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

%I #23 Aug 12 2019 10:16:39

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

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

%U 4,9,1,7,6,9,6,7,0,7,4,1,3,3,6,9,7,4,6,9,8,8,5,9,0,0,1,5,4,0,5,4

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

%H Seiichi Manyama, <a href="/A309645/b309645.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) = 9, 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 9^3 == 9 (mod 10).

%e 59^3 == 79 (mod 10^2).

%e 859^3 == 779 (mod 10^3).

%e 6859^3 == 7779 (mod 10^4).

%e 66859^3 == 77779 (mod 10^5).

%e 566859^3 == 777779 (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 A309645(n)

%o ary = [9]

%o a = 9

%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 A309645(100)

%Y Cf. A309600, A309603.

%K nonn,base

%O 0,1

%A _Seiichi Manyama_, Aug 11 2019