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%I #54 Aug 13 2019 08:36:44
%S 9,2,5,1,1,7,1,3,6,2,6,3,3,8,2,1,4,1,0,2,7,1,2,2,4,6,1,6,0,1,0,1,2,7,
%T 2,8,2,8,8,3,6,7,0,7,7,7,2,2,6,2,6,9,9,6,8,1,3,2,1,5,4,3,7,4,7,7,6,9,
%U 6,1,4,0,2,0,9,6,3,6,6,1,9,1,9,9,7,4,9,8,8,7,7,3,0,8,7,7,8,8,0,8
%N Digits of the 10-adic integer (1/9)^(1/3).
%C This is the 10's complement of A153042.
%C Equivalently, the 10-adic cube root of 1/9, i.e., x such that 9*x^3 = 1 (mod 10^n) for all n. - _M. F. Hasler_, Jan 02 2019
%H Seiichi Manyama, <a href="/A225412/b225412.txt">Table of n, a(n) for n = 0..10000</a>
%F p = ...711529, q = A153042 = ...288471, 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) + 7 * (9 * b(n-1)^3 - 1) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - _Seiichi Manyama_, Aug 13 2019
%e 9^3 == -1 (mod 10).
%e 29^3 == -11 (mod 10^2).
%e 529^3 == -111 (mod 10^3).
%e 1529^3 == -1111 (mod 10^4).
%e 11529^3 == -11111 (mod 10^5).
%e 711529^3 == -111111 (mod 10^6).
%p op([1,3],padic:-rootp(9*x^3 -1, 10, 101)); # _Robert Israel_, Aug 04 2019
%o (PARI) n=0; for(i=1, 100, m=(8*(10^i-1)/9)+1; 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) upto(N=100, m=1/3)=Vecrev(digits(lift(chinese(Mod((1/9+O(5^N))^m, 5^N), Mod((1/9+O(2^N))^m, 2^N)))), N) \\ Following _Andrew Howroyd_'s code for A319740. - _M. F. Hasler_, Jan 02 2019
%o (PARI) Vecrev(digits(truncate(-(-1/9+O(10^100))^(1/3)))) \\ _Seiichi Manyama_, Aug 04 2019
%o (Ruby)
%o def A225412(n)
%o ary = [9]
%o a = 9
%o n.times{|i|
%o b = (a + 7 * (9 * a ** 3 - 1)) % (10 ** (i + 2))
%o ary << (b - a) / (10 ** (i + 1))
%o a = b
%o }
%o ary
%o end
%o p A225412(100) # _Seiichi Manyama_, Aug 13 2019
%Y Cf. A309600, A319740 (10-adic cube root of 1/11).
%Y Digits of 10-adic integers:
%Y A153042 ((-1/9)^(1/3));
%Y A225406 ( 9^(1/3));
%Y A225409 ( (-9)^(1/3)).
%K nonn,base
%O 0,1
%A _Aswini Vaidyanathan_, May 07 2013
%E Name edited by _Seiichi Manyama_, Aug 04 2019
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