 # Thread: Pattern: 1 2 6 14 27 46 72

1. ## Pattern: 1 2 6 14 27 46 72

What could be the formula for the following sequence of numbers: 1 2 6 14 27 46 72  Reply With Quote

2. ## Re: Pattern: 1 2 6 14 27 46 72

What do you think the formula should look like?
My initial stab at generating the series with code would be (using VB6):
Code:
```Option Explicit

Private Sub Command1_Click()
Dim a As Integer, b As Integer, c As Integer, d As Integer

a = 1
b = 2
Debug.Print a
Debug.Print b

For c = 3 To 7
d = (b - a) + c
a = b
b = b + d
Debug.Print b
Next

End Sub```
I don't know if that helps with determining a formula, or whether the above code could be simplified.  Reply With Quote

3. ## Re: Pattern: 1 2 6 14 27 46 72

First step is to see if the differences of terms gives any insight into any formula used to generate the terms:

1 2 6 14 27 46 72

1st differences: 1 4 8 13 19 26
2nd differences: 3 4 5 6 7
3rd differences: 1 1 1 1

Since the 3rd differences are all equal, that means the formula used to generate the terms is of 3rd degree in the form of:

A(x) = a*x^3 + b*x^2 + c*x + d

You have been given A(1) = 1, A(2) = 2, A(3) = 6, A(4) = 14, ...

Using those given values, you can build a series of simultaneous equations that allow you to solve for the values of the coefficients a, b, c, and d.

a + b + c + d = 1
8*a + 4*b + 2*c + d = 2
27*a + 9*b + 3*c + d = 6
64*a + 16*b + 4*c + d = 14

That's as far as I'll go for now. Good luck!  Reply With Quote

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