%I #13 Mar 26 2019 11:45:56
%S 3,5,6,4,1,9,3,2,8,7,4,0,8,3,6,5,7,7,0,9,8,2,7,5,1,4,8,0,9,5,1,6,0,6,
%T 2,1,3,2,2,6,4,2,7,0,6,8,6,1,3,3,2,2,0,0,1,5,6,7,9,6,2,7,8,4,2,6,3,6,
%U 3,0,1,0,4,5,5,6,6,1,3,5,4,3,3,3,1,7,0
%N Expansion of the 10-adic cube root of 1/13, that is, the 10-adic integer solution to x^3 = 1/13.
%C 10's complement of A306555.
%H Robert Israel, <a href="/A306554/b306554.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = 9 - A306555(n) for n >= 2.
%e 3^3 == 7 == 1/13 (mod 10).
%e 53^3 == 77 == 1/13 (mod 100).
%e 653^3 == 77 == 1/13 (mod 1000).
%e 4653^3 == 3077 == 1/13 (mod 10000).
%e ...
%e ...047823914653^3 = ...923076923077 = 1 + (...999999999999)*(12/13) = 1 - 12/13 = 1/13.
%p op([1,3],padic:-rootp(13*x^3-1,10,100)); # _Robert Israel_, Mar 24 2019
%o (PARI) seq(n)={Vecrev(digits(lift(chinese( Mod((1/13 + O(5^n))^(1/3), 5^n), Mod((1/13 + 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: A306553 (p=-1), A319740 (p=1);
%Y q=13: A306555 (p=-1), this sequence (p=1).
%K nonn,base
%O 1,1
%A _Jianing Song_, Feb 23 2019