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A309601 Digits of the 10-adic integer (53/9)^(1/3).

%I A309601 #22 Aug 11 2019 14:38:19
%S A309601 3,7,3,8,6,7,0,5,3,0,8,5,3,4,8,1,3,0,9,0,3,2,9,2,3,6,3,2,4,3,5,1,5,2,
%T A309601 9,8,0,7,6,0,3,9,9,4,2,5,3,0,3,2,0,3,2,8,2,8,7,8,3,1,0,0,4,6,4,1,8,9,
%U A309601 4,8,5,3,5,3,7,3,1,6,7,9,1,1,8,5,0,2,5,7,6,3,8,9,4,2,7,3,0,3,6,6
%N A309601 Digits of the 10-adic integer (53/9)^(1/3).
%H A309601 Seiichi Manyama, <a href="/A309601/b309601.txt">Table of n, a(n) for n = 0..10000</a>
%F A309601 Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, 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 A309601        3^3 == 7      (mod 10).
%e A309601       73^3 == 17     (mod 10^2).
%e A309601      373^3 == 117    (mod 10^3).
%e A309601     8373^3 == 1117   (mod 10^4).
%e A309601    68373^3 == 11117  (mod 10^5).
%e A309601   768373^3 == 111117 (mod 10^6).
%o A309601 (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 A309601 (Ruby)
%o A309601 def A309601(n)
%o A309601   ary = [3]
%o A309601   a = 3
%o A309601   n.times{|i|
%o A309601     b = (a + 3 * (9 * a ** 3 - 53)) % (10 ** (i + 2))
%o A309601     ary << (b - a) / (10 ** (i + 1))
%o A309601     a = b
%o A309601   }
%o A309601   ary
%o A309601 end
%o A309601 p A309601(100)
%Y A309601 Cf. A309600.
%K A309601 nonn,base
%O A309601 0,1
%A A309601 _Seiichi Manyama_, Aug 09 2019