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Digits of the 10-adic integer 9^(1/3).
8

%I #42 Aug 13 2019 08:35:22

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

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

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

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

%H Robert Israel, <a href="/A225406/b225406.txt">Table of n, a(n) for n = 0..10000</a>

%F p = ...660569, q = A225409 = ...339431, p + q = 0. - _Seiichi Manyama_, Aug 04 2019

%F Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 9, b(n) = b(n-1) + 3 * (b(n-1)^3 - 9) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - _Seiichi Manyama_, Aug 13 2019

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

%e 69^3 == 9 (mod 10^2).

%e 569^3 == 9 (mod 10^3).

%e 569^3 == 9 (mod 10^4).

%e 60569^3 == 9 (mod 10^5).

%e 660569^3 == 9 (mod 10^6).

%p op([1,3],padic:-rootp((x)^3 -9, 10, 101)); # _Robert Israel_, Aug 04 2019

%o (PARI) n=0; for(i=1, 100, m=9; for(x=0, 9, if(((n+(x*10^(i-1)))^3)%(10^i)==m, n=n+(x*10^(i-1)); print1(x", "); break)))

%o (PARI) Vecrev(digits(truncate(-(-9+O(10^100))^(1/3)))) \\ _Seiichi Manyama_, Aug 04 2019

%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((9+O(2^N))^(1/3), 2^N), Mod((9+O(5^N))^(1/3), 5^N)))), N) \\ _Seiichi Manyama_, Aug 04 2019

%o (Ruby)

%o def A225406(n)

%o ary = [9]

%o a = 9

%o n.times{|i|

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

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

%o a = b

%o }

%o ary

%o end

%o p A225406(100) # _Seiichi Manyama_, Aug 13 2019

%Y Cf. A309600.

%Y Digits of 10-adic integers:

%Y A153042 ((-1/9)^(1/3));

%Y A225409 ( (-9)^(1/3));

%Y A225412 ( (1/9)^(1/3)).

%K nonn,base

%O 0,1

%A _Aswini Vaidyanathan_, May 07 2013