

A271223


Digits of one of the two 3adic integers sqrt(2).


10



1, 1, 2, 0, 0, 2, 0, 1, 0, 0, 0, 2, 1, 2, 0, 1, 0, 1, 1, 0, 2, 2, 0, 1, 1, 1, 2, 2, 2, 0, 0, 0, 2, 1, 0, 1, 2, 0, 2, 2, 0, 2, 0, 1, 2, 0, 1, 2, 2, 2, 1, 0, 2, 0, 1, 2, 0, 2, 0, 0, 1, 1, 2, 1, 0, 1, 2, 1, 1, 2, 0
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OFFSET

0,3


COMMENTS

This is the scaled first difference sequence of A268924. See the formula.
The digits of the other 3adic integer sqrt(2), are given in A271224. See also A268924 for the two 3adic numbers sqrt(2), called there u and u.
a(n) is the unique solution of the linear congruence 2*A268924(n)*a(n) + A271225(n) == 0 (mod 3), n>=1. Therefore only the values 0, 1, and 2 appear. See the Nagell reference given in A268922, eq. (6) on p. 86, adapted to this case. a(0) = 1 follows from the formula given below.
For details see the Wolfdieter Lang link under A268992.


REFERENCES

Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, pp. 86 and 7778.


LINKS

Seiichi Manyama, Table of n, a(n) for n = 0..10000
BCMATH Congruence Programs, Finding a padic square root of a quadratic residue (mod p), p an odd prime.


FORMULA

a(n) = (b(n+1)  b(n))/3^n, n >= 0, with b(n) = A268924(n), n >= 0.
a(n) =  A271225(n)*2*A268924(n) (mod 3), n >= 1. Solution of the linear congruence given above in a comment. See, e.g., Nagell, Theorem 38 pp. 7778.
A268924(n+1) = sum(a(k)*3^k, k=0..n), n >= 0.


EXAMPLE

a(4) = 0 because 2*22*3 + 6 = 138 == 0 (mod 3).
a(4) =  6*(2*22) (mod 3) = 0*(2*1) (mod 3) = 0.
A268924(5) = 22 = 1*3^0 + 1*3^1 + 2*3^2 + 0*3^3 + 0*3^4.


PROG

(PARI) a(n) = truncate(sqrt(2+O(3^(n+1))))\3^n; \\ Michel Marcus, Apr 09 2016


CROSSREFS

Cf. A268924, A268992, A271224, A271225.
Sequence in context: A061897 A047919 A272624 * A260944 A101670 A219491
Adjacent sequences: A271220 A271221 A271222 * A271224 A271225 A271226


KEYWORD

nonn,base,easy


AUTHOR

Wolfdieter Lang, Apr 05 2016


STATUS

approved



