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Digits of the 10-adic integer (71/9)^(1/3).
3

%I #20 Aug 12 2019 02:40:52

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

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

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

%N Digits of the 10-adic integer (71/9)^(1/3).

%H Seiichi Manyama, <a href="/A309602/b309602.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 - 71) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.

%e 9^3 == 9 (mod 10).

%e 39^3 == 19 (mod 10^2).

%e 639^3 == 119 (mod 10^3).

%e 8639^3 == 1119 (mod 10^4).

%e 38639^3 == 11119 (mod 10^5).

%e 238639^3 == 111119 (mod 10^6).

%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((71/9+O(2^N))^(1/3), 2^N), Mod((71/9+O(5^N))^(1/3), 5^N)))), N)

%o (Ruby)

%o def A309602(n)

%o ary = [9]

%o a = 9

%o n.times{|i|

%o b = (a + 7 * (9 * a ** 3 - 71)) % (10 ** (i + 2))

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

%o a = b

%o }

%o ary

%o end

%o p A309602(100)

%Y Cf. A165247, A309600, A309646.

%K nonn

%O 0,1

%A _Seiichi Manyama_, Aug 09 2019