# How can I figure out eqn to this sequence? 8, 10, 14, 22, 38, 70,...

#### Liftamanjaro

##### New member
My sequence starts at 4. Double 4 to get 8. 8 minus 3 to get 5. Double 5 to get 10. 10 minus 3 to get 7 double 7 to get 14 and so forth.

4, 8, 5, 10, 7, 14, 11, 22, 19, 38........

It can be simplified for my specific needs, as I only care about every other number, starting with 8.

8, 10, 14, 22, 38, 70........

I double the difference with every number; since 8 + 2 = 10, 10 + 4 = 14, 14 + 8 = 22, 22 + 16 = 38 and 38 + 32 = 70

a. How can I figure out the equation for this sequence?

b. Is there a way to figure out the equation for the original sequence?

#### Subhotosh Khan

##### Super Moderator
Staff member
My sequence starts at 4. Double 4 to get 8. 8 minus 3 to get 5. Double 5 to get 10. 10 minus 3 to get 7 double 7 to get 14 and so forth.

4, 8, 5, 10, 7, 14, 11, 22, 19, 38........

It can be simplified for my specific needs, as I only care about every other number, starting with 8.

8, 10, 14, 22, 38, 70........

I double the difference with every number; since 8 + 2 = 10, 10 + 4 = 14, 14 + 8 = 22, 22 + 16 = 38 and 38 + 32 = 70

a. How can I figure out the equation for this sequence?

b. Is there a way to figure out the equation for the original sequence?
So we have:

term 1 = 8 = a[SUB]1[/SUB]

term 2 = 10 = 8 + 2 = a[SUB]2 [/SUB]= a[SUB]1[/SUB] + 2 = a[SUB]1[/SUB] + 2[SUP](2-1)[/SUP]

term 3 = 14 = 10 + 4 = a[SUB]3 [/SUB]= a[SUB]2[/SUB] + 2*2 = a[SUB]2[/SUB] + 2[SUP](3 - 1)[/SUP]

term 4 = 22 = 14 + 8 = a[SUB]4 [/SUB]= a[SUB]3[/SUB] + 2[SUP](4 - 1)

[/SUP]Do you see the pattern.....

#### pka

##### Elite Member
My sequence starts at 4. Double 4 to get 8. 8 minus 3 to get 5. Double 5 to get 10. 10 minus 3 to get 7 double 7 to get 14 and so forth.
4, 8, 5, 10, 7, 14, 11, 22, 19, 38........
I like to use recursive definitions:
\displaystyle \begin{align*}a_1&=4 \\\text{if }n\ge 2,~a_n&= (2\cdot a_{n-1}[n \mod 2])+(a_{n-1}-3)[n+1 \mod 2] \end{align*}
For $$\displaystyle \ \ n \ge 1,$$
$$\displaystyle a_n \ = \ 2^{(n - 1)/2} \ + \ 3, \ \ \ if \ \ n \ \ odd$$
$$\displaystyle a_n \ = \ \ \ \ \ \ 2^{n/2} \ + \ 6, \ \ \ if \ \ n \ \ even$$