A309699 Digits of the 6-adic integer 5^(1/5).

5, 4, 0, 3, 1, 5, 0, 0, 3, 3, 2, 1, 3, 0, 0, 3, 4, 3, 1, 1, 1, 1, 1, 4, 3, 4, 0, 5, 3, 1, 1, 5, 3, 3, 0, 2, 2, 2, 5, 3, 5, 5, 2, 5, 2, 2, 2, 3, 4, 2, 0, 5, 4, 3, 3, 2, 0, 0, 4, 1, 1, 5, 5, 5, 0, 0, 1, 4, 3, 5, 4, 5, 1, 5, 5, 0, 5, 4, 0, 4, 4, 4, 4, 3, 4, 4, 0, 4, 3, 4, 0, 5, 4, 4

Offset: 0

Author

Seiichi Manyama, Aug 13 2019

Keywords

• nonn
• base

Comments

x = ...513045,

x^2 = ...433521,

x^3 = ...051525,

x^4 = ...354241,

x^5 = ...000005.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..10000
• Wikipedia, Hensel's Lemma.

Crossrefs

Digits of the k-adic integer (k-1)^(1/(k-1)): A309698 (k=4), this sequence (k=6), A309700 (k=8), A225458 (k=10).

Cf. A309448.

Programs

PARI

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

Ruby

def A309699(n)
  ary = [5]
  a = 5
  n.times{|i|
    b = (a + a ** 5 - 5) % (6 ** (i + 2))
    ary << (b - a) / (6 ** (i + 1))
    a = b
  }
  ary
end
p A309699(100)

Formula

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