%I #31 Aug 12 2019 10:17:29
%S 1,8,0,1,1,8,6,6,4,1,6,0,3,9,9,0,9,3,8,0,7,1,6,5,5,2,0,8,8,4,6,7,9,8,
%T 3,0,6,7,0,7,4,0,5,9,9,5,2,0,6,7,8,9,7,8,7,2,1,2,0,7,4,8,8,4,3,6,0,6,
%U 8,2,1,4,2,8,6,7,0,5,7,4,9,7,7,5,8,4,5,9,5,7,8,4,7,9,4,4,3,7,9,1
%N Digits of the 10-adic integer (-31/9)^(1/3).
%H Seiichi Manyama, <a href="/A309595/b309595.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) = 1, b(n) = b(n-1) + 7 * (9 * b(n-1)^3 + 31) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.
%e 1^3 == 1 (mod 10).
%e 81^3 == 41 (mod 10^2).
%e 81^3 == 441 (mod 10^3).
%e 1081^3 == 4441 (mod 10^4).
%e 11081^3 == 44441 (mod 10^5).
%e 811081^3 == 444441 (mod 10^6).
%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((-31/9+O(2^N))^(1/3), 2^N), Mod((-31/9+O(5^N))^(1/3), 5^N)))), N)
%o (Ruby)
%o def A309595(n)
%o ary = [1]
%o a = 1
%o n.times{|i|
%o b = (a + 7 * (9 * a ** 3 + 31)) % (10 ** (i + 2))
%o ary << (b - a) / (10 ** (i + 1))
%o a = b
%o }
%o ary
%o end
%o p A309595(100)
%Y Cf. A173768, A309600, A309614.
%K nonn,base
%O 0,2
%A _Seiichi Manyama_, Aug 10 2019
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