A309643 Digits of the 10-adic integer (-61/9)^(1/3).

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

Offset: 0

Author

Seiichi Manyama, Aug 11 2019

Keywords

• nonn
• base

Examples

			       1^3 == 1      (mod 10).
      91^3 == 71     (mod 10^2).
     491^3 == 771    (mod 10^3).
    9491^3 == 7771   (mod 10^4).
   49491^3 == 77771  (mod 10^5).
  549491^3 == 777771 (mod 10^6).
		

Links

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

Crossrefs

Cf. A309600, A309605.

Programs

PARI

N=100; Vecrev(digits(lift(chinese(Mod((-61/9+O(2^N))^(1/3), 2^N), Mod((-61/9+O(5^N))^(1/3), 5^N)))), N)

Ruby

def A309643(n)
  ary = [1]
  a = 1
  n.times{|i|
    b = (a + 7 * (9 * a ** 3 + 61)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309643(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 + 61) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.