%I #14 Aug 19 2019 10:13:47
%S 4,4,10,4,0,9,5,8,7,7,3,9,7,4,1,8,10,5,10,8,10,2,5,6,10,10,3,10,2,2,1,
%T 6,6,5,7,10,0,6,9,8,3,1,6,5,3,1,9,0,5,8,8,10,4,8,9,1,9,7,1,2,1,0,5,0,
%U 2,9,1,5,2,1,3,1,7,10,3,5,0,4,10,5,1,2,10,2,1,3,0
%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 4. The other, A321079, ends with digit 7.
%H Seiichi Manyama, <a href="/A321078/b321078.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) = (A321076(n+1) - A321076(n))/11^n.
%F For n > 0, a(n) = 10 - A321079(n).
%e ...8960A7566122A3AA652A8A5A8147937785904A44.
%o (PARI) a(n) = truncate(sqrt(5+O(11^(n+1))))\11^n
%Y Cf. A321076, A321079.
%K nonn,base
%O 0,1
%A _Jianing Song_, Oct 27 2018