%I #21 Aug 12 2019 10:17:20

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

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

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

%N Digits of the 10-adic integer (23/9)^(1/3).

%H Seiichi Manyama, <a href="/A309609/b309609.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 - 23) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n

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

%e 63^3 == 47 (mod 10^2).

%e 263^3 == 447 (mod 10^3).

%e 9263^3 == 4447 (mod 10^4).

%e 79263^3 == 44447 (mod 10^5).

%e 679263^3 == 444447 (mod 10^6).

%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((23/9+O(2^N))^(1/3), 2^N), Mod((23/9+O(5^N))^(1/3), 5^N)))), N)

%o (Ruby)

%o def A309609(n)

%o ary = [3]

%o a = 3

%o n.times{|i|

%o b = (a + 3 * (9 * a ** 3 - 23)) % (10 ** (i + 2))

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

%o a = b

%o }

%o ary

%o end

%o p A309609(100)

%Y Cf. A173772, A309600, A309612.

%K nonn,base

%O 0,1

%A _Seiichi Manyama_, Aug 10 2019