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A225410 10-adic integer x such that x^3 = 7/9.

%I A225410 #18 Aug 13 2019 08:12:42
%S A225410 7,4,2,9,3,3,0,0,1,6,6,7,2,0,5,8,6,3,0,4,4,6,0,9,7,1,9,4,2,6,8,7,9,6,
%T A225410 8,0,5,7,1,0,6,6,9,8,6,4,9,0,9,8,5,9,0,5,9,6,5,2,1,5,3,4,6,7,2,4,4,1,
%U A225410 6,3,2,6,1,4,1,0,2,7,0,0,5,4,1,7,9,6,4,1,3,2,1,0,4,6,1,5,6,1,5,2
%N A225410 10-adic integer x such that x^3 = 7/9.
%C A225410 This is the 10's complement of A225401.
%H A225410 Seiichi Manyama, <a href="/A225410/b225410.txt">Table of n, a(n) for n = 0..10000</a>
%F A225410 Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 7, b(n) = b(n-1) + 3 * (9 * b(n-1)^3 - 7) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - _Seiichi Manyama_, Aug 13 2019
%e A225410        7^3 == 3      (mod 10).
%e A225410       47^3 == 23     (mod 10^2).
%e A225410      247^3 == 223    (mod 10^3).
%e A225410     9247^3 == 2223   (mod 10^4).
%e A225410    39247^3 == 22223  (mod 10^5).
%e A225410   339247^3 == 222223 (mod 10^6).
%o A225410 (PARI) n=0; for(i=1, 100, m=(2*(10^i-1)/9)+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 A225410 (PARI) N=100; Vecrev(digits(lift(chinese(Mod((7/9+O(2^N))^(1/3), 2^N), Mod((7/9+O(5^N))^(1/3), 5^N)))), N) \\ _Seiichi Manyama_, Aug 05 2019
%o A225410 (Ruby)
%o A225410 def A225410(n)
%o A225410   ary = [7]
%o A225410   a = 7
%o A225410   n.times{|i|
%o A225410     b = (a + 3 * (9 * a ** 3 - 7)) % (10 ** (i + 2))
%o A225410     ary << (b - a) / (10 ** (i + 1))
%o A225410     a = b
%o A225410   }
%o A225410   ary
%o A225410 end
%o A225410 p A225410(100) # _Seiichi Manyama_, Aug 13 2019
%Y A225410 Cf. A225401, A309600.
%K A225410 nonn,base
%O A225410 0,1
%A A225410 _Aswini Vaidyanathan_, May 07 2013