal to the 5-adic limit of the sequence {L(5^n,2)}, where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - _Peter Bala_, Dec 02 2022
%H A210850 Robert Israel, <a href="/A210850/b210850.txt">Table of n, a(n) for n = 0..10000</a>
%H A210850 Peter Bala, <a href="/A210850/a210850.pdf">Using Lucas polynomials to find the p-adic square roots of -1, -2 and -3</a>, Dec 2022.
%F A210850 a(n) = (b(n+1) - b(n))/5^n, n >= 0, with b(n):=A048898(n) computed from its recurrence. A Maple program for b(n) is given there.
%F A210850 A048898(n+1) = Sum_{k=0..n} a(k)*5^k, n >= 0.
%e A210850 a(4) = 3 because 2*182*3 + 53 = 1145 == 0 (mod 5).
%e A210850 A048898(5) = 2057 = 2*5^0 + 1*5^1 + 2*5^2 + 1*5^3 + 3*5^4.
%e A210850 a(8) = 0, therefore A048898(9) = A048898(8) = Sum_{k=0..7} a(k)*5^k = 280182.
%p A210850 R:= select(t -> padic:-ratvaluep(t,1)=2,[padic:-rootp(x^2+1,5,10001)]):
%p A210850 op([1,1,3],R); # _Robert Israel_, Mar 04 2016
%t A210850 Table[Floor[First@Select[PowerModList[-1,1/2,5^(k+1)],Mod[#,5]==2&]/5^k],{k,0,99}] (* _Giorgos Kalogeropoulos_, Feb 28 2023 *)
%o A210850 (PARI) a(n) = truncate(sqrt(-1+O(5^(n+1))))\5^n; \\ _Michel Marcus_, Mar 05 2016
%Y A210850 Cf. A048898, A048899, A114525, A210848, A210849, A210851.
%K A210850 nonn,base,easy
%O A210850 0,1
%A A210850 _Wolfdieter Lang_, Apr 30 2012
%E A210850 Keyword "base" added by _Jianing Song_, Feb 17 2021