|
%I #18 Aug 12 2019 10:17:01
%S 9,1,9,8,8,1,3,3,5,8,3,9,6,0,0,9,0,6,1,9,2,8,3,4,4,7,9,1,1,5,3,2,0,1,
%T 6,9,3,2,9,2,5,9,4,0,0,4,7,9,3,2,1,0,2,1,2,7,8,7,9,2,5,1,1,5,6,3,9,3,
%U 1,7,8,5,7,1,3,2,9,4,2,5,0,2,2,4,1,5,4,0,4,2,1,5,2,0,5,5,6,2,0,8
%N Digits of the 10-adic integer (31/9)^(1/3).
%H Seiichi Manyama, <a href="/A309614/b309614.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) + 7 * (9 * b(n-1)^3 - 31) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.
%e 9^3 == 9 (mod 10).
%e 19^3 == 59 (mod 10^2).
%e 919^3 == 559 (mod 10^3).
%e 8919^3 == 5559 (mod 10^4).
%e 88919^3 == 55559 (mod 10^5).
%e 188919^3 == 555559 (mod 10^6).
%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((31/9+O(2^N))^(1/3), 2^N), Mod((31/9+O(5^N))^(1/3), 5^N)))), N)
%o (Ruby)
%o def A309614(n)
%o ary = [9]
%o a = 9
%o n.times{|i|
%o b = (a + 7 * (9 * a ** 3 - 31)) % (10 ** (i + 2))
%o ary << (b - a) / (10 ** (i + 1))
%o a = b
%o }
%o ary
%o end
%o p A309614(100)
%Y Cf. A309595, A309600.
%K nonn,base
%O 0,1
%A _Seiichi Manyama_, Aug 10 2019
|
