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Digits of the 10-adic integer (-11/9)^(1/3).
3

%I #22 Aug 12 2019 02:40:31

%S 1,4,1,3,3,4,9,2,3,9,4,8,9,1,0,5,1,4,1,4,3,7,7,4,7,8,1,2,3,0,0,1,7,6,

%T 1,8,9,4,1,4,2,9,9,0,3,2,5,7,9,3,3,2,2,8,7,5,8,2,0,8,7,4,5,1,2,2,6,5,

%U 5,0,8,2,3,0,3,2,9,2,5,8,6,6,3,0,2,5,3,0,1,1,4,0,9,9,8,4,5,9,4,5

%N Digits of the 10-adic integer (-11/9)^(1/3).

%H Seiichi Manyama, <a href="/A309603/b309603.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 + 11) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.

%e 1^3 == 1 (mod 10).

%e 41^3 == 21 (mod 10^2).

%e 141^3 == 221 (mod 10^3).

%e 3141^3 == 2221 (mod 10^4).

%e 33141^3 == 22221 (mod 10^5).

%e 433141^3 == 222221 (mod 10^6).

%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((-11/9+O(2^N))^(1/3), 2^N), Mod((-11/9+O(5^N))^(1/3), 5^N)))), N)

%o (Ruby)

%o def A309603(n)

%o ary = [1]

%o a = 1

%o n.times{|i|

%o b = (a + 7 * (9 * a ** 3 + 11)) % (10 ** (i + 2))

%o ary << (b - a) / (10 ** (i + 1))

%o a = b

%o }

%o ary

%o end

%o p A309603(100)

%Y Cf. A165402, A309600, A309645.

%K nonn

%O 0,2

%A _Seiichi Manyama_, Aug 09 2019