How to determine the formula to generate following numbers?

[Guest]

Does anyone have any suggestions on how to determine the formula to generate
following number?

1.333, 1.5, 1.667, 1.75, 2, 3
If I insert 1 in cell A1, then 1.333 will be returned in cell B1.
Does anyone have any idea on what formula will generate this rhythm for the
list of numbers?
Thank for any suggestions
Eric

 

[Rick Rothstein \(MVP - VB\)]

Does anyone have any suggestions on how to determine the formula to

generate
following number?

1.333, 1.5, 1.667, 1.75, 2, 3
If I insert 1 in cell A1, then 1.333 will be returned in cell B1.
Does anyone have any idea on what formula will generate this rhythm for
the
list of numbers?

There is none. If your 4th number were 1.833, then there would be a
relationship covering the first 5 numbers, but nothing would handle the jump
from 2 to 3 as the end of your series. On the off chance there is a
mistyping in your sequence, let's look at the first 5 numbers in more
detail. Although you stop the numbers at 3 decimal places, the .333 and .667
are obviously shortenings of the fractions 1/3 and 2/3. Let look at these
numbers as mixed numbers (whole number plus fraction). They are...

1-1/3 1-1/2 1-2/3 1-3/4 2

Changing them to pure fractions, they become...

4/3 3/2 5/3 7/4 2/1

Lets reformat them so they each have a denominator of 12. The changed
fractions become...

16/12 18/12 20/12 21/12 24/12

Look at the progression of the numerators... 16, 18, 20, 21, 24. If that 4th
number were 22 instead of 21, then we would have a series that increases by
2. But, in order for that to happen, the 4th number would have to be 22/12
and not 21/12. The fraction 22/12 is, in decimal form, 1.833 rounded to 3
decimal places. But even with that, there is no way to account for the huge
jump from 2 to 3 from your original sequence. The number 3 is the same as
36/12 and its numerator, 36, is 12 greater than 2's numerator when converted
to a numerator of 12, not the 26 one would expect if the series were to
continue.

Hence, there is no formula relationship that can cover the progression of
numbers you showed originally.

Rick

 

[Guest]

Rick: I went through the same analysis you did. Is Eric looking for an exact
results. Maybe if Eric plotted the data and then added a trend line he could
get an approximate result.

A 3rd order polynomial give a good approximation

y = 0.0417x3 - 0.3575x2 + 1.043x + 0.582

 

[Rick Rothstein \(MVP - VB\)]

Rick: I went through the same analysis you did. Is Eric looking for an

exact
results. Maybe if Eric plotted the data and then added a trend line he
could
get an approximate result.

A 3rd order polynomial give a good approximation

y = 0.0417x3 - 0.3575x2 + 1.043x + 0.582

You may be right, but my gut tells me there are probably several other
polynomials, of completely different shapes, that could also pass through or
near those points. Perhaps if Eric provides more details (for example, as to
where these numbers come from or what they are supposed to be modeling),
then maybe we can make better guesses for him.

Rick

 

[Guest]

Thank everyone for suggestions
Those number is based on Arithmetic rhythm
1.333, 1.500, 1.667, 1.750, 2.000, 3.000
There is the given list from articles, and not much description about it.
Does anyone have any idea about the structure for this Arithmetic rhythm?
Thank for any suggestions
Eric

 

[Rick Rothstein \(MVP - VB\)]

Can you provide a link for a web page showing that sequence of numbers? I
did a Google search on them (and the term "arithmetic rhythm" and could find
nothing.

Rick

 

[Dana DeLouis]

1.333, 1.5, 1.667, 1.75, 2, 3

I didn't find any sequence in a lookup, even for variations.
One remote database find was this "off the wall guess."

If n^(n + 2) + (n + 1)^(n + 3) is Prime, return (n+15)/12

Let's test it (using another program)

Table[If[PrimeQ[n^(n+2)+(n+1)^(n+3)],(n+15)/12.,],{n,21}]

{1.3333333333333333,,1.5,,1.6666666666666665,1.75,,,2.,,,,,,,,,,,,3.}

Yeah. I know. it's a wild guess!!
The next two solutions would be 3.25, and 4.1666666666

 

[Rick Rothstein \(MVP - VB\)]

1.333, 1.5, 1.667, 1.75, 2, 3

I didn't find any sequence in a lookup, even for variations.
One remote database find was this "off the wall guess."

If n^(n + 2) + (n + 1)^(n + 3) is Prime, return (n+15)/12

Let's test it (using another program)

Table[If[PrimeQ[n^(n+2)+(n+1)^(n+3)],(n+15)/12.,],{n,21}]

{1.3333333333333333,,1.5,,1.6666666666666665,1.75,,,2.,,,,,,,,,,,,3.}

Yeah. I know. it's a wild guess!!
The next two solutions would be 3.25, and 4.1666666666

Given the complexity of the prime test.. I'm impressed. Unfortunately, I
don't think this method will qualify as the "formula" Eric was looking for.

Rick

 

[Dana DeLouis]

If I select the data table, another program I have converted it to the
following function.
However, it's not valid outside the range 1-6. It's just designed to remove
a lookup table.
I'm sure this is not what the op wanted either.

Function Fx(n)
Fx = (215376 Mod (n + 31)) / 12
End Function
'= = = = = = = = =

Sub TestIt()
Dim j
For j = 1 To 6
Debug.Print j; Fx(j)
Next j
End Sub

1 1.33333333333333
2 1.5
3 1.66666666666667
4 1.75
5 2
6 3
--
HTH :>)
Dana DeLouis
Windows XP & Excel 2007

<snip>