|
%I #13 Aug 19 2019 10:20:00
%S 7,6,0,6,10,1,5,2,3,3,7,1,3,6,9,2,0,5,0,2,0,8,5,4,0,0,7,0,8,8,9,4,4,5,
%T 3,0,10,4,1,2,7,9,4,5,7,9,1,10,5,2,2,0,6,2,1,9,1,3,9,8,9,10,5,10,8,1,
%U 9,5,8,9,7,9,3,0,7,5,10,6,0,5,9,8,0,8,9,7,10
%N Digits of one of the two 11-adic integers sqrt(5).
%C This square root of 5 in the 11-adic field ends with digit 7. The other, A321078, ends with digit 4.
%H Seiichi Manyama, <a href="/A321079/b321079.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) = (A321077(n+1) - A321077(n))/11^n.
%F For n > 0, a(n) = 10 - A321078(n).
%e ...214A0354498807004580205029631733251A6067.
%o (PARI) a(n) = truncate(-sqrt(5+O(11^(n+1))))\11^n
%Y Cf. A321077, A321078.
%K nonn,base
%O 0,1
%A _Jianing Song_, Oct 27 2018
|
