A309569 Digits of the 10-adic integer (11/3)^(1/3).

3, 3, 2, 2, 5, 8, 6, 8, 3, 1, 7, 3, 2, 6, 1, 0, 0, 1, 3, 2, 5, 3, 3, 5, 5, 5, 0, 8, 8, 9, 0, 9, 1, 7, 3, 2, 9, 4, 3, 9, 9, 8, 3, 3, 0, 1, 4, 2, 7, 6, 9, 5, 1, 5, 9, 3, 2, 5, 3, 7, 3, 1, 4, 8, 9, 7, 0, 1, 9, 1, 1, 4, 1, 4, 7, 4, 9, 0, 7, 7, 1, 2, 4, 9, 3, 4, 3, 8, 0, 8, 1, 8, 9, 8, 3, 5, 5, 1, 9, 2

Offset: 0

Author

Seiichi Manyama, Aug 10 2019

Keywords

• nonn
• base

Examples

			       3^3 == 7      (mod 10).
      33^3 == 37     (mod 10^2).
     233^3 == 337    (mod 10^3).
    2233^3 == 3337   (mod 10^4).
   52233^3 == 33337  (mod 10^5).
  852233^3 == 333337 (mod 10^6).
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A173766, A309600, A309641.

Programs

PARI

N=100; Vecrev(digits(lift(chinese(Mod((11/3+O(2^N))^(1/3), 2^N), Mod((11/3+O(5^N))^(1/3), 5^N)))), N)

Ruby

def A309569(n)
  ary = [3]
  a = 3
  n.times{|i|
    b = (a + 9 * (3 * a ** 3 - 11)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309569(100)

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 9 * (3 * b(n-1)^3 - 11) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.