%I #18 Aug 12 2019 02:40:18
%S 3,0,6,8,5,0,7,1,6,9,9,9,1,7,3,8,5,6,2,9,8,1,0,9,6,8,3,0,5,1,5,1,5,7,
%T 7,1,1,5,9,9,9,9,1,2,9,9,2,1,0,3,6,9,9,5,9,4,0,5,3,0,3,0,7,9,8,1,4,6,
%U 7,9,8,7,9,4,2,0,6,6,0,5,4,3,7,9,6,8,6,4,8,5,9,4,1,7,4,2,7,3,5,0
%N Digits of the 10-adic integer (43/9)^(1/3).
%H Seiichi Manyama, <a href="/A309604/b309604.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) + 3 * (9 * b(n-1)^3 - 43) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.
%e 3^3 == 7 (mod 10).
%e 3^3 == 27 (mod 10^2).
%e 603^3 == 227 (mod 10^3).
%e 8603^3 == 2227 (mod 10^4).
%e 58603^3 == 22227 (mod 10^5).
%e 58603^3 == 222227 (mod 10^6).
%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((43/9+O(2^N))^(1/3), 2^N), Mod((43/9+O(5^N))^(1/3), 5^N)))), N)
%o (Ruby)
%o def A309604(n)
%o ary = [3]
%o a = 3
%o n.times{|i|
%o b = (a + 3 * (9 * a ** 3 - 43)) % (10 ** (i + 2))
%o ary << (b - a) / (10 ** (i + 1))
%o a = b
%o }
%o ary
%o end
%o p A309604(100)
%Y Cf. A309600, A309644.
%K nonn
%O 0,1
%A _Seiichi Manyama_, Aug 09 2019