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A271222
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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).
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9
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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
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OFFSET
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0,2
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COMMENTS
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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.
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REFERENCES
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Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, p. 87.
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LINKS
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FORMULA
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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}.
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EXAMPLE
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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).
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MAPLE
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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);
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PROG
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(PARI) a(n) = if (n, 3^n - truncate(sqrt(-2+O(3^(n)))), 0); \\ Michel Marcus, Apr 09 2016
(Ruby)
ary = [0]
a, mod = 2, 3
n.times{
b = a % mod
ary << b
a = 2 * b * b + b + 4
mod *= 3
}
ary
end
(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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