A063006 Coefficients in a 10-adic square root of 1.

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, 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, 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

Offset

0,2

Comments

10-adic integer x=.....86760045215487480163574218751 satisfying x^3=x.

A "bug" in the decimal enumeration system: another square root of 1.

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

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

References

K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973.

Links

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

Formula

(a_0 + a_1*10 + a_2*10^2 + a_3*10^3 + ... )^2 = 1 + 0*10 + 0*10^2 + 0*10^3 + ...

For n > 0, a(n) = 9 - A091661(n).

Example

...4218751^2 = ...0000001

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

Crossrefs

Another 10-adic root of 1 is given by A091661.

Cf. A075693.

Sequence in context: A266712 A021177 A091662 * A135096 A153104 A233527

Adjacent sequences: A063003 A063004 A063005 * A063007 A063008 A063009

Keyword

base,nonn,nice,easy

Author

Robert Lozyniak (11(AT)onna.com), Aug 03 2001

Extensions

More terms from Vladeta Jovovic, Aug 11 2001

Status

approved