A309989 - OEIS

%I #15 Aug 26 2019 12:51:10

%S 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 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 5,1,11,16,15,7,5,6,14,3,12,0,0,11,12,13,9,5,4,16,13

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

%C 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 Seiichi Manyama, <a href="/A309989/b309989.txt">Table of n, a(n) for n = 0..10000</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>

%F a(n) = (A286877(n+1) - A286877(n))/17^n.

%F For n > 0, a(n) = 16 - A309990(n).

%e 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 (PARI) a(n) = truncate(sqrt(-1+O(17^(n+1))))\17^n

%Y Cf. A286877, A286878.

%Y Digits of p-adic square roots:

%Y A318962, A318963 (2-adic, sqrt(-7));

%Y A271223, A271224 (3-adic, sqrt(-2));

%Y A269591, A269592 (5-adic, sqrt(-4));

%Y A210850, A210851 (5-adic, sqrt(-1));

%Y A290794, A290795 (7-adic, sqrt(-6));

%Y A290798, A290799 (7-adic, sqrt(-5));

%Y A290796, A290797 (7-adic, sqrt(-3));

%Y A051277, A290558 (7-adic, sqrt(2));

%Y A321074, A321075 (11-adic, sqrt(3));

%Y A321078, A321079 (11-adic, sqrt(5));

%Y A322091, A322092 (13-adic, sqrt(-3));

%Y A286838, A286839 (13-adic, sqrt(-1));

%Y A322087, A322088 (13-adic, sqrt(3));

%Y this sequence, A309990 (17-adic, sqrt(-1)).

%K nonn,base

%O 0,1

%A _Jianing Song_, Aug 26 2019