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A225409 Digits of the 10-adic integers (-9)^(1/3).

%I A225409 #34 Aug 13 2019 08:36:06
%S A225409 1,3,4,9,3,3,6,5,1,3,3,9,5,1,4,5,4,0,5,4,8,8,0,5,9,3,9,1,8,6,2,9,3,3,
%T A225409 0,5,1,6,0,0,6,9,7,5,7,9,6,4,0,1,3,4,4,9,0,3,2,2,5,1,9,2,5,3,8,9,6,7,
%U A225409 0,1,4,1,7,8,4,2,9,0,9,0,1,1,8,3,9,3,1,3,9,6,0,4,9,0,0,4,3,4,6,2
%N A225409 Digits of the 10-adic integers (-9)^(1/3).
%C A225409 This is the 10's complement of A225406.
%H A225409 Seiichi Manyama, <a href="/A225409/b225409.txt">Table of n, a(n) for n = 0..10000</a>
%F A225409 Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 1, 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 A225409        1^3 == -9 (mod 10).
%e A225409       31^3 == -9 (mod 10^2).
%e A225409      431^3 == -9 (mod 10^3).
%e A225409     9431^3 == -9 (mod 10^4).
%e A225409    39431^3 == -9 (mod 10^5).
%e A225409   339431^3 == -9 (mod 10^6).
%o A225409 (PARI) n=0; for(i=1, 100, m=(10^i-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 A225409 (PARI) Vecrev(digits(truncate((-9+O(10^100))^(1/3)))) \\ _Seiichi Manyama_, Aug 04 2019
%o A225409 (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 A225409 (Ruby)
%o A225409 def A225409(n)
%o A225409   ary = [1]
%o A225409   a = 1
%o A225409   n.times{|i|
%o A225409     b = (a + 3 * (a ** 3 + 9)) % (10 ** (i + 2))
%o A225409     ary << (b - a) / (10 ** (i + 1))
%o A225409     a = b
%o A225409   }
%o A225409   ary
%o A225409 end
%o A225409 p A225409(100) # _Seiichi Manyama_, Aug 13 2019
%Y A225409 Cf. A309600,
%Y A225409 Digits of 10-adic integers:
%Y A225409 A153042 ((-1/9)^(1/3));
%Y A225409 A225406 (     9^(1/3));
%Y A225409 A225412 ( (1/9)^(1/3)).
%K A225409 nonn,base
%O A225409 0,2
%A A225409 _Aswini Vaidyanathan_, May 07 2013