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%I #19 Aug 12 2019 10:17:13
%S 1,5,3,4,0,3,5,3,0,3,6,3,0,2,6,6,9,7,3,0,6,0,1,5,2,1,1,3,8,4,4,2,8,1,
%T 5,5,9,6,3,8,1,7,8,4,0,7,9,6,1,0,4,3,5,8,4,2,7,9,8,1,0,1,6,8,8,2,6,3,
%U 1,3,8,6,2,7,4,2,8,2,2,6,2,8,2,3,0,2,8,2,9,6,8,1,2,6,8,9,3,7,6,6
%N Digits of the 10-adic integer (-41/9)^(1/3).
%H Seiichi Manyama, <a href="/A309611/b309611.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 + 41) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n
%e 1^3 == 1 (mod 10).
%e 51^3 == 51 (mod 10^2).
%e 351^3 == 551 (mod 10^3).
%e 4351^3 == 5551 (mod 10^4).
%e 4351^3 == 55551 (mod 10^5).
%e 304351^3 == 555551 (mod 10^6).
%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((-41/9+O(2^N))^(1/3), 2^N), Mod((-41/9+O(5^N))^(1/3), 5^N)))), N)
%o (Ruby)
%o def A309611(n)
%o ary = [1]
%o a = 1
%o n.times{|i|
%o b = (a + 7 * (9 * a ** 3 + 41)) % (10 ** (i + 2))
%o ary << (b - a) / (10 ** (i + 1))
%o a = b
%o }
%o ary
%o end
%o p A309611(100)
%Y Cf. A173804, A309600, A309610.
%K nonn,base
%O 0,2
%A _Seiichi Manyama_, Aug 10 2019
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