A225458 10-adic integer x such that x^9 = 9.

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

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

Example

       9^9 == 9 (mod 10).
      89^9 == 9 (mod 10^2).
     289^9 == 9 (mod 10^3).
    1289^9 == 9 (mod 10^4).
   21289^9 == 9 (mod 10^5).
  921289^9 == 9 (mod 10^6).

Prog

(PARI) n=0; for(i=1, 100, m=9; 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((9+O(2^N))^(1/9), 2^N), Mod((9+O(5^N))^(1/9), 5^N)))), N) \\ Seiichi Manyama, Aug 06 2019
(Ruby)
def A225458(n)
  ary = [9]
  a = 9
  n.times{|i|
    b = (a + a ** 9 - 9) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225458(100) # Seiichi Manyama, Aug 14 2019

Crossrefs

Digits of the k-adic integer (k-1)^(1/(k-1)): A309698 (k=4), A309699 (k=6), A309700 (k=8), this sequence (k=10).

Sequence in context: A094135 A021897 A378205 * A092172 A133619 A276499

Adjacent sequences: A225455 A225456 A225457 * A225459 A225460 A225461

Keyword

nonn,base

Author

Aswini Vaidyanathan, May 11 2013

Status

approved