%I #21 Aug 13 2019 08:12:35

%S 3,1,4,5,6,8,4,0,1,2,1,9,9,4,1,6,7,5,9,3,8,4,3,3,7,0,5,1,8,6,9,6,3,8,

%T 1,5,1,5,3,8,0,7,7,8,6,6,1,3,1,2,1,1,0,9,3,4,0,3,6,4,6,9,7,5,7,6,9,0,

%U 9,4,7,8,3,9,2,9,8,0,0,5,2,4,2,0,6,5,1,3,3,6,6,8,8,5,6,6,3,3,7,5

%N 10-adic integer x such that x^3 = -3.

%C This is the 10's complement of A225404.

%H Seiichi Manyama, <a href="/A225407/b225407.txt">Table of n, a(n) for n = 0..10000</a>

%F 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 + 3) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - _Seiichi Manyama_, Aug 13 2019

%e 3^3 == -7 (mod 10).

%e 13^3 == -7 (mod 10^2).

%e 413^3 == -7 (mod 10^3).

%e 5413^3 == -7 (mod 10^4).

%e 65413^3 == -7 (mod 10^5).

%e 865413^3 == -7 (mod 10^6).

%o (PARI) n=0; for(i=1, 100, m=(10^i-3); for(x=0, 9, if(((n+(x*10^(i-1)))^3)%(10^i)==m, n=n+(x*10^(i-1)); print1(x", "); break)))

%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((-3+O(2^N))^(1/3), 2^N), Mod((-3+O(5^N))^(1/3), 5^N)))), N) \\ _Seiichi Manyama_, Aug 06 2019

%o (Ruby)

%o def A225407(n)

%o ary = [3]

%o a = 3

%o n.times{|i|

%o b = (a + 7 * (a ** 3 + 3)) % (10 ** (i + 2))

%o ary << (b - a) / (10 ** (i + 1))

%o a = b

%o }

%o ary

%o end

%o p A225407(100) # _Seiichi Manyama_, Aug 13 2019

%Y Cf. A225402, A225404, A309600.

%K nonn,base

%O 0,1

%A _Aswini Vaidyanathan_, May 07 2013