|
%I #13 Apr 01 2019 03:02:16
%S 3,9,5,5,4,0,3,8,3,9,1,6,4,7,2,6,5,2,9,6,2,4,5,7,0,0,9,0,6,1,9,3,8,2,
%T 5,1,4,1,8,4,1,0,2,4,4,7,8,5,0,6,2,4,3,8,4,2,0,2,4,7,3,3,4,7,1,9,9,3,
%U 5,3,9,7,0,4,4,6,3,7,7,1,7,6,3,5,5,9,6
%N Expansion of the 10-adic cube root of -1/7, that is, the 10-adic integer solution to x^3 = -1/7.
%C 10's complement of A319739.
%H Robert Israel, <a href="/A306552/b306552.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = 9 - A319739(n) for n >= 2.
%e 3^3 == 7 == -1/7 (mod 10).
%e 93^3 == 57 == -1/7 (mod 100).
%e 593^3 == 857 == -1/7 (mod 1000).
%e 5593^3 == 2857 == -1/7 (mod 10000).
%e ...
%e ...619383045593^3 = ...142857142857 = ...999999999999/7 = -1/7.
%p op([1, 3], padic:-rootp(7*x^3+1, 10, 100)); # _Robert Israel_, Mar 31 2019
%o (PARI) seq(n)={Vecrev(digits(lift(chinese( Mod((-1/7 + O(5^n))^(1/3), 5^n), Mod((-1/7 + 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: this sequence (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), A306554 (p=1).
%K nonn,base
%O 1,1
%A _Jianing Song_, Feb 23 2019
|
