A309606 Digits of the 10-adic integer (-7/3)^(1/3).

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

Offset: 0

Author

Seiichi Manyama, Aug 09 2019

Keywords

• nonn

Examples

			       1^3 == 1      (mod 10).
      11^3 == 31     (mod 10^2).
      11^3 == 331    (mod 10^3).
    4011^3 == 3331   (mod 10^4).
   64011^3 == 33331  (mod 10^5).
  764011^3 == 333331 (mod 10^6).
		

Links

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

Crossrefs

Cf. A309600, A309642.

Programs

Maple

op([1,3],padic:-rootp(x^3+7/3,10,100)); # Robert Israel, Aug 09 2019

PARI

N=100; Vecrev(digits(lift(chinese(Mod((-7/3+O(2^N))^(1/3), 2^N), Mod((-7/3+O(5^N))^(1/3), 5^N)))), N)

Ruby

def A309606(n)
  ary = [1]
  a = 1
  n.times{|i|
    b = (a + 3 * a ** 3 + 7) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309606(100)

Formula

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