%I #20 Aug 12 2019 02:40:52
%S 9,3,6,8,3,2,7,4,5,4,1,1,9,2,2,9,0,0,3,4,5,8,1,0,7,1,6,4,6,5,3,0,3,1,
%T 5,6,9,7,3,2,4,2,4,2,0,6,2,2,0,2,3,6,7,8,4,6,5,1,5,7,5,0,9,4,4,0,9,5,
%U 5,1,9,0,2,4,7,7,6,6,4,0,1,0,6,2,9,6,8,3,9,7,9,6,0,2,3,4,6,8,6,8
%N Digits of the 10-adic integer (71/9)^(1/3).
%H Seiichi Manyama, <a href="/A309602/b309602.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) = 9, b(n) = b(n-1) + 7 * (9 * b(n-1)^3 - 71) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.
%e 9^3 == 9 (mod 10).
%e 39^3 == 19 (mod 10^2).
%e 639^3 == 119 (mod 10^3).
%e 8639^3 == 1119 (mod 10^4).
%e 38639^3 == 11119 (mod 10^5).
%e 238639^3 == 111119 (mod 10^6).
%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((71/9+O(2^N))^(1/3), 2^N), Mod((71/9+O(5^N))^(1/3), 5^N)))), N)
%o (Ruby)
%o def A309602(n)
%o ary = [9]
%o a = 9
%o n.times{|i|
%o b = (a + 7 * (9 * a ** 3 - 71)) % (10 ** (i + 2))
%o ary << (b - a) / (10 ** (i + 1))
%o a = b
%o }
%o ary
%o end
%o p A309602(100)
%Y Cf. A165247, A309600, A309646.
%K nonn
%O 0,1
%A _Seiichi Manyama_, Aug 09 2019