A309603 Digits of the 10-adic integer (-11/9)^(1/3).

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

Offset

0,2

Links

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

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 + 11) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.

Example

       1^3 == 1      (mod 10).
      41^3 == 21     (mod 10^2).
     141^3 == 221    (mod 10^3).
    3141^3 == 2221   (mod 10^4).
   33141^3 == 22221  (mod 10^5).
  433141^3 == 222221 (mod 10^6).

Prog

(PARI) N=100; Vecrev(digits(lift(chinese(Mod((-11/9+O(2^N))^(1/3), 2^N), Mod((-11/9+O(5^N))^(1/3), 5^N)))), N)
(Ruby)
def A309603(n)
  ary = [1]
  a = 1
  n.times{|i|
    b = (a + 7 * (9 * a ** 3 + 11)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309603(100)

Crossrefs

Cf. A165402, A309600, A309645.

Sequence in context: A177951 A222130 A061903 * A328611 A084118 A046071

Adjacent sequences: A309600 A309601 A309602 * A309604 A309605 A309606

Keyword

nonn

Author

Seiichi Manyama, Aug 09 2019

Status

approved