A309640 Digits of the 10-adic integer (-17/3)^(1/3).

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

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) + 3 * b(n-1)^3 + 17 mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.

Example

       1^3 == 1      (mod 10).
      21^3 == 61     (mod 10^2).
     821^3 == 661    (mod 10^3).
    3821^3 == 6661   (mod 10^4).
   73821^3 == 66661  (mod 10^5).
  173821^3 == 666661 (mod 10^6).

Prog

(PARI) N=100; Vecrev(digits(lift(chinese(Mod((-17/3+O(2^N))^(1/3), 2^N), Mod((-17/3+O(5^N))^(1/3), 5^N)))), N)
(Ruby)
def A309640(n)
  ary = [1]
  a = 1
  n.times{|i|
    b = (a + 3 * a ** 3 + 17) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309640(100)

Crossrefs

Cf. A309570, A309600.

Sequence in context: A008908 A050077 A261715 * A185576 A256921 A133840

Adjacent sequences: A309637 A309638 A309639 * A309641 A309642 A309643

Keyword

nonn,base

Author

Seiichi Manyama, Aug 11 2019

Status

approved