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%I #14 Aug 11 2019 11:59:56
%S 7,2,6,1,3,2,9,4,6,9,1,4,6,5,1,8,6,9,0,9,6,7,0,7,6,3,6,7,5,6,4,8,4,7,
%T 0,1,9,2,3,9,6,0,0,5,7,4,6,9,6,7,9,6,7,1,7,1,2,1,6,8,9,9,5,3,5,8,1,0,
%U 5,1,4,6,4,6,2,6,8,3,2,0,8,8,1,4,9,7,4,2,3,6,1,0,5,7,2,6,9,6,3,3
%N Digits of the 10-adic integer (-53/9)^(1/3).
%H Seiichi Manyama, <a href="/A309647/b309647.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) = 7, b(n) = b(n-1) + 3 * (9 * b(n-1)^3 + 53) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.
%e 7^3 == 3 (mod 10).
%e 27^3 == 83 (mod 10^2).
%e 627^3 == 883 (mod 10^3).
%e 1627^3 == 8883 (mod 10^4).
%e 31627^3 == 88883 (mod 10^5).
%e 231627^3 == 888883 (mod 10^6).
%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((-53/9+O(2^N))^(1/3), 2^N), Mod((-53/9+O(5^N))^(1/3), 5^N)))), N)
%o (Ruby)
%o def A309647(n)
%o ary = [7]
%o a = 7
%o n.times{|i|
%o b = (a + 3 * (9 * a ** 3 + 53)) % (10 ** (i + 2))
%o ary << (b - a) / (10 ** (i + 1))
%o a = b
%o }
%o ary
%o end
%o p A309647(100)
%Y Cf. A309600, A309601.
%K nonn,base
%O 0,1
%A _Seiichi Manyama_, Aug 11 2019
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