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A309990 Digits of one of the two 17-adic integers sqrt(-1).

%I A309990 #18 Aug 26 2019 12:51:20
%S A309990 13,14,6,11,4,0,4,8,3,13,2,16,10,15,16,1,15,8,2,11,9,0,2,15,11,3,7,10,
%T A309990 11,4,0,1,7,0,2,4,0,15,13,10,12,6,1,11,0,4,14,15,11,12,16,1,14,5,2,7,
%U A309990 11,15,5,0,1,9,11,10,2,13,4,16,16,5,4,3,7,11,12,0
%N A309990 Digits of one of the two 17-adic integers sqrt(-1).
%C A309990 This square root of -1 in the 17-adic field ends with digit 13 (D when written as a 17-adic number). The other, A309989, ends with digit 4.
%H A309990 Seiichi Manyama, <a href="/A309990/b309990.txt">Table of n, a(n) for n = 0..10000</a>
%H A309990 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A309990 a(n) = (A286878(n+1) - A286878(n))/17^n.
%F A309990 For n > 0, a(n) = 16 - A309989(n).
%e A309990 The solution to x^2 == -1 (mod 17^4) such that x == 13 (mod 17) is x == 56028 (mod 17^4), and 56028 is written as B6ED in heptadecimal, so the first four terms are 13, 14, 6 and 11.
%o A309990 (PARI) a(n) = truncate(-sqrt(-1+O(17^(n+1))))\17^n
%Y A309990 Cf. A286877, A286878.
%Y A309990 Digits of p-adic square roots:
%Y A309990 A318962, A318963 (2-adic, sqrt(-7));
%Y A309990 A271223, A271224 (3-adic, sqrt(-2));
%Y A309990 A269591, A269592 (5-adic, sqrt(-4));
%Y A309990 A210850, A210851 (5-adic, sqrt(-1));
%Y A309990 A290794, A290795 (7-adic, sqrt(-6));
%Y A309990 A290798, A290799 (7-adic, sqrt(-5));
%Y A309990 A290796, A290797 (7-adic, sqrt(-3));
%Y A309990 A051277, A290558 (7-adic, sqrt(2));
%Y A309990 A321074, A321075 (11-adic, sqrt(3));
%Y A309990 A321078, A321079 (11-adic, sqrt(5));
%Y A309990 A322091, A322092 (13-adic, sqrt(-3));
%Y A309990 A286838, A286839 (13-adic, sqrt(-1));
%Y A309990 A322087, A322088 (13-adic, sqrt(3));
%Y A309990 A309989, this sequence (17-adic, sqrt(-1)).
%K A309990 nonn,base
%O A309990 0,1
%A A309990 _Jianing Song_, Aug 26 2019