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%I #14 Aug 12 2019 02:38:52
%S 7,9,3,1,4,9,2,8,3,0,0,0,8,2,6,1,4,3,7,0,1,8,9,0,3,1,6,9,4,8,4,8,4,2,
%T 2,8,8,4,0,0,0,0,8,7,0,0,7,8,9,6,3,0,0,4,0,5,9,4,6,9,6,9,2,0,1,8,5,3,
%U 2,0,1,2,0,5,7,9,3,3,9,4,5,6,2,0,3,1,3,5,1,4,0,5,8,2,5,7,2,6,4,9
%N Digits of the 10-adic integer (-43/9)^(1/3).
%H Seiichi Manyama, <a href="/A309644/b309644.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 + 43) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.
%e 7^3 == 3 (mod 10).
%e 97^3 == 73 (mod 10^2).
%e 397^3 == 773 (mod 10^3).
%e 1397^3 == 7773 (mod 10^4).
%e 41397^3 == 77773 (mod 10^5).
%e 941397^3 == 777773 (mod 10^6).
%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((-43/9+O(2^N))^(1/3), 2^N), Mod((-43/9+O(5^N))^(1/3), 5^N)))), N)
%o (Ruby)
%o def A309644(n)
%o ary = [7]
%o a = 7
%o n.times{|i|
%o b = (a + 3 * (9 * a ** 3 + 43)) % (10 ** (i + 2))
%o ary << (b - a) / (10 ** (i + 1))
%o a = b
%o }
%o ary
%o end
%o p A309644(100)
%Y Cf. A309600, A309604.
%K nonn,base
%O 0,1
%A _Seiichi Manyama_, Aug 11 2019
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