A309647 Digits of the 10-adic integer (-53/9)^(1/3).

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

Offset: 0

Author

Seiichi Manyama, Aug 11 2019

Keywords

• nonn
• base

Examples

			       7^3 == 3      (mod 10).
      27^3 == 83     (mod 10^2).
     627^3 == 883    (mod 10^3).
    1627^3 == 8883   (mod 10^4).
   31627^3 == 88883  (mod 10^5).
  231627^3 == 888883 (mod 10^6).
		

Links

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

Crossrefs

Cf. A309600, A309601.

Programs

PARI

N=100; Vecrev(digits(lift(chinese(Mod((-53/9+O(2^N))^(1/3), 2^N), Mod((-53/9+O(5^N))^(1/3), 5^N)))), N)

Ruby

def A309647(n)
  ary = [7]
  a = 7
  n.times{|i|
    b = (a + 3 * (9 * a ** 3 + 53)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309647(100)

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 7, b(n) = b(n-1) + 3 * (9 * b(n-1)^3 + 53) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.