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A225408 10-adic integer x=.....8457 satisfying x^3 = -7.

%I A225408 #19 Aug 13 2019 08:35:45
%S A225408 7,5,4,8,4,9,2,8,7,7,7,5,5,7,0,3,9,2,6,4,5,4,1,1,1,9,5,8,1,4,8,5,9,9,
%T A225408 3,8,6,4,5,5,1,8,6,2,5,9,2,5,1,4,4,8,3,2,5,4,4,9,9,5,0,9,5,2,9,1,3,1,
%U A225408 5,2,5,5,7,7,9,6,7,7,9,8,3,4,4,5,6,9,6,5,0,2,8,4,8,7,6,9,7,4,3,1
%N A225408 10-adic integer x=.....8457 satisfying x^3 = -7.
%C A225408 This is the 10's complement of A225405.
%H A225408 Seiichi Manyama, <a href="/A225408/b225408.txt">Table of n, a(n) for n = 0..10000</a>
%F A225408 Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 7, 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 A225408        7^3 == -7 (mod 10).
%e A225408       57^3 == -7 (mod 10^2).
%e A225408      457^3 == -7 (mod 10^3).
%e A225408     8457^3 == -7 (mod 10^4).
%e A225408    48457^3 == -7 (mod 10^5).
%e A225408   948457^3 == -7 (mod 10^6).
%o A225408 (PARI) n=0; for(i=1, 100, m=(10^i-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 A225408 (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 04 2019
%o A225408 (Ruby)
%o A225408 def A225408(n)
%o A225408   ary = [7]
%o A225408   a = 7
%o A225408   n.times{|i|
%o A225408     b = (a + 7 * (a ** 3 + 7)) % (10 ** (i + 2))
%o A225408     ary << (b - a) / (10 ** (i + 1))
%o A225408     a = b
%o A225408   }
%o A225408   ary
%o A225408 end
%o A225408 p A225408(100) # _Seiichi Manyama_, Aug 13 2019
%Y A225408 Cf. A091663, A225405, A309600.
%K A225408 nonn,base
%O A225408 0,1
%A A225408 _Aswini Vaidyanathan_, May 07 2013