How to express sum as triple summation I am trying to express the following sequences as summations:
$$
1+2^2+3^2+4^4+5^4+6^4+7^4
$$
and 
$$
1+(2+3)^2 + (4+5+6+7)^4
$$
as summations. I think they will likely be triple summations, so something like this:
$$
\sum\sum\sum n^k
$$
but I can't think of what to put on the top and bottom of each sum.
Any help would be appreciated. Thanks!
 A: To me,
it looks like
at each successive iteration,
the number of terms doubles
((1), (2, 3), (4, 5, 6, 7))
and the exponent
doubles 
(1, 2, 4).
To get the elements in the terms,
since
$1+2+4+...+2^n
=2^{n+1}-1
$,
if stage 1 is just $(1)$,
the terms added at stage $n$
are from
$2^{n-1}$
to
$2^n-1$
and the exponent
is $2^{n-1}$.
As a check,
$n=1$ gives $(1)$ with exponent $1$,
$n=2$ gives $(2, 3)$ with exponent $2$,
and
$n=3$ gives $(4, 5, 6, 7)$ with exponent $4$.
So the next stage,
$n=4$,
would be
$(8, 9, 10,11, 12, 13, 14, 15)$
 with exponent $8$.
The the $n$-th stage summation
for the first sum
would be
$\sum_{k=2^{n-1}}^{2^n-1} k^{2^{n-1}}$
and the sum for all stages 
up to $N$
would be
$\sum_{n=1}^N\sum_{k=2^{n-1}}^{2^n-1} k^{2^{n-1}}$.
Since,
if $2^{n-1} \le k \le 2^n-1$,
$n
=\lfloor \log_2 k \rfloor +1
$,
this last sum can be written as
$\sum_{k=1}^{2^N-1} k^{2^{\lfloor \log_2 k \rfloor}}
$.
For the second sum,
arguing similarly,
the total up to stage $N$
would be
$\sum_{n=1}^N\left(\sum_{k=2^{n-1}}^{2^n-1} k \right)^{2^{n-1}}$.
Because
$\begin{array}\\
\sum_{k=2^{n-1}}^{2^n-1} k
&=\frac{(2^n-1)2^n}{2}-\frac{(2^{n-1}-1)2^{n-1}}{2}\\
&=\frac{2^{2n}-2^n-2^{2n-2}+2^{n-1}}{2}\\
&=\frac{3\cdot 2^{2n-2}-2^{n-1}}{2}\\
&=3\cdot 2^{2n-3}-2^{n-2}\\
&=2^{n-2}(3\cdot 2^{n-1}-1)\\
\end{array}
$
the last sum simplifies to
$\sum_{n=1}^N\left(2^{n-2}(3\cdot 2^{n-1}-1)\right)^{2^{n-1}}
=\sum_{n=1}^N2^{(n-2)2^{n-1}}\left(3\cdot 2^{n-1}-1\right)^{2^{n-1}}
$.
A: Interesting question.
$$\large1+(2^2+3^2)+(4^4+5^4+6^4+7^4)=\sum_{r=0}^2\;\sum_{j=2^r}^{2^{r+1}-1}\;j^{2^r}$$
and
$$\large 1+(2+3)^2+(4+5+6+7)^4=\sum_{r=0}^2\;\left(\sum_{j=2^r}^{2^{r+1}-1}\;j\right)^{2^r}$$
(Typeset in large for greater clarity as the equations contain exponents of exponents)
NB - Only a double summation is required, not triple.
A: I'll give a partial answer to the first one. I think that you should ask about the second one in a separate question, but that is my opinion. I say partial answer because I couldn't manage to include that $1$ in the summation, but here it goes:
$$\begin{align} 2^2+3^2+4^4+5^4+6^4+7^4+\cdots &=\sum_{k=2}^3k^2+\sum_{k=4}^7k^4+\sum_{k=7}^12 k^6 +\cdots\\ &=\sum_{k=2}^{2+2\cdot 2 -1}k^{2\cdot 1}+\sum_{k=4}^{4+2\cdot 3-1}k^{2 \cdot 2}+\sum_{k=7}^{7+2\cdot 3 -1} k^{2 \cdot 3}+\cdots \\ &=  \sum_{n \geq 1} \sum_{k=F(n)}^{F(n)+2n-1}k^{2n},\end{align}$$  where $F(n)$ is defined by $F(0) = 1$ and $F(n)= n+F(n-1)$.
