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A225401 10-adic integer x such that x^3 = -7/9.

%I A225401 #22 Apr 14 2022 10:51:56
%S A225401 3,5,7,0,6,6,9,9,8,3,3,2,7,9,4,1,3,6,9,5,5,3,9,0,2,8,0,5,7,3,1,2,0,3,
%T A225401 1,9,4,2,8,9,3,3,0,1,3,5,0,9,0,1,4,0,9,4,0,3,4,7,8,4,6,5,3,2,7,5,5,8,
%U A225401 3,6,7,3,8,5,8,9,7,2,9,9,4,5,8,2,0,3,5,8,6,7,8,9,5,3,8,4,3,8,4,7
%N A225401 10-adic integer x such that x^3 = -7/9.
%H A225401 Seiichi Manyama, <a href="/A225401/b225401.txt">Table of n, a(n) for n = 0..10000</a>
%F A225401 Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, 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 A225401        3^3 == 7      (mod 10).
%e A225401       53^3 == 77     (mod 10^2).
%e A225401      753^3 == 777    (mod 10^3).
%e A225401      753^3 == 7777   (mod 10^4).
%e A225401    60753^3 == 77777  (mod 10^5).
%e A225401   660753^3 == 777777 (mod 10^6).
%p A225401 b:= proc(n) option remember; `if`(n<2, 3*n,
%p A225401       irem(b(n-1)+3*(9*b(n-1)^3+7), 10^n))
%p A225401     end:
%p A225401 a:= n-> (b(n+1)-b(n))/10^n:
%p A225401 seq(a(n), n=0..100);  # _Alois P. Heinz_, Apr 14 2022
%o A225401 (PARI) n=0;for(i=1,100,m=7*(10^i-1)/9;for(x=0,9,if(((n+(x*10^(i-1)))^3)%(10^i)==m,n=n+(x*10^(i-1));print1(x", ");break)))
%o A225401 (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 A225401 (Ruby)
%o A225401 def A225401(n)
%o A225401   ary = [3]
%o A225401   a = 3
%o A225401   n.times{|i|
%o A225401     b = (a + 3 * (9 * a ** 3 + 7)) % (10 ** (i + 2))
%o A225401     ary << (b - a) / (10 ** (i + 1))
%o A225401     a = b
%o A225401   }
%o A225401   ary
%o A225401 end
%o A225401 p A225401(100) # _Seiichi Manyama_, Aug 13 2019
%Y A225401 Cf. A225410, A309600.
%K A225401 nonn,base
%O A225401 0,1
%A A225401 _Aswini Vaidyanathan_, May 07 2013