A271222 One of the two successive approximations up to 3^n for the 3-adic integer sqrt(-2). These are the numbers congruent to 2 mod 3 (except for the initial 0).

0, 2, 5, 5, 59, 221, 221, 1679, 3866, 16988, 56354, 174452, 174452, 705893, 705893, 10271831, 24620738, 110714180, 239854343, 627274832, 2951797766, 2951797766, 2951797766, 65713916984, 159857095811, 442286632292

Offset: 0

Author

Wolfdieter Lang, Apr 05 2016

Keywords

• nonn
• easy

Comments

The other approximation for the 3-adic integer sqrt(-2) with numbers 1 (mod 3) is given in A268924.

For the digits of this 3-adic integer sqrt(-2), that is the scaled first differences, see A271224. This 3-adic number has the digits read from the right to the left ... 20020121022200011120021121201022212022012 = -u. For the digits of u see A271223.

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_2 = 2 and z_2 = 2.

Examples

			n=2: 5^2 + 2 = 27 == 0 (mod 3^2), and 5 is the only solution from {0, 1, ..., 8} which is congruent to 2 modulo 3.
n=3: the only solution of X^2 + 2 == 0 (mod 3^3) with X from {0, ..., 26} and X == 2(mod 3) is 5. The number 22 = A268924(3) also satisfies the first congruence but not the second one: 22  == 1 (mod 3).
n=4: the only solution of X^2 + 2 == 0 (mod 3^4) with X from {0, ..., 80} and X == 2 (mod 3) is 59. The number 22 = A268924(4) also satisfies the first congruence but not the second one: 59  == 1 (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..2095
• Peter Bala, A note on A268924 and A271222, Nov 28 2022.
• Wikipedia, Hensel's Lemma.

Crossrefs

Cf. A002293, A268922, A268924, A271223, A271224.

Programs

Maple

with(padic):D2:=op(3,op([evalp(RootOf(x^2+2),3,20)][2])): 0,seq(sum('D2[k]*3^(k-1)','k'=1..n), n=1..20);

PARI

a(n) = if (n, 3^n - truncate(sqrt(-2+O(3^(n)))), 0); \\ Michel Marcus, Apr 09 2016

Python

def a271222(n):
      ary=[0]
      a, mod = 2, 3
      for i in range(n):
          b=a%mod
          ary.append(b)
          a=2*b**2 + b + 4
          mod*=3
      return ary
print(a271222(100)) # Indranil Ghosh, Aug 04 2017, after Ruby

Ruby

def A271222(n)
  ary = [0]
  a, mod = 2, 3
  n.times{
    b = a % mod
    ary << b
    a = 2 * b * b + b + 4
    mod *= 3
  }
  ary
end
p A271222(100) # Seiichi Manyama, Aug 03 2017

Formula

a(n)^2 + 2 == 0 (mod 3^n), and a(n) == 2 (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) + 2*(a(n-1)^2 + 2), 3^n), n >= 2, with a(1) = 2. 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 - A268924(n), n >= 1.

a(n) == A002203(3^n) (mod 3^n). - Peter Bala, Nov 10 2022