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.
%I A210851 #39 Jan 15 2025 04:18:48
%S A210851 3,3,2,3,1,0,2,1,4,1,2,2,4,0,3,1,2,0,4,0,1,0,4,0,3,2,0,3,0,3,3,1,3,0,
%T A210851 3,0,2,4,3,3,1,1,2,2,0,4,0,2,0,4,1,3,2,0,4,1,1,4,1,4,4,4,1,3,1,3,3,4,
%U A210851 1,4,4,1,0,3,1,1,1,0,4,2,2,4,2,4,3,4,0,3,3,0,0,2,3,4,2,4,4,1,4,0
%N A210851 Digits of one of the two 5-adic integers sqrt(-1).
%C A210851 See A048899 for the successive approximations to this 5-adic integer, called -u in a comment on A048898.
%C A210851 The digits of u, the other 5-adic integer sqrt(-1), are given in A210850.
%C A210851 a(n) is the (unique) solution of the linear congruence 2*A048899(n)*a(n) + A210849(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)=3 follows from the formula given below.
%C A210851 If n>0, a(n) == -(A210849(n)) (mod 5), since A048899(n) == 3 (mod 5). - _Álvar Ibeas_, Feb 21 2017
%C A210851 If a(n)=0 then A048899(n+1) and A048899(n) coincide.
%C A210851 From _Jianing Song_, Sep 06 2022: (Start)
%C A210851 With a(0) = 2, 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 2 modulo 5.
%C A210851 With a(0) = 4, 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 4 modulo 5. (End)
%C A210851 This square root of -1 in the 5-adic integers is equal to the 5-adic limit of the sequence {L(5^n,3)}, where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - _Peter Bala_, Dec 02 2022
%H A210851 Robert Israel, <a href="/A210851/b210851.txt">Table of n, a(n) for n = 0..10000</a>
%H A210851 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 A210851 a(n) = (b(n+1) - b(n))/5^n, n >= 0, with b(n):=A048899(n) computed from its recurrence. A Maple program for b(n) is given there.
%F A210851 A048899(n+1) = Sum_{k=0..n} a(k)*5^k, n >= 0.
%e A210851 a(3) = 3 because 2*68*3 + 37 == 0 (mod 5).
%e A210851 A048899(4) = 443 = 3*5^0 + 3*5^1 + 2*5^2 + 3*5^3.
%e A210851 a(5) = 0 because A048899(6) = A048899(5) = 3*5^0 + 3*5^1 + 2*5^2 + 3*5^3 + 1*5^4 = 1068.
%p A210851 R:= select(t -> padic:-ratvaluep(t,1)=3,[padic:-rootp(x^2+1,5,200)]):
%p A210851 op([1,1,3],R); # _Robert Israel_, Mar 04 2016
%t A210851 Join[{3}, MapIndexed[#/5^#2[[1]] &, Differences[FoldList[PowerMod[#, 5, 5^#2] &, 3, Range[2, 100]]]]] (* _Paolo Xausa_, Jan 15 2025 *)
%o A210851 (PARI) a(n) = truncate(-sqrt(-1+O(5^(n+1))))\5^n; \\ _Michel Marcus_, Mar 05 2016
%Y A210851 Cf. A048898, A048899, A114525, A210848, A210849, A210850.
%K A210851 nonn,base,easy
%O A210851 0,1
%A A210851 _Wolfdieter Lang_, Apr 30 2012
%E A210851 Keyword "base" added by _Jianing Song_, Feb 17 2021