A309642 Digits of the 10-adic integer (7/3)^(1/3).

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

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

Example

       9^3 == 9      (mod 10).
      89^3 == 69     (mod 10^2).
     989^3 == 669    (mod 10^3).
    5989^3 == 6669   (mod 10^4).
   35989^3 == 66669  (mod 10^5).
  235989^3 == 666669 (mod 10^6).

Prog

(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 A309642(n)
  ary = [9]
  a = 9
  n.times{|i|
    b = (a + 3 * a ** 3 - 7) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309642(100)

Crossrefs

Cf. A309600, A309606.

Sequence in context: A243277 A200003 A159590 * A146484 A116695 A109940

Adjacent sequences: A309639 A309640 A309641 * A309643 A309644 A309645

Keyword

nonn

Author

Seiichi Manyama, Aug 11 2019

Status

approved