|
%I #22 Aug 12 2019 10:17:05
%S 3,9,6,9,6,6,2,5,6,6,5,7,4,8,2,6,1,5,2,3,5,4,9,5,1,2,1,5,3,6,9,7,1,6,
%T 7,5,5,3,7,6,4,9,5,3,1,7,9,8,0,4,4,5,7,1,2,1,3,3,4,9,1,0,1,7,6,1,9,8,
%U 0,4,1,3,7,6,7,1,2,2,0,1,1,5,4,9,2,5,2,8,9,7,5,0,9,1,4,5,4,9,7,3
%N Digits of the 10-adic integer (13/9)^(1/3).
%H Seiichi Manyama, <a href="/A309613/b309613.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) = 3, b(n) = b(n-1) + 3 * (9 * b(n-1)^3 - 13) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.
%e 3^3 == 7 (mod 10).
%e 93^3 == 57 (mod 10^2).
%e 693^3 == 557 (mod 10^3).
%e 9693^3 == 5557 (mod 10^4).
%e 69693^3 == 55557 (mod 10^5).
%e 669693^3 == 555557 (mod 10^6).
%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((13/9+O(2^N))^(1/3), 2^N), Mod((13/9+O(5^N))^(1/3), 5^N)))), N)
%o (Ruby)
%o def A309613(n)
%o ary = [3]
%o a = 3
%o n.times{|i|
%o b = (a + 3 * (9 * a ** 3 - 13)) % (10 ** (i + 2))
%o ary << (b - a) / (10 ** (i + 1))
%o a = b
%o }
%o ary
%o end
%o p A309613(100)
%Y Cf. A178769, A309600, A309608.
%K nonn,base
%O 0,1
%A _Seiichi Manyama_, Aug 10 2019
|
