%I #14 Mar 26 2019 11:45:49
%S 9,6,8,2,3,8,1,4,2,0,2,0,6,9,8,3,8,9,4,5,4,0,6,0,0,9,6,8,6,1,3,4,7,8,
%T 0,6,6,7,1,6,5,5,3,6,4,9,9,0,2,7,1,7,4,2,6,5,1,4,0,6,9,0,7,0,8,7,8,1,
%U 4,1,2,6,6,9,4,2,5,3,5,7,4,9,6,4,4,0,5
%N Expansion of the 10-adic cube root of -1/11, that is, the 10-adic integer solution to x^3 = -1/11.
%C 10's complement of A319740.
%H Robert Israel, <a href="/A306553/b306553.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = 9 - A319740(n) for n >= 2.
%e 9^3 == 9 == -1/11 (mod 10).
%e 69^3 == 9 == -1/11 (mod 100).
%e 869^3 == 909 == -1/11 (mod 1000).
%e 2869^3 == 909 == -1/11 (mod 10000).
%e ...
%e ...020241832869^3 = ...090909090909 = ...999999999999/11 = -1/11.
%p op([1,3],padic:-rootp(11*x^3+1,10,100)); # _Robert Israel_, Mar 24 2019
%o (PARI) seq(n)={Vecrev(digits(lift(chinese( Mod((-1/11 + O(5^n))^(1/3), 5^n), Mod((-1/11 + O(2^n))^(1/3), 2^n)))), n)} \\ Following _Andrew Howroyd_'s code for A319740.
%Y 10-adic cube root of p/q:
%Y q=1: A225409 (p=-9), A225408 (p=-7), A225407 (p=-3), A225404 (p=3), A225405 (p=7), A225406 (p=9);
%Y q=3: A225402 (p=-1), A225411 (p=1);
%Y q=7: A306552 (p=-1), A319739 (p=1);
%Y q=9: A225401 (p=-7), A153042 (p=-1), A225412 (p=1), A225410 (p=7);
%Y q=11: this sequence (p=-1), A319740 (p=1);
%Y q=13: A306555 (p=-1), A306554 (p=1).
%K nonn,base
%O 1,1
%A _Jianing Song_, Feb 23 2019