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%I #22 Aug 12 2019 10:17:09
%S 7,3,7,0,2,3,9,5,2,5,7,6,5,0,9,7,8,3,4,4,5,4,0,2,6,6,7,3,5,0,3,9,9,3,
%T 5,0,4,6,7,6,8,0,3,6,6,9,4,3,8,8,5,2,7,6,8,3,7,4,2,0,0,2,6,4,8,9,1,5,
%U 7,9,7,3,6,8,3,1,7,3,5,1,5,6,5,4,0,4,6,1,0,1,3,4,2,0,8,2,7,2,3,8
%N Digits of the 10-adic integer (-23/9)^(1/3).
%H Seiichi Manyama, <a href="/A309612/b309612.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 + 23) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n
%e 7^3 == 3 (mod 10).
%e 37^3 == 53 (mod 10^2).
%e 737^3 == 553 (mod 10^3).
%e 737^3 == 5553 (mod 10^4).
%e 20737^3 == 55553 (mod 10^5).
%e 320737^3 == 555553 (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 A309612(n)
%o ary = [7]
%o a = 7
%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 A309612(100)
%Y Cf. A173802, A309600, A309609.
%K nonn,base
%O 0,1
%A _Seiichi Manyama_, Aug 10 2019
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