How to find a mathematical formula for these points?

[transgalactic]

how to find a mathematical formula for these points??

i got these points for this function "b" :
b(0,0)=1
b(1,0)=3
b(1,1)=4
b(2,0)=8

b(0)=1
b(1)=2+1=3
b(2)=4+1+3=8

b(n)=2^n +b(0) +b(1) +b(2) +.. + b(n-2)+b(n-1)

b(n-1)=2^n +b(0) +b(1) +b(2) +.. + b(n-2)

b(n)=b(n-1)+b(n-1) =2*b(n-1)which by the way is wrong because b(1) doesn't equal to 2* b(0)

b(n)=2*b(n-1)how to transform it to the formula
b(n,c)=c+(n-2)*2^(n-1) ?

 

[HallsofIvy]

You have both "b(0,0)" and "b(0)". Is b a function of one variable or two?

Are you just trying to find a function that fits the points (and if so which points: (0,0,0) or (0,0)?) or do you have some other reason to believe that 2^n comes into the equataion?

 

[transgalactic]

every point ends with zero

b(0,0)=b(0)
b(1,0)=b(1)

the reason i put 2^n into the equation is

b(0)=1 b(0)=1
b(1)=2+1=3 b(1)=2+b(0)
b(2)=4+1+3=8 b(2)=4+b(0)+b(1)so each ne member is the sum of the previous members and 2^n

unfortunately i don't know how build a standard formula from that series expression

b(n,c)=c+(n-2)*2^(n-1) ?

 

[HallsofIvy]

So this is NOT a problem of finding a formula that will give those points (there are an infinite number of formulas to fit any finite number of points) but of finding a "closed" formula for that iteration:
b_n= 2ⁿ+ b₀+ b₁+ ...+ b_n-1 with b₀= 1.

 

[transgalactic]

how do i find a "closed" formula for that iteration:
b_n= 2ⁿ+ b₀+ b₁+ ...+ b_n-1 with b₀= 1.

so ill get

b(n)=(n-2)*2^(n-1)

??

 

[transgalactic]

i simplified it into
b(n)=2^(n-1) +2*b(n-1) b0=1

what to do now