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A225411 10-adic integer x such that x^3 == (6*(10^(n+1)-1)/9)+1 mod 10^(n+1) for all n.

%I A225411 #30 Aug 13 2019 08:12:50
%S A225411 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 A225411 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 A225411 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 A225411 10-adic integer x such that x^3 == (6*(10^(n+1)-1)/9)+1 mod 10^(n+1) for all n.
%C A225411 This is the 10's complement of A225402.
%C A225411 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 A225411 Robert Israel, <a href="/A225411/b225411.txt">Table of n, a(n) for n = 0..10000</a>
%F A225411 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 A225411        3^3 ==      7 (mod 10).
%e A225411       23^3 ==     67 (mod 10^2).
%e A225411      523^3 ==    667 (mod 10^3).
%e A225411     3523^3 ==   6667 (mod 10^4).
%e A225411    53523^3 ==  66667 (mod 10^5).
%e A225411   553523^3 == 666667 (mod 10^6).
%p A225411 op([1,3],padic:-rootp(3*x^3  -1,  10, 101)); # _Robert Israel_, Aug 04 2019
%o A225411 (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 A225411 (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 A225411 (Ruby)
%o A225411 def A225411(n)
%o A225411   ary = [3]
%o A225411   a = 3
%o A225411   n.times{|i|
%o A225411     b = (a + 9 * (3 * a ** 3 - 1)) % (10 ** (i + 2))
%o A225411     ary << (b - a) / (10 ** (i + 1))
%o A225411     a = b
%o A225411   }
%o A225411   ary
%o A225411 end
%o A225411 p A225411(100) # _Seiichi Manyama_, Aug 12 2019
%Y A225411 Cf. A225402, A309600, A319740 (10-adic cube root of 1/11).
%K A225411 nonn,base
%O A225411 0,1
%A A225411 _Aswini Vaidyanathan_, May 07 2013