A225464 10-adic integer x such that x^9 = 1/9.

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

Offset

0,1

Comments

This is the 10's complement of A225455.

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) + 9 * (9 * b(n-1)^9 - 1) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 14 2019

Example

       9^9 == -1      (mod 10).
       9^9 == -11     (mod 10^2).
     609^9 == -111    (mod 10^3).
    5609^9 == -1111   (mod 10^4).
   15609^9 == -11111  (mod 10^5).
  415609^9 == -111111 (mod 10^6).

Prog

(PARI) n=0; for(i=1, 100, m=(8*(10^i-1)/9)+1; for(x=0, 9, if(((n+(x*10^(i-1)))^9)%(10^i)==m, n=n+(x*10^(i-1)); print1(x", "); break)))
(PARI) N=100; Vecrev(digits(lift(chinese(Mod((1/9+O(2^N))^(1/9), 2^N), Mod((1/9+O(5^N))^(1/9), 5^N)))), N) \\ Seiichi Manyama, Aug 07 2019
(Ruby)
def A225464(n)
  ary = [9]
  a = 9
  n.times{|i|
    b = (a + 9 * (9 * a ** 9 - 1)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225464(100) # Seiichi Manyama, Aug 14 2019

Crossrefs

Cf. A225455.

Sequence in context: A068467 A372392 A131223 * A296566 A198213 A093766

Adjacent sequences: A225461 A225462 A225463 * A225465 A225466 A225467

Keyword

nonn,base

Author

Aswini Vaidyanathan, May 11 2013

Status

approved