A225410 10-adic integer x such that x^3 = 7/9.

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

Offset

0,1

Comments

This is the 10's complement of A225401.

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) = 7, b(n) = b(n-1) + 3 * (9 * b(n-1)^3 - 7) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 13 2019

Example

       7^3 == 3      (mod 10).
      47^3 == 23     (mod 10^2).
     247^3 == 223    (mod 10^3).
    9247^3 == 2223   (mod 10^4).
   39247^3 == 22223  (mod 10^5).
  339247^3 == 222223 (mod 10^6).

Prog

(PARI) n=0; for(i=1, 100, m=(2*(10^i-1)/9)+1; for(x=0, 9, if(((n+(x*10^(i-1)))^3)%(10^i)==m, n=n+(x*10^(i-1)); print1(x", "); break)))
(PARI) N=100; Vecrev(digits(lift(chinese(Mod((7/9+O(2^N))^(1/3), 2^N), Mod((7/9+O(5^N))^(1/3), 5^N)))), N) \\ Seiichi Manyama, Aug 05 2019
(Ruby)
def A225410(n)
  ary = [7]
  a = 7
  n.times{|i|
    b = (a + 3 * (9 * a ** 3 - 7)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225410(100) # Seiichi Manyama, Aug 13 2019

Crossrefs

Cf. A225401, A309600.

Sequence in context: A245055 A335020 A377796 * A248750 A071875 A200687

Adjacent sequences: A225407 A225408 A225409 * A225411 A225412 A225413

Keyword

nonn,base

Author

Aswini Vaidyanathan, May 07 2013

Status

approved