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For a natural number $n$ and non-negative integers $i,j$ I want to evaluate:

$$\sum_{k=n\left(i+j\right)+j}2^k$$

For this, I need to know how many possible combinations of $i$ and $j$ can give a certain value $k$, but I can't find any solution. Where $n$ is some natural number constant and $i$ and $j$ can get any non-negative integer value. Obviously, $k{\geq}n$ or $k=0$.

I've played a little with the numbers, and I reached the expression $\left\lfloor\frac{k}{2^n}\right\rfloor+1$ which seems to be an upper limit on the value, but I can't prove it analytically, see graph below.

enter image description here

Number indicates the value of $n$, "upper lim" is the graph of $\left\lfloor\frac{k}{2^n}\right\rfloor+1$

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Since $$k=n(i+j)+j\iff \frac{k-ni}{n+1}=j$$ for any given $k$ and $n$, the maximum possible value of $j$ occurs when $i=0$ and it is $\frac{k}{n+1}$.

Since $j$ must be a nonnegative integer, it follows that $j$ will lie in $\{0,1,2,\dots,\left\lfloor \frac{k}{n+1}\right \rfloor\}$.

We perform the substitution $p=i+j$. It will suffice to find the number of pairs of values of $p$ and $j$ s.t. $k=np+j$. Since $k$ is fixed, it is enough that we find the number of possible values of $p$ (for which constraints on both $p$ and $j$ are satisfied).

Since $j\in \{0,1,2,\dots,\left\lfloor \frac{k}{n+1}\right \rfloor\}$, $k-j\in \{k,k-1,k-2,\dots,k-\left\lfloor\frac{k}{n+1}\right\rfloor\}$. But $k-j=np$ assuming that $j$ and $p$ satisfy the equation.

This means that the number of possible values of $p$ (i.e. the number of pairs of values for $i$ and $j$) equals the number of multiples of $n$ in $\{k,k-1,k-2,\dots,k-\left\lfloor\frac{k}{n+1}\right\rfloor\}$.

This gives us that the number of pairs of values of $i$ and $j$ satisfying the conditions (for nonzero $k$) is $$\left\lfloor\frac{k}{n}\right\rfloor-\left\lfloor\frac{k-\left\lfloor\frac{k}{n+1}\right\rfloor-1}{n}\right\rfloor$$

I think this can be simplified to $$\left\lfloor\frac{k}{n}\right\rfloor-\left\lfloor\frac{k-1}{n+1}\right\rfloor$$

(I'm not entirely sure why this is well-bounded by $\left\lfloor\frac{k}{2^n}\right\rfloor$+1, but I did notice that the 'number-of-pairs' function grows very slowly with $k$ - perhaps the large denominator of $2^n$ is a significant limiting factor in the size of $\left\lfloor\frac{k}{2^n}\right\rfloor$+1, rendering its growth comparable to that of the 'number-of-pairs' function initially but for very large values the upper bound and the function grow disparate.)

Note: If there's anything wrong or confusing in my answer, feel free to edit it (and comment please) or just comment

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  • $\begingroup$ Where does the $-1$ in $\left\lfloor\frac{k-\left\lfloor\frac{k}{n+1}\right\rfloor-1}{n}\right\rfloor$ come from? $\endgroup$ – SIMEL Jan 30 at 22:39
  • $\begingroup$ Say it was just $\left\lfloor\frac{k-\left\lfloor\frac{k}{n+1}\right\rfloor}{n}\right\rfloor$. In $\left\lfloor\frac{k}{n}\right\rfloor-\left\lfloor\frac{k-\left\lfloor\frac{k}{n+1}\right\rfloor}{n}\right\rfloor$, we'd be subtracting the no. of multiples of $n+1$ $\leq k-\left\lfloor\frac{k}{n+1}\right\rfloor$ from the no. that are $\leq k$. But this means that even if $k-\left\lfloor\frac{k}{n+1}\right\rfloor$ was a multiple of $n$ it wouldn't be counted, even though the interval $\{k,k-1,k-2,\dots,k-\left\lfloor\frac{k}{n+1}\right\rfloor\}$ contains $k-\left\lfloor\frac{k}{n+1}\right\rfloor$ $\endgroup$ – Cardioid_Ass_22 Jan 31 at 11:28

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