A210851 Digits of one of the two 5-adic integers sqrt(-1).

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, 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, 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

Offset: 0

Author

Wolfdieter Lang, Apr 30 2012

Keywords

• nonn
• base
• easy

Comments

See A048899 for the successive approximations to this 5-adic integer, called -u in a comment on A048898.

The digits of u, the other 5-adic integer sqrt(-1), are given in A210850.

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.

If n>0, a(n) == -(A210849(n)) (mod 5), since A048899(n) == 3 (mod 5). - Álvar Ibeas, Feb 21 2017

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

From Jianing Song, Sep 06 2022: (Start)

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.

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)

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

Examples

			a(3) = 3 because 2*68*3 + 37 == 0 (mod 5).
A048899(4) = 443 = 3*5^0 + 3*5^1 + 2*5^2 + 3*5^3.
a(5) = 0 because A048899(6) = A048899(5) = 3*5^0 + 3*5^1 + 2*5^2 + 3*5^3 + 1*5^4 = 1068.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10000
• Peter Bala, Using Lucas polynomials to find the p-adic square roots of -1, -2 and -3, Dec 2022.

Crossrefs

Cf. A048898, A048899, A114525, A210848, A210849, A210850.

Programs

Maple

R:= select(t -> padic:-ratvaluep(t,1)=3,[padic:-rootp(x^2+1,5,200)]):
op([1,1,3],R); # Robert Israel, Mar 04 2016

Mathematica

Join[{3}, MapIndexed[#/5^#2[[1]] &, Differences[FoldList[PowerMod[#, 5, 5^#2] &, 3, Range[2, 100]]]]] (* Paolo Xausa, Jan 15 2025 *)

PARI

a(n) = truncate(-sqrt(-1+O(5^(n+1))))\5^n; \\ Michel Marcus, Mar 05 2016

Formula

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.

A048899(n+1) = Sum_{k=0..n} a(k)*5^k, n >= 0.

Extensions

Keyword "base" added by Jianing Song, Feb 17 2021