A091661 Coefficients in a 10-adic square root of 1.

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

Offset

0,1

Comments

10-adic integer x=.....239954784512519836425781249 satisfying x^3 = x.

Let a,b be integers defined in A018247, A018248 satisfying a^2=a, b^2=b, obviously a^3=a, b^3=b; let c,d,e,f be integers defined in A091661, A063006, A091663, A091664 then c^3=c, d^3=d, e^3=e, f^3=f, c+d=1, a+e=1, b+f=1, b+c=a, d+f=e, a+f=c, a=f+1, b=e+1, cd=-1, af=-1, gh=-1 where -1=.....999999999.

Links

Seiichi Manyama, Table of n, a(n) for n = 0..9999

Formula

For n>0, a(n) = 9 - A063006(n).

Mathematica

To calculate c, d, e, f use Mathematica algorithms for a, b and equations: c=a-b, d=1-c, e=b-1, f=a-1.

Prog

(Ruby)
def A(s, n)
  n.times{|i|
    m = 10 ** (i + 1)
    (0..9).each{|j|
      k = j * m + s
      if (k ** 2 - k) % (m * 10) == 0
        s = k
        break
      end
    }
  }
  s
end
def A091661(n)
  str = (10 ** (n + 1) + A(5, n) - A(6, n)).to_s.reverse
  (0..n).map{|i| str[i].to_i}
end
p A091661(100) # Seiichi Manyama, Jul 31 2017

Crossrefs

Another 10-adic root of 1 is given by A063006.

Cf. A018247, A018248.

Sequence in context: A330274 A248197 A199291 * A362943 A011313 A319530

Adjacent sequences: A091658 A091659 A091660 * A091662 A091663 A091664

Keyword

base,nonn

Author

Edoardo Gueglio (egueglio(AT)yahoo.it), Jan 28 2004

Status

approved