This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A271224 #17 Nov 29 2022 11:53:05 %S A271224 2,1,0,2,2,0,2,1,2,2,2,0,1,0,2,1,2,1,1,2,0,0,2,1,1,1,0,0,0,2,2,2,0,1, %T A271224 2,1,0,2,0,0,2,0,2,1,0,2,1,0,0,0,1,2,0,2,1,0,2,0,2,2,1 %N A271224 Digits of one of the two 3-adic integers sqrt(-2). Here the sequence with first digit 2. %C A271224 This is the scaled first difference sequence of A271222. See the formula. %C A271224 The digits of the other 3-adic integer sqrt(-2), are given in A271223. See also a comment on A268924 for the two 3-adic numbers sqrt(-2), called there u and -u. %C A271224 a(n) is the unique solution of the linear congruence 2*A271222(n)*a(n) + A271226(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. %C A271224 a(0) = 2 follows from the formula given below. %C A271224 For details see the Wolfdieter Lang link under A268992. %C A271224 The first k digits in the base 3 representation of A002203(3^k) = A006266(k) give the first k terms of the sequence. - _Peter Bala_, Nov 26 2022 %D A271224 Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, pp. 86 and 77-78. %H A271224 Seiichi Manyama, <a href="/A271224/b271224.txt">Table of n, a(n) for n = 0..10000</a> %H A271224 Peter Bala, <a href="/A268924/a268924.pdf">A note on A268924 and A271222</a>, Nov 28 2022. %H A271224 BCMATH Congruence Programs, <a href="http://www.numbertheory.org/php/p-adic.html">Finding a p-adic square root of a quadratic residue (mod p), p an odd prime.</a> %F A271224 a(n) = (b(n+1) - b(n))/3^n, n >= 0, with b(n) = A271222(n), n >= 0. %F A271224 a(n) = - A271226(n)*2*A271222(n) (mod 3), n >= 1. Solution of the linear congruence given above in a comment. See, e.g., Nagell, Theorem 38 pp. 77-78. %F A271224 A271222(n+1) = sum(a(k)*3^k, k=0..n), n >= 0. %e A271224 a(4) = 2 because 2*59*2 + 43 = 279 == 0 (mod 3). %e A271224 a(4) = - 43*(2*59) (mod 3) = -1*(2*(-1)) (mod 3) = 2. %e A271224 A271222(5) = 221 = 2*3^0 + 1*3^1 + 0*3^2 + 2*3^3 + 2*3^4. %o A271224 (PARI) a(n) = truncate(-sqrt(-2+O(3^(n+1))))\3^n; \\ _Michel Marcus_, Apr 09 2016 %Y A271224 Cf. A268924, A268992, A271222, A271226, A271223 (companion). %K A271224 nonn,base,easy %O A271224 0,1 %A A271224 _Wolfdieter Lang_, Apr 05 2016