%I #16 Aug 17 2019 05:06:43

%S 5,0,4,5,5,1,5,4,1,2,0,3,5,3,2,5,0,0,5,4,5,3,0,3,4,2,0,0,1,4,0,0,1,2,

%T 2,3,3,4,5,4,4,5,3,1,1,5,2,0,5,4,2,2,5,4,0,2,5,5,4,2,1,1,4,2,2,4,5,4,

%U 1,5,0,1,0,4,2,1,3,4,0,1,2,0,0,3,3,1,3,4,2,3,4,2,5,0,3,3,0,1,2,0,4

%N Digits of the 6-adic integer (1/5)^(1/5).

%H Seiichi Manyama, <a href="/A309723/b309723.txt">Table of n, a(n) for n = 0..10000</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Hensel%27s_lemma">Hensel's Lemma</a>.

%F Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 5, b(n) = b(n-1) + 5 * (5 * b(n-1)^5 - 1) mod 6^n for n > 1, then a(n) = (b(n+1) - b(n))/6^n.

%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((1/5+O(2^N))^(1/5), 2^N), Mod((1/5+O(3^N))^(1/5), 3^N))), 6), N)

%o (Ruby)

%o def A309723(n)

%o ary = [5]

%o a = 5

%o n.times{|i|

%o b = (a + 5 * (5 * a ** 5 - 1)) % (6 ** (i + 2))

%o ary << (b - a) / (6 ** (i + 1))

%o a = b

%o }

%o ary

%o end

%o p A309723(100)

%Y Digits of the k-adic integer (1/(k-1))^(1/(k-1)): A309722 (k=4), this sequence (k=6), A309724 (k=8), A225464 (k=10).

%Y Cf. A309699.

%K nonn,base

%O 0,1

%A _Seiichi Manyama_, Aug 14 2019