A322931 Digits of the 8-adic integer 3^(1/3).

3, 7, 5, 0, 7, 3, 0, 1, 7, 1, 7, 6, 4, 2, 3, 6, 7, 7, 7, 0, 3, 1, 2, 0, 1, 7, 2, 6, 1, 2, 5, 4, 1, 1, 1, 2, 3, 5, 5, 2, 3, 5, 4, 7, 3, 6, 0, 0, 3, 4, 7, 1, 3, 3, 6, 4, 6, 0, 0, 4, 4, 6, 0, 5, 6, 4, 1, 5, 5, 6, 0, 0, 0, 6, 2, 6, 1, 0, 7, 1, 6, 0, 0, 4, 6, 5, 0, 7, 0, 1, 3, 7, 3, 7, 0, 0, 2, 0, 6, 1

Offset: 0

Author

Patrick A. Thomas, Dec 31 2018

Keywords

• nonn
• base
• easy

Comments

The octal version of A225404.

Examples

			4671710370573^3 == 3 (mod 8^13) in octal.
		

Links

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

Crossrefs

Cf. A225404 (decimal version), A290563, A309698, A322932, A322933.

Programs

PARI

N=100; Vecrev(digits(lift((3+O(2^(3*N)))^(1/3)), 8), N) \\ Seiichi Manyama, Aug 14 2019

Ruby

def A322931(n)
  ary = [3]
  a = 3
  n.times{|i|
    b = (a + 5 * (a ** 3 - 3)) % (8 ** (i + 2))
    ary << (b - a) / (8 ** (i + 1))
    a = b
  }
  ary
end
p A322931(100) # Seiichi Manyama, Aug 14 2019

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 5 * (b(n-1)^3 - 3) mod 8^n for n > 1, then a(n) = (b(n+1) - b(n))/8^n. - Seiichi Manyama, Aug 14 2019