A(b) = A(b/2) + A(b/4) for b is an even power of 4 > 0
A(1) = 1.
Paying well for explanation.

This post was edited by MsRailgun on Oct 5 2013 05:38pm

just dont forget about dre

3.

does not compute, please rephrase.

not a good form of rephrasing the problem
without any further information the op does not allow to provide you with a solution
so please spell out the complete question you have been asked to answer

if you wrote the question wrong or mistook parts of the question for something else, and there's no solution is probably why you're asking for help lol

recheck +1

This post was edited by Sefira on Oct 5 2013 07:48pm

It's not 'recurrence'. The phrase you want is mathematical induction.

To prove something by mathematical induction, you need to first prove the "base case", meaning the very lowest number that fits the definition given in your problem. What I can't understand, though, is what you mean by "b is an even power of 4 > 0".

The second step is to first assume that you have some arbitrary (and valid) value for b that the equation works for. Assume A(k) == A(k/2) + A(k/4). Now from there, prove that the equation is also true for A("k+1"). I said k+1, but in reality that depends on how you're defining your allowed b values. If it's even powers of 4, then assume A(k) and prove A(16k) is also true. It's whatever the next highest allowed value for b is. For even powers of 4, that means you would have to multiple by 4^2 to get to the next even power of 4.

I think the OP is describing a recurrence relation though...

and I don't think it's a proof; the problem is to find the A function probably

this is what i think as well^^^
but I can't quite understand this part: