A309610 Digits of the 10-adic integer (41/9)^(1/3).

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

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) = 9, b(n) = b(n-1) + 7 * (9 * b(n-1)^3 - 41) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n

Example

       9^3 == 9      (mod 10).
      49^3 == 49     (mod 10^2).
     649^3 == 449    (mod 10^3).
    5649^3 == 4449   (mod 10^4).
   95649^3 == 44449  (mod 10^5).
  695649^3 == 444449 (mod 10^6).

Prog

(PARI) N=100; Vecrev(digits(lift(chinese(Mod((41/9+O(2^N))^(1/3), 2^N), Mod((41/9+O(5^N))^(1/3), 5^N)))), N)
(Ruby)
def A309610(n)
  ary = [9]
  a = 9
  n.times{|i|
    b = (a + 7 * (9 * a ** 3 - 41)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309610(100)

Crossrefs

Cf. A173776, A309600, A309611.

Sequence in context: A139720 A248177 A194826 * A198989 A255013 A088539

Adjacent sequences: A309607 A309608 A309609 * A309611 A309612 A309613

Keyword

nonn,base

Author

Seiichi Manyama, Aug 10 2019

Status

approved