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A309989 Digits of one of the two 17-adic integers sqrt(-1).

%I A309989 #15 Aug 26 2019 12:51:10
%S A309989 4,2,10,5,12,16,12,8,13,3,14,0,6,1,0,15,1,8,14,5,7,16,14,1,5,13,9,6,5,
%T A309989 12,16,15,9,16,14,12,16,1,3,6,4,10,15,5,16,12,2,1,5,4,0,15,2,11,14,9,
%U A309989 5,1,11,16,15,7,5,6,14,3,12,0,0,11,12,13,9,5,4,16,13
%N A309989 Digits of one of the two 17-adic integers sqrt(-1).
%C A309989 This square root of -1 in the 17-adic field ends with digit 4. The other, A309990, ends with digit 13 (D when written as a 17-adic number).
%H A309989 Seiichi Manyama, <a href="/A309989/b309989.txt">Table of n, a(n) for n = 0..10000</a>
%H A309989 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A309989 a(n) = (A286877(n+1) - A286877(n))/17^n.
%F A309989 For n > 0, a(n) = 16 - A309990(n).
%e A309989 The solution to x^2 == -1 (mod 17^4) such that x == 4 (mod 17) is x == 27493 (mod 17^4), and 27493 is written as 5A24 in heptadecimal, so the first four terms are 4, 2, 10 and 5.
%o A309989 (PARI) a(n) = truncate(sqrt(-1+O(17^(n+1))))\17^n
%Y A309989 Cf. A286877, A286878.
%Y A309989 Digits of p-adic square roots:
%Y A309989 A318962, A318963 (2-adic, sqrt(-7));
%Y A309989 A271223, A271224 (3-adic, sqrt(-2));
%Y A309989 A269591, A269592 (5-adic, sqrt(-4));
%Y A309989 A210850, A210851 (5-adic, sqrt(-1));
%Y A309989 A290794, A290795 (7-adic, sqrt(-6));
%Y A309989 A290798, A290799 (7-adic, sqrt(-5));
%Y A309989 A290796, A290797 (7-adic, sqrt(-3));
%Y A309989 A051277, A290558 (7-adic, sqrt(2));
%Y A309989 A321074, A321075 (11-adic, sqrt(3));
%Y A309989 A321078, A321079 (11-adic, sqrt(5));
%Y A309989 A322091, A322092 (13-adic, sqrt(-3));
%Y A309989 A286838, A286839 (13-adic, sqrt(-1));
%Y A309989 A322087, A322088 (13-adic, sqrt(3));
%Y A309989 this sequence, A309990 (17-adic, sqrt(-1)).
%K A309989 nonn,base
%O A309989 0,1
%A A309989 _Jianing Song_, Aug 26 2019