%I #23 Aug 12 2019 10:16:39
%S 9,5,8,6,6,5,0,7,6,0,5,1,0,8,9,4,8,5,8,5,6,2,2,5,2,1,8,7,6,9,9,8,2,3,
%T 8,1,0,5,8,5,7,0,0,9,6,7,4,2,0,6,6,7,7,1,2,4,1,7,9,1,2,5,4,8,7,7,3,4,
%U 4,9,1,7,6,9,6,7,0,7,4,1,3,3,6,9,7,4,6,9,8,8,5,9,0,0,1,5,4,0,5,4
%N Digits of the 10-adic integer (11/9)^(1/3).
%H Seiichi Manyama, <a href="/A309645/b309645.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 - 11) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.
%e 9^3 == 9 (mod 10).
%e 59^3 == 79 (mod 10^2).
%e 859^3 == 779 (mod 10^3).
%e 6859^3 == 7779 (mod 10^4).
%e 66859^3 == 77779 (mod 10^5).
%e 566859^3 == 777779 (mod 10^6).
%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((11/9+O(2^N))^(1/3), 2^N), Mod((11/9+O(5^N))^(1/3), 5^N)))), N)
%o (Ruby)
%o def A309645(n)
%o ary = [9]
%o a = 9
%o n.times{|i|
%o b = (a + 7 * (9 * a ** 3 - 11)) % (10 ** (i + 2))
%o ary << (b - a) / (10 ** (i + 1))
%o a = b
%o }
%o ary
%o end
%o p A309645(100)
%Y Cf. A309600, A309603.
%K nonn,base
%O 0,1
%A _Seiichi Manyama_, Aug 11 2019