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A225405 10-adic integer x such that x^3 = 7.

%I A225405 #19 Aug 13 2019 08:12:29
%S A225405 3,4,5,1,5,0,7,1,2,2,2,4,4,2,9,6,0,7,3,5,4,5,8,8,8,0,4,1,8,5,1,4,0,0,
%T A225405 6,1,3,5,4,4,8,1,3,7,4,0,7,4,8,5,5,1,6,7,4,5,5,0,0,4,9,0,4,7,0,8,6,8,
%U A225405 4,7,4,4,2,2,0,3,2,2,0,1,6,5,5,4,3,0,3,4,9,7,1,5,1,2,3,0,2,5,6,8
%N A225405 10-adic integer x such that x^3 = 7.
%H A225405 Seiichi Manyama, <a href="/A225405/b225405.txt">Table of n, a(n) for n = 0..10000</a>
%F A225405 Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 7 * (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 A225405       3^3 == 7 (mod 10).
%e A225405      43^3 == 7 (mod 10^2).
%e A225405     543^3 == 7 (mod 10^3).
%e A225405    1543^3 == 7 (mod 10^4).
%e A225405   51543^3 == 7 (mod 10^5).
%e A225405   51543^3 == 7 (mod 10^6).
%o A225405 (PARI) n=0; for(i=1, 100, m=7; for(x=0, 9, if(((n+(x*10^(i-1)))^3)%(10^i)==m, n=n+(x*10^(i-1)); print1(x", "); break)))
%o A225405 (PARI) N=100; Vecrev(digits(lift(chinese(Mod((7+O(2^N))^(1/3), 2^N), Mod((7+O(5^N))^(1/3), 5^N)))), N) \\ _Seiichi Manyama_, Aug 05 2019
%o A225405 (Ruby)
%o A225405 def A225405(n)
%o A225405   ary = [3]
%o A225405   a = 3
%o A225405   n.times{|i|
%o A225405     b = (a + 7 * (a ** 3 - 7)) % (10 ** (i + 2))
%o A225405     ary << (b - a) / (10 ** (i + 1))
%o A225405     a = b
%o A225405   }
%o A225405   ary
%o A225405 end
%o A225405 p A225405(100) # _Seiichi Manyama_, Aug 13 2019
%Y A225405 Cf. A225408, A309600.
%K A225405 nonn,base
%O A225405 0,1
%A A225405 _Aswini Vaidyanathan_, May 07 2013