A309609 Digits of the 10-adic integer (23/9)^(1/3).

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

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

Example

       3^3 == 7      (mod 10).
      63^3 == 47     (mod 10^2).
     263^3 == 447    (mod 10^3).
    9263^3 == 4447   (mod 10^4).
   79263^3 == 44447  (mod 10^5).
  679263^3 == 444447 (mod 10^6).

Prog

(PARI) N=100; Vecrev(digits(lift(chinese(Mod((23/9+O(2^N))^(1/3), 2^N), Mod((23/9+O(5^N))^(1/3), 5^N)))), N)
(Ruby)
def A309609(n)
  ary = [3]
  a = 3
  n.times{|i|
    b = (a + 3 * (9 * a ** 3 - 23)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309609(100)

Crossrefs

Cf. A173772, A309600, A309612.

Sequence in context: A099506 A205001 A154204 * A266971 A257106 A360255

Adjacent sequences: A309606 A309607 A309608 * A309610 A309611 A309612

Keyword

nonn,base

Author

Seiichi Manyama, Aug 10 2019

Status

approved