Does anyone have any suggestions on how to determine the formula for
following series?

The 1 st number is n
Ther 2 nd number is Power(Sqrt(n)+2,2),
The 3 rd number is Power(Sqrt(Power(Sqrt(n)+2,2))+2,2), then substitute the
previous value into the same formula format.
The 4 th number is Power(Sqrt(Power(Sqrt(Power(Sqrt(n)+2,2))+2,2))+2, 2),
The 5 th number is Power(Sqrt([Previous value])+2,2),

Does anyone have any suggestions on how to determine this repeated structure
in a formula? therefore, if the any number is given, then I can determine the
result directly rather than calculate a list of values in order to get the
result.
Thank in advance for any suggestions
Eric

An example might help because I am confused by then substitute the
previous value into the same formula format.

n=2
first = 3
second =
etc
best wishes
--
Bernard V Liengme
Microsoft Excel MVP
www.stfx.ca/people/bliengme
remove caps from email

"Eric" wrote in message
...
Does anyone have any suggestions on how to determine the formula for
following series?

The 1 st number is n
Ther 2 nd number is Power(Sqrt(n)+2,2),
The 3 rd number is Power(Sqrt(Power(Sqrt(n)+2,2))+2,2), then substitute
the
previous value into the same formula format.
The 4 th number is Power(Sqrt(Power(Sqrt(Power(Sqrt(n)+2,2))+2,2))+2, 2),
The 5 th number is Power(Sqrt([Previous value])+2,2),

Does anyone have any suggestions on how to determine this repeated
structure
in a formula? therefore, if the any number is given, then I can determine
the
result directly rather than calculate a list of values in order to get the
result.
Thank in advance for any suggestions
Eric

Try this small UDF:

Function eric(v As Variant, n As Integer) As Double
Dim tw As Double
tw = 2
eric = v
For i = 1 To n
eric = (eric ^ 0.5 + tw) ^ tw
Next
End Function

So
=eric(3,2) displays
32.85640646

--
Gary''s Student - gsnu200745


"Eric" wrote:

Does anyone have any suggestions on how to determine the formula for
following series?

The 1 st number is n
Ther 2 nd number is Power(Sqrt(n)+2,2),
The 3 rd number is Power(Sqrt(Power(Sqrt(n)+2,2))+2,2), then substitute the
previous value into the same formula format.
The 4 th number is Power(Sqrt(Power(Sqrt(Power(Sqrt(n)+2,2))+2,2))+2, 2),
The 5 th number is Power(Sqrt([Previous value])+2,2),

Does anyone have any suggestions on how to determine this repeated structure
in a formula? therefore, if the any number is given, then I can determine the
result directly rather than calculate a list of values in order to get the
result.
Thank in advance for any suggestions
Eric

Does anyone have any suggestions on how to determine this repeated
structure

I believe this is correct:

Function Fx(n, k)
' n Number
' k Number of iterations

Fx = (2 * k + Sqrt(n) - 2) ^ 2
End Function

--
HTH :)
Dana DeLouis
Windows XP & Excel 2007


"Gary''s Student" wrote in message
...
Try this small UDF:

Function eric(v As Variant, n As Integer) As Double
Dim tw As Double
tw = 2
eric = v
For i = 1 To n
eric = (eric ^ 0.5 + tw) ^ tw
Next
End Function

So
=eric(3,2) displays
32.85640646

--
Gary''s Student - gsnu200745


"Eric" wrote:

Does anyone have any suggestions on how to determine the formula for
following series?

The 1 st number is n
Ther 2 nd number is Power(Sqrt(n)+2,2),
The 3 rd number is Power(Sqrt(Power(Sqrt(n)+2,2))+2,2), then substitute
the
previous value into the same formula format.
The 4 th number is Power(Sqrt(Power(Sqrt(Power(Sqrt(n)+2,2))+2,2))+2, 2),
The 5 th number is Power(Sqrt([Previous value])+2,2),

Does anyone have any suggestions on how to determine this repeated
structure
in a formula? therefore, if the any number is given, then I can determine
the
result directly rather than calculate a list of values in order to get
the
result.
Thank in advance for any suggestions
Eric

Oops. Try this instead.

Function Fx(n, k)
' n Number
' k Number of iterations

Fx = (2 * k + Sqr(n) - 2) ^ 2
End Function

--
Dana DeLouis
Windows XP & Excel 2007


"Gary''s Student" wrote in message
...
Try this small UDF:

Function eric(v As Variant, n As Integer) As Double
Dim tw As Double
tw = 2
eric = v
For i = 1 To n
eric = (eric ^ 0.5 + tw) ^ tw
Next
End Function

So
=eric(3,2) displays
32.85640646

--
Gary''s Student - gsnu200745


"Eric" wrote:

Does anyone have any suggestions on how to determine the formula for
following series?

The 1 st number is n
Ther 2 nd number is Power(Sqrt(n)+2,2),
The 3 rd number is Power(Sqrt(Power(Sqrt(n)+2,2))+2,2), then substitute
the
previous value into the same formula format.
The 4 th number is Power(Sqrt(Power(Sqrt(Power(Sqrt(n)+2,2))+2,2))+2, 2),
The 5 th number is Power(Sqrt([Previous value])+2,2),

Does anyone have any suggestions on how to determine this repeated
structure
in a formula? therefore, if the any number is given, then I can determine
the
result directly rather than calculate a list of values in order to get
the
result.
Thank in advance for any suggestions
Eric

The 1 st number is n

Hi. I had k=1 return the original number. To match Gary's solution, just
drop the -2

Function Fx(n, k)
' n Start Number
' k Number of iterations
'= = = = = = = = = = = = = =
Fx = (2 * k + Sqr(n)) ^ 2
'= = = = = = = = = = = = = =
End Function

Hence:
?fx(3,2)
32.856406460551

As a worksheet function
=POWER(2 * B1 + SQRT(A1),2)

(A1 = Number, B1 = # of Iterations)
--
HTH :)
Dana DeLouis


"Gary''s Student" wrote in message
...
Try this small UDF:

Function eric(v As Variant, n As Integer) As Double
Dim tw As Double
tw = 2
eric = v
For i = 1 To n
eric = (eric ^ 0.5 + tw) ^ tw
Next
End Function

So
=eric(3,2) displays
32.85640646

--
Gary''s Student - gsnu200745


"Eric" wrote:

Does anyone have any suggestions on how to determine the formula for
following series?

The 1 st number is n
Ther 2 nd number is Power(Sqrt(n)+2,2),
The 3 rd number is Power(Sqrt(Power(Sqrt(n)+2,2))+2,2), then substitute
the
previous value into the same formula format.
The 4 th number is Power(Sqrt(Power(Sqrt(Power(Sqrt(n)+2,2))+2,2))+2, 2),
The 5 th number is Power(Sqrt([Previous value])+2,2),

Does anyone have any suggestions on how to determine this repeated
structure
in a formula? therefore, if the any number is given, then I can determine
the
result directly rather than calculate a list of values in order to get
the
result.
Thank in advance for any suggestions
Eric

Thank everyone for suggestions
Eric

"Dana DeLouis" wrote:

The 1 st number is n

Hi. I had k=1 return the original number. To match Gary's solution, just
drop the -2

Function Fx(n, k)
' n Start Number
' k Number of iterations
'= = = = = = = = = = = = = =
Fx = (2 * k + Sqr(n)) ^ 2
'= = = = = = = = = = = = = =
End Function

Hence:
?fx(3,2)
32.856406460551

As a worksheet function
=POWER(2 * B1 + SQRT(A1),2)

(A1 = Number, B1 = # of Iterations)
--
HTH :)
Dana DeLouis


"Gary''s Student" wrote in message
...
Try this small UDF:

Function eric(v As Variant, n As Integer) As Double
Dim tw As Double
tw = 2
eric = v
For i = 1 To n
eric = (eric ^ 0.5 + tw) ^ tw
Next
End Function

So
=eric(3,2) displays
32.85640646

--
Gary''s Student - gsnu200745


"Eric" wrote:

Does anyone have any suggestions on how to determine the formula for
following series?

The 1 st number is n
Ther 2 nd number is Power(Sqrt(n)+2,2),
The 3 rd number is Power(Sqrt(Power(Sqrt(n)+2,2))+2,2), then substitute
the
previous value into the same formula format.
The 4 th number is Power(Sqrt(Power(Sqrt(Power(Sqrt(n)+2,2))+2,2))+2, 2),
The 5 th number is Power(Sqrt([Previous value])+2,2),

Does anyone have any suggestions on how to determine this repeated
structure
in a formula? therefore, if the any number is given, then I can determine
the
result directly rather than calculate a list of values in order to get
the
result.
Thank in advance for any suggestions
Eric