%I #30 Aug 13 2019 08:12:50
%S 3,2,5,3,5,5,4,1,6,5,5,3,1,6,8,4,8,9,4,8,3,0,3,8,6,8,6,2,7,7,5,3,5,5,
%T 6,1,6,4,6,6,4,3,7,4,7,6,7,1,3,6,5,8,1,0,3,2,9,5,1,6,3,2,8,8,3,3,4,7,
%U 9,2,5,9,0,7,1,9,6,9,6,9,2,6,0,2,0,2,6,5,3,4,6,8,6,9,9,1,2,5,4,2
%N 10-adic integer x such that x^3 == (6*(10^(n+1)-1)/9)+1 mod 10^(n+1) for all n.
%C This is the 10's complement of A225402.
%C Equivalently, the 10-adic cube root of 1/3, i.e., solution to 3*x^3 = 1 (mod 10^n) for all n. - _M. F. Hasler_, Jan 02 2019
%H Robert Israel, <a href="/A225411/b225411.txt">Table of n, a(n) for n = 0..10000</a>
%F Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 9 * (3 * b(n-1)^3 - 1) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - _Seiichi Manyama_, Aug 12 2019
%e 3^3 == 7 (mod 10).
%e 23^3 == 67 (mod 10^2).
%e 523^3 == 667 (mod 10^3).
%e 3523^3 == 6667 (mod 10^4).
%e 53523^3 == 66667 (mod 10^5).
%e 553523^3 == 666667 (mod 10^6).
%p op([1,3],padic:-rootp(3*x^3 -1, 10, 101)); # _Robert Israel_, Aug 04 2019
%o (PARI) n=0; for(i=1, 100, m=(2*(10^i-1)/3)+1; for(x=0, 9, if(((n+(x*10^(i-1)))^3)%(10^i)==m, n=n+(x*10^(i-1)); print1(x", "); break)))
%o (PARI) upto(N=100,m=1/3)=Vecrev(digits(lift(chinese(Mod((m+O(5^N))^m, 5^N), Mod((m+O(2^N))^m, 2^N)))),N) \\ Following _Andrew Howroyd_'s code for A319740. - _M. F. Hasler_, Jan 02 2019
%o (Ruby)
%o def A225411(n)
%o ary = [3]
%o a = 3
%o n.times{|i|
%o b = (a + 9 * (3 * a ** 3 - 1)) % (10 ** (i + 2))
%o ary << (b - a) / (10 ** (i + 1))
%o a = b
%o }
%o ary
%o end
%o p A225411(100) # _Seiichi Manyama_, Aug 12 2019
%Y Cf. A225402, A309600, A319740 (10-adic cube root of 1/11).
%K nonn,base
%O 0,1
%A _Aswini Vaidyanathan_, May 07 2013