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%I #19 Aug 12 2019 02:40:04
%S 9,0,5,0,5,4,7,1,9,1,6,0,9,8,5,7,1,0,7,3,1,0,9,5,1,4,9,9,5,7,9,3,0,1,
%T 1,9,0,1,4,1,4,0,6,4,4,1,8,0,0,1,7,6,9,1,5,3,8,1,4,2,6,7,1,3,3,9,8,0,
%U 4,5,3,7,2,5,2,7,5,5,4,6,1,0,0,2,2,3,2,0,7,3,4,2,7,7,1,0,3,1,0,9
%N Digits of the 10-adic integer (61/9)^(1/3).
%H Seiichi Manyama, <a href="/A309605/b309605.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 - 61) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.
%e 9^3 == 9 (mod 10).
%e 9^3 == 29 (mod 10^2).
%e 509^3 == 229 (mod 10^3).
%e 509^3 == 2229 (mod 10^4).
%e 50509^3 == 22229 (mod 10^5).
%e 450509^3 == 222229 (mod 10^6).
%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((61/9+O(2^N))^(1/3), 2^N), Mod((61/9+O(5^N))^(1/3), 5^N)))), N)
%o (Ruby)
%o def A309605(n)
%o ary = [9]
%o a = 9
%o n.times{|i|
%o b = (a + 7 * (9 * a ** 3 - 61)) % (10 ** (i + 2))
%o ary << (b - a) / (10 ** (i + 1))
%o a = b
%o }
%o ary
%o end
%o p A309605(100)
%Y Cf. A309600, A309643.
%K nonn
%O 0,1
%A _Seiichi Manyama_, Aug 09 2019
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