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

%I A091661 #22 Jul 31 2017 08:32:27
%S A091661 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,
%T A091661 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,
%U A091661 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
%N A091661 Coefficients in a 10-adic square root of 1.
%C A091661 10-adic integer x=.....239954784512519836425781249 satisfying x^3 = x.
%C A091661 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.
%H A091661 Seiichi Manyama, <a href="/A091661/b091661.txt">Table of n, a(n) for n = 0..9999</a>
%F A091661 For n>0, a(n) = 9 - A063006(n).
%t A091661 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.
%o A091661 (Ruby)
%o A091661 def A(s, n)
%o A091661   n.times{|i|
%o A091661     m = 10 ** (i + 1)
%o A091661     (0..9).each{|j|
%o A091661       k = j * m + s
%o A091661       if (k ** 2 - k) % (m * 10) == 0
%o A091661         s = k
%o A091661         break
%o A091661       end
%o A091661     }
%o A091661   }
%o A091661   s
%o A091661 end
%o A091661 def A091661(n)
%o A091661   str = (10 ** (n + 1) + A(5, n) - A(6, n)).to_s.reverse
%o A091661   (0..n).map{|i| str[i].to_i}
%o A091661 end
%o A091661 p A091661(100) # _Seiichi Manyama_, Jul 31 2017
%Y A091661 Another 10-adic root of 1 is given by A063006.
%Y A091661 Cf. A018247, A018248.
%K A091661 base,nonn
%O A091661 0,1
%A A091661 Edoardo Gueglio (egueglio(AT)yahoo.it), Jan 28 2004