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%I #15 Aug 12 2019 02:39:03
%S 1,9,4,9,4,5,2,8,0,8,3,9,0,1,4,2,8,9,2,6,8,9,0,4,8,5,0,0,4,2,0,6,9,8,
%T 8,0,9,8,5,8,5,9,3,5,5,8,1,9,9,8,2,3,0,8,4,6,1,8,5,7,3,2,8,6,6,0,1,9,
%U 5,4,6,2,7,4,7,2,4,4,5,3,8,9,9,7,7,6,7,9,2,6,5,7,2,2,8,9,6,8,9,0
%N Digits of the 10-adic integer (-61/9)^(1/3).
%H Seiichi Manyama, <a href="/A309643/b309643.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) = 1, b(n) = b(n-1) + 7 * (9 * b(n-1)^3 + 61) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.
%e 1^3 == 1 (mod 10).
%e 91^3 == 71 (mod 10^2).
%e 491^3 == 771 (mod 10^3).
%e 9491^3 == 7771 (mod 10^4).
%e 49491^3 == 77771 (mod 10^5).
%e 549491^3 == 777771 (mod 10^6).
%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((-61/9+O(2^N))^(1/3), 2^N), Mod((-61/9+O(5^N))^(1/3), 5^N)))), N)
%o (Ruby)
%o def A309643(n)
%o ary = [1]
%o a = 1
%o n.times{|i|
%o b = (a + 7 * (9 * a ** 3 + 61)) % (10 ** (i + 2))
%o ary << (b - a) / (10 ** (i + 1))
%o a = b
%o }
%o ary
%o end
%o p A309643(100)
%Y Cf. A309600, A309605.
%K nonn,base
%O 0,2
%A _Seiichi Manyama_, Aug 11 2019
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