

A244658


a(n) = x/(x1floor(x1)) where x = sqrt(n)  floor(sqrt(n)), x1 = 1/x, a(n) = 1 if division by zero, a(n) = 0 for nonintegers.


2



1, 1, 2, 1, 1, 2, 0, 4, 1, 1, 2, 3, 0, 0, 6, 1, 1, 2, 0, 4, 0, 0, 0, 8, 1, 1, 2, 0, 0, 5, 0, 0, 0, 0, 10, 1, 1, 2, 3, 4, 0, 6, 0, 0, 0, 0, 0, 12, 1, 1, 2, 0, 0, 0, 0, 7, 0, 0, 0, 0, 0, 0, 14, 1, 1, 2, 0, 4, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 0, 16, 1, 1, 2, 3, 0, 0, 6, 0, 0, 9, 0, 0, 0, 0, 0, 0, 0, 0, 18, 1, 1, 2, 0, 4, 5
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OFFSET

1,3


COMMENTS

Inspired by the square root of three poem in "Harold and Kumar" the movie. This operation can be done in iterations for which some n seem to give all integer results after iteration k >= 2. Some of them have periodic properties similar to that of the continued fraction method (but not exactly the same). When a(n) is arranged as a table read by rows, the row sums would be A062731. See illustrations in links.


LINKS



EXAMPLE

For n = 3, x = sqrt(3)  floor(sqrt(3)) = 0.732050807..., x1 = 1/x = 1/0.732050807... = 1.366025403..., x1  floor(x1) = 0.366025403..., a(3) = 0.732050807.../0.366025402... = 2.


PROG

(Small Basic)
For n=1 To 500
x=math.Power(n, .5)
y=xmath.Floor(x)
If y<>0 Then
x1=1/y
x2=1/(x1math.Floor(x1))
a1=x2/x1
a2=a1math.floor(a1)
If a2 > 0.999999 or a2 < 0.0000001 then
a=math.Round(a1)
TextWindow.Write(a+", ")
Else
TextWindow.Write(0+", ")'noninteger
Endif
Else
TextWindow.Write(1+", ")'zero division, Square number
EndIf
EndFor


CROSSREFS



KEYWORD

sign


AUTHOR



STATUS

approved



