Calculating the Value of $f(m,n)$

[juantheron]

Let $f(m,n) = 3m+n+(m+n)^2.$ Then value of $\displaystyle \sum_{n=0}^{\infty}\;\; \sum_{m=0}^{\infty}2^{-f(m,n)}=$

 

[Krizalid1]

Take a look at $f(n,m).$

 

[CaptainBlack]

Let $f(m,n) = 3m+n+(m+n)^2.$ Then value of $\displaystyle \sum_{n=0}^{\infty}\;\; \sum_{m=0}^{\infty}2^{-f(m,n)}=$

Might I ask where this question comes from?

The sum converges very quickly and can be evaluated numerically with 4 terms of each summation (it is \(\approx 1.33333\), which is very suggestive ... ) to good accuracy.

CB

(why the previous calc got the wrong answer I still have no idea)

 

[Opalg]

I hope Krizalid will not object if I add a bit to his very helpful suggestion. In problems like this, my advice is always to start by looking at what happens for small values of the variables. In this case, if you make a table of the values of $f(m,n)$ for small values of $m$ and $n$, it looks like this:

$$\begin{array}{cc}&\;\;\;\;n \\ \rlap{m} & \begin{array}{c|cccc} &0&1&2&3 \\ \hline 0&0&2&6&12 \\ 1&4&8&14&. \\ 2&10&16&.&. \\ 3&18&.&.&. \end{array} \end{array}$$

Doesn't that suggest something very interesting about the range of the function $f(m,n)$?

 

[Krizalid1]

Might I ask where this question comes from?

It's from a Putnam. I don't remember the year though.