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Digits of the 10-adic integer (43/9)^(1/3).
3

%I #18 Aug 12 2019 02:40:18

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

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

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

%N Digits of the 10-adic integer (43/9)^(1/3).

%H Seiichi Manyama, <a href="/A309604/b309604.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) = 3, b(n) = b(n-1) + 3 * (9 * b(n-1)^3 - 43) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.

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

%e 3^3 == 27 (mod 10^2).

%e 603^3 == 227 (mod 10^3).

%e 8603^3 == 2227 (mod 10^4).

%e 58603^3 == 22227 (mod 10^5).

%e 58603^3 == 222227 (mod 10^6).

%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((43/9+O(2^N))^(1/3), 2^N), Mod((43/9+O(5^N))^(1/3), 5^N)))), N)

%o (Ruby)

%o def A309604(n)

%o ary = [3]

%o a = 3

%o n.times{|i|

%o b = (a + 3 * (9 * a ** 3 - 43)) % (10 ** (i + 2))

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

%o a = b

%o }

%o ary

%o end

%o p A309604(100)

%Y Cf. A309600, A309644.

%K nonn

%O 0,1

%A _Seiichi Manyama_, Aug 09 2019