

A268924


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


11



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
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OFFSET

0,3


COMMENTS

The other approximation for the 3adic integer sqrt(2) with numbers 2 (mod 3) is given in A271222.
For the digits of this 3adic integer sqrt(2), that is the scaled first differences, see A271223. This 3adic number has the digits read from the right to the left ...2202101200022211102201101021200010200211 = u.
The companion 3adic number has digits ...20020121022200011120021121201022212022012 = u. See A271224.
For details see the W. Lang link ``Note on a Recurrence or Approximation Sequences of padic 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.


REFERENCES

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


LINKS



FORMULA

a(n)^2 + 2 == 0 (mod 3^n), and a(n) == 1 (mod 3). Representatives of the complete residue system {0, 1, ..., 3^n1} are taken.
Recurrence for n >= 1: a(n) = modp(a(n1) + a(n1)^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, ..., m1}.
a(n) == Lucas(3^n) (mod 3^n).  Peter Bala, Nov 10 2022


EXAMPLE

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).


MAPLE

with(padic):D1:=op(3, op([evalp(RootOf(x^2+2), 3, 20)][1])): 0, seq(sum('D1[k]*3^(k1)', '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(n1)^3 + 3*a(n1), 3^n) end if; end: seq(a(n), n = 1..22); # Peter Bala, Nov 15 2022


PROG

(PARI) a(n) = truncate(sqrt(2+O(3^(n)))); \\ Michel Marcus, Apr 09 2016
(Ruby)
ary = [0]
a, mod = 1, 3
n.times{
b = a % mod
ary << b
a = b * b + b + 2
mod *= 3
}
ary
end
(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


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



