A309601 Digits of the 10-adic integer (53/9)^(1/3).

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

Offset

0,1

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) = 3, b(n) = b(n-1) + 3 * (9 * b(n-1)^3 - 53) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.

Example

       3^3 == 7      (mod 10).
      73^3 == 17     (mod 10^2).
     373^3 == 117    (mod 10^3).
    8373^3 == 1117   (mod 10^4).
   68373^3 == 11117  (mod 10^5).
  768373^3 == 111117 (mod 10^6).

Prog

(PARI) N=100; Vecrev(digits(lift(chinese(Mod((53/9+O(2^N))^(1/3), 2^N), Mod((53/9+O(5^N))^(1/3), 5^N)))), N)
(Ruby)
def A309601(n)
  ary = [3]
  a = 3
  n.times{|i|
    b = (a + 3 * (9 * a ** 3 - 53)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309601(100)

Crossrefs

Cf. A309600.

Sequence in context: A118452 A286090 A225447 * A201903 A133056 A131917

Adjacent sequences: A309598 A309599 A309600 * A309602 A309603 A309604

Keyword

nonn,base

Author

Seiichi Manyama, Aug 09 2019

Status

approved