A268924 One of the two successive approximations up to 3^n for the 3-adic integer sqrt(-2). These are the numbers congruent to 1 mod 3 (except for n = 0).

0, 1, 4, 22, 22, 22, 508, 508, 2695, 2695, 2695, 2695, 356989, 888430, 4077076, 4077076, 18425983, 18425983, 147566146, 534986635, 534986635, 7508555437, 28429261843, 28429261843, 122572440670, 405001977151

Offset: 0

Author

Wolfdieter Lang, Apr 05 2016

Keywords

• nonn
• easy

Comments

The other approximation for the 3-adic integer sqrt(-2) with numbers 2 (mod 3) is given in A271222.

For the digits of this 3-adic integer sqrt(-2), that is the scaled first differences, see A271223. This 3-adic number has the digits read from the right to the left ...2202101200022211102201101021200010200211 = u.

The companion 3-adic number has digits ...20020121022200011120021121201022212022012 = -u. See A271224.

For details see the W. Lang link ``Note on a Recurrence or Approximation Sequences of p-adic Square Roots'' given under A268922, also for the Nagell reference and Hensel lifting. Here p = 3, b = 2, x_1 = 1 and z_1 = 1.

Examples

			n=2: 4^2 + 2 = 18 == 0 (mod 3^2), and 4 is the only solution from {0, 1, ..., 8} which is congruent to 1 modulo 3.
n=3: the only solution of  X^2 + 2 == 0 (mod 3^3) with X from {0, ..., 26} and X == 1 (mod 3) is 22. The number 5 = A271222(3)  also satisfies the first congruence but not the second one: 5  == 2 (mod 3).
n=4: the only solution of X^2 + 2 == 0 (mod 3^4) with X from {0, ..., 80} and X == 1 (mod 3) is also 22. The number 59 = A271222(4) also satisfies the first congruence but not the second one: 59  == 2 (mod 3).
		

References

• Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, p. 87.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..2096
• Peter Bala, A note on A268924 and A271222, Nov 28 2022.
• Wikipedia, Hensel's Lemma.

Crossrefs

Cf. A000032, A006267, A268922, A271222, A271223, A271224.

Programs

Maple

with(padic):D1:=op(3,op([evalp(RootOf(x^2+2),3,20)][1])): 0,seq(sum('D1[k]*3^(k-1)','k'=1..n), n=1..20);
# alternative program based on the Lucas numbers L(3^n) = A006267(n)
a := proc(n) option remember; if n = 1 then 1 else irem(a(n-1)^3 + 3*a(n-1), 3^n) end if; end: seq(a(n), n = 1..22); # Peter Bala, Nov 15 2022

PARI

a(n) = truncate(sqrt(-2+O(3^(n)))); \\ Michel Marcus, Apr 09 2016

Python

def a268924(n):
    ary=[0]
    a, mod = 1, 3
    for i in range(n):
          b=a%mod
          ary.append(b)
          a=b**2 + b + 2
          mod*=3
    return ary
print(a268924(100)) # Indranil Ghosh, Aug 04 2017, after Ruby

Ruby

def A268924(n)
  ary = [0]
  a, mod = 1, 3
  n.times{
    b = a % mod
    ary << b
    a = b * b + b + 2
    mod *= 3
  }
  ary
end
p A268924(100) # Seiichi Manyama, Aug 03 2017

Formula

a(n)^2 + 2 == 0 (mod 3^n), and a(n) == 1 (mod 3). Representatives of the complete residue system {0, 1, ..., 3^n-1} are taken.

Recurrence for n >= 1: a(n) = modp(a(n-1) + a(n-1)^2 + 2, 3^n), n >= 2, with a(1) = 1. 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}.

a(n) = 3^n - A271222(n), n >= 1.

a(n) == Lucas(3^n) (mod 3^n). - Peter Bala, Nov 10 2022