A309648 Digits of the 10-adic integer (-17/9)^(1/3).

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

Offset: 0

Author

Seiichi Manyama, Aug 11 2019

Keywords

• nonn
• base

Examples

			       3^3 == 7      (mod 10).
      83^3 == 87     (mod 10^2).
     383^3 == 887    (mod 10^3).
    1383^3 == 8887   (mod 10^4).
   21383^3 == 88887  (mod 10^5).
  921383^3 == 888887 (mod 10^6).
		

Links

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

Crossrefs

Cf. A309600

Programs

PARI

N=100; Vecrev(digits(lift(chinese(Mod((-17/9+O(2^N))^(1/3), 2^N), Mod((-17/9+O(5^N))^(1/3), 5^N)))), N)

Ruby

def A309648(n)
  ary = [3]
  a = 3
  n.times{|i|
    b = (a + 3 * (9 * a ** 3 + 17)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309648(100)

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 3 * (9 * b(n-1)^3 + 17) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.