A309646 Digits of the 10-adic integer (-71/9)^(1/3).

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

Offset: 0

Author

Seiichi Manyama, Aug 11 2019

Keywords

• nonn
• base

Examples

			       1^3 == 1      (mod 10).
      61^3 == 81     (mod 10^2).
     361^3 == 881    (mod 10^3).
    1361^3 == 8881   (mod 10^4).
   61361^3 == 88881  (mod 10^5).
  761361^3 == 888881 (mod 10^6).
		

Links

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

Crossrefs

Cf. A309600, A309602.

Programs

PARI

N=100; Vecrev(digits(lift(chinese(Mod((-71/9+O(2^N))^(1/3), 2^N), Mod((-71/9+O(5^N))^(1/3), 5^N)))), N)

Ruby

def A309646(n)
  ary = [1]
  a = 1
  n.times{|i|
    b = (a + 7 * (9 * a ** 3 + 71)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309646(100)

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 1, b(n) = b(n-1) + 7 * (9 * b(n-1)^3 + 71) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.