A210850 - OEIS

%I #40 Mar 16 2023 06:00:39

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

%C See A048898 for the successive approximations to this 5-adic integer, called there u.

%C The digits of -u, the other 5-adic integer sqrt(-1), are given in A210851.

%C 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 If n>0, a(n) == A210848(n) (mod 5), since A048898(n) == 2 (mod 5). - _Álvar Ibeas_, Feb 21 2017

%C If a(n)=0 then A048899(n+1) and A048899(n) coincide.

%C a(n) + A210851(n) = 4 for n >= 1. - _Robert Israel_, Mar 04 2016

%C From _Jianing Song_, Sep 06 2022: (Start)

%C 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 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 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 Robert Israel, <a href="/A210850/b210850.txt">Table of n, a(n) for n = 0..10000</a>

%H 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 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 A048898(n+1) = Sum_{k=0..n} a(k)*5^k, n >= 0.

%e a(4) = 3 because 2*182*3 + 53 = 1145 == 0 (mod 5).

%e A048898(5) = 2057 = 2*5^0 + 1*5^1 + 2*5^2 + 1*5^3 + 3*5^4.

%e a(8) = 0, therefore A048898(9) = A048898(8) = Sum_{k=0..7} a(k)*5^k = 280182.

%p R:= select(t -> padic:-ratvaluep(t,1)=2,[padic:-rootp(x^2+1,5,10001)]):

%p op([1,1,3],R); # _Robert Israel_, Mar 04 2016

%t 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 (PARI) a(n) = truncate(sqrt(-1+O(5^(n+1))))\5^n; \\ _Michel Marcus_, Mar 05 2016

%Y Cf. A048898, A048899, A114525, A210848, A210849, A210851.

%K nonn,base,easy

%O 0,1

%A _Wolfdieter Lang_, Apr 30 2012

%E Keyword "base" added by _Jianing Song_, Feb 17 2021