A309611 Digits of the 10-adic integer (-41/9)^(1/3).

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

Offset: 0

Author

Seiichi Manyama, Aug 10 2019

Keywords

• nonn
• base

Examples

			       1^3 == 1      (mod 10).
      51^3 == 51     (mod 10^2).
     351^3 == 551    (mod 10^3).
    4351^3 == 5551   (mod 10^4).
    4351^3 == 55551  (mod 10^5).
  304351^3 == 555551 (mod 10^6).
		

Links

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

Crossrefs

Cf. A173804, A309600, A309610.

Programs

PARI

N=100; Vecrev(digits(lift(chinese(Mod((-41/9+O(2^N))^(1/3), 2^N), Mod((-41/9+O(5^N))^(1/3), 5^N)))), N)

Ruby

def A309611(n)
  ary = [1]
  a = 1
  n.times{|i|
    b = (a + 7 * (9 * a ** 3 + 41)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309611(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 + 41) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n