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

%I #13 Aug 12 2019 02:38:20

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

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

%U 0,8,8,5,8,5,2,5,0,9,2,2,8,7,5,0,6,5,6,1,9,1,8,1,0,1,6,4,4,8,0,7

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

%H Seiichi Manyama, <a href="/A309641/b309641.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) + 9 * (3 * b(n-1)^3 + 11) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.

%e 7^3 == 3 (mod 10).

%e 67^3 == 63 (mod 10^2).

%e 767^3 == 663 (mod 10^3).

%e 7767^3 == 6663 (mod 10^4).

%e 47767^3 == 66663 (mod 10^5).

%e 147767^3 == 666663 (mod 10^6).

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

%o (Ruby)

%o def A309641(n)

%o ary = [7]

%o a = 7

%o n.times{|i|

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

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

%o a = b

%o }

%o ary

%o end

%o p A309641(100)

%Y Cf. A309569, A309600.

%K nonn,base

%O 0,1

%A _Seiichi Manyama_, Aug 11 2019