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%I #20 Aug 14 2019 08:34:06
%S 9,0,6,5,1,4,8,8,0,6,4,1,3,1,2,9,1,6,1,5,4,7,3,1,2,7,1,7,4,1,0,1,7,9,
%T 4,9,1,2,5,6,3,3,0,5,5,6,3,7,7,1,9,7,7,6,2,4,7,9,4,4,3,0,7,1,7,2,8,2,
%U 8,9,1,9,3,6,9,1,1,1,3,2,3,2,4,2,5,0,1,8,4,4,7,8,2,9,6,8,3,9,3,0
%N 10-adic integer x such that x^9 = 1/9.
%C This is the 10's complement of A225455.
%H Seiichi Manyama, <a href="/A225464/b225464.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) + 9 * (9 * b(n-1)^9 - 1) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - _Seiichi Manyama_, Aug 14 2019
%e 9^9 == -1 (mod 10).
%e 9^9 == -11 (mod 10^2).
%e 609^9 == -111 (mod 10^3).
%e 5609^9 == -1111 (mod 10^4).
%e 15609^9 == -11111 (mod 10^5).
%e 415609^9 == -111111 (mod 10^6).
%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)))^9)%(10^i)==m,n=n+(x*10^(i-1));print1(x", ");break)))
%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((1/9+O(2^N))^(1/9), 2^N), Mod((1/9+O(5^N))^(1/9), 5^N)))), N) \\ _Seiichi Manyama_, Aug 07 2019
%o (Ruby)
%o def A225464(n)
%o ary = [9]
%o a = 9
%o n.times{|i|
%o b = (a + 9 * (9 * a ** 9 - 1)) % (10 ** (i + 2))
%o ary << (b - a) / (10 ** (i + 1))
%o a = b
%o }
%o ary
%o end
%o p A225464(100) # _Seiichi Manyama_, Aug 14 2019
%Y Cf. A225455.
%K nonn,base
%O 0,1
%A _Aswini Vaidyanathan_, May 11 2013
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