%I #23 May 18 2024 00:34:24
%S 1,5,7,8,1,2,4,7,5,3,6,1,0,8,4,7,8,4,5,1,2,5,4,0,0,6,7,6,8,7,1,9,9,1,
%T 8,7,7,0,2,8,3,5,3,5,1,3,5,1,5,8,8,8,9,9,7,7,3,2,7,2,8,3,8,0,8,9,6,6,
%U 6,5,7,8,9,1,2,0,8,9,2,2,1,4,9,3,0,6,6,3,8,7,1,6,3,5,8,9,3,9,0,2,9,1,2,7,4
%N Coefficients in a 10-adic square root of 1.
%C 10-adic integer x=.....86760045215487480163574218751 satisfying x^3=x.
%C A "bug" in the decimal enumeration system: another square root of 1.
%C 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. - Edoardo Gueglio (egueglio(AT)yahoo.it), Jan 28 2004
%C What about the 10-adic square roots of -1, -2, -3, 2, 3, 4, ...? They do not exist, unless the square really is a square (+1, +4, +9, +16, ...). Then there's a nontrivial square root; for example, sqrt(4)=...44002229693692923584436016426479909569025039672851562498. - _Don Reble_, Apr 25 2006
%D K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973.
%H Seiichi Manyama, <a href="/A063006/b063006.txt">Table of n, a(n) for n = 0..9999</a>
%F (a_0 + a_1*10 + a_2*10^2 + a_3*10^3 + ... )^2 = 1 + 0*10 + 0*10^2 + 0*10^3 + ...
%F For n > 0, a(n) = 9 - A091661(n).
%e ...4218751^2 = ...0000001
%t 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. - Edoardo Gueglio (egueglio(AT)yahoo.it), Jan 28 2004
%Y Another 10-adic root of 1 is given by A091661.
%Y Cf. A075693.
%K base,nonn,nice,easy
%O 0,2
%A Robert Lozyniak (11(AT)onna.com), Aug 03 2001
%E More terms from _Vladeta Jovovic_, Aug 11 2001