TITLE: Sum of sum of $k$th power of first $n$ natural numbers.
QUESTION [13 upvotes]: I was working on a problem which involves computation of $k$-th power of first $n$ natural numbers.
Say $f(n) = 1^k+2^k+3^k+\cdots+n^k$ we can compute $f(n)$ by using Faulhaber's Triangle also by fast computation of Bernoulli numbers.
Here we have to compute $G(n)= f(1)+f(2)+f(3)+\cdots+f(n)$
I did some work on this to simplify but unable to come up with any easy equation.
Can someone help me how to compute $G(n)$ efficiently , I can efficiently compute $f(n)$.
Here $n<123456789$
and $k<321$.
Thanks.

REPLY [4 votes]: One more type of answer is to use the Faulhaber-matrix (with a little tweak) and iterate.
Let's denote one vector of consecutive powers of $x$ as $V(x)=[1,x,x^2,x^3,...]^T$ either up to required dimension (see below) or simply assume as of infinite size.
Then the dotproduct with the (tweaked) Faulhabermatrix with that vector gives the vector of sums-of-like powers.                
The top-left of the Faulhabermatrix, with one empty column prepended:
$$ G= \small \begin{bmatrix} 
 . & 1 & . & . & . & . & . & . \\ 
 . & 1/2 & 1/2 & . & . & . & . & . \\ 
 . & 1/6 & 1/2 & 1/3 & . & . & . & . \\ 
 . & . & 1/4 & 1/2 & 1/4 & . & . & . \\ 
 . & -1/30 & . & 1/3 & 1/2 & 1/5 & . & . \\ 
 . & . & -1/12 & . & 5/12 & 1/2 & 1/6 & . \\ 
 . & 1/42 & . & -1/6 & . & 1/2 & 1/2 & 1/7 \\ 
 . & . & 1/12 & . & -7/24 & . & 7/12 & 1/2
 \end{bmatrix} \tag 1
$$
Then $$G \cdot V(x)=S(x) \tag 2$$ where due to Faulhabers'analysis $$S(x)=V(1)+V(2)+V(3)+...+V(x) \\ \qquad \qquad \quad=[x, 1+2+...x,1^2+2^2+...+x^2,...]^T \tag 3$$
Iterating this gives a fairly intuitive method do arrive at the "sum-of-sums" $SS(x)$ 
$$ \begin{array} {} G \cdot S(x) &= SS(x) \\
 &= G\cdot V(1)+ G\cdot V(2)+...+G \cdot V(x)  \\
 &=  V(1)+ \left(V(1)+V(2)\right)+...+\left(V(1)+V(2)+...+V(x)\right) \\
 &=  [1+1+1+...+1, \\
 & \qquad 1+(1+2)+(1+2+3)+...(1+2+3+...+x) , \\
 & \qquad 1+(1+2^2)+(1+2^2+3^2)+...(1+2^2+3^2+...+x^2) , \\
 & \qquad ... , \\
 & \qquad1+(1+2^k)+(1+2^k+3^k)+...(1+2^k+3^k+...+x^k), \\
 & \qquad ...]^T
 \\\end{array} \tag 4$$
This can be rewritten as
$$ G^2 \cdot V(x) = SS(x) \tag 5$$ and thus that square of the Faulhaber-matrix is of interest. We find the top-left of $G^2$ :
$$G^2 = \small \begin{bmatrix} 
 . & 1/2 & 1/2 & . & . & . & . & . \\ 
 . & 1/3 & 1/2 & 1/6 & . & . & . & . \\ 
 . & 1/6 & 5/12 & 1/3 & 1/12 & . & . & . \\ 
 . & 1/30 & 1/4 & 5/12 & 1/4 & 1/20 & . & . \\ 
 . & -1/30 & 1/20 & 1/3 & 5/12 & 1/5 & 1/30 & . \\ 
 . & -1/42 & -1/12 & 1/12 & 5/12 & 5/12 & 1/6 & 1/42 \\ 
 . & 1/42 & -5/84 & -1/6 & 1/8 & 1/2 & 5/12 & 1/7 \\ 
 . & 1/30 & 1/12 & -5/36 & -7/24 & 7/40 & 7/12 & 5/12
 \end{bmatrix} \tag 6
$$ and indeed, the dot-product with $V(3)$ looks like
$$ G^2 \cdot V(3) = SS(3) = \small \begin{bmatrix} 
 6 \\ 
 10 \\ 
 20 \\ 
 46 \\ 
 116 \\ 
 310 \\ 
 860 \\ 
 2446
 \end{bmatrix}  \tag 7
$$
where in the $k$'th row is the required sum with summands to the $k$'th power.    
Your example:
Using dimension $8$ only: 
$$ G^2 \cdot V(123456789) = \Tiny \begin{bmatrix} 
 7620789436823655 \\ 
 313612736252315226397035 \\ 
 19358810860413733959550603308825 \\ 
 1433985988226290537213517145779502623643 \\ 
 118023538007597075768233960814021641290827487705 \\ 
 10407719390614949736958778538255639230572384078931269755 \\ 
 963677720389558291637495115742280128808534343727190032990523625 \\ 
 92534211741854093396678577789655310126697035147135144478399613856675643
 \end{bmatrix}$$
Using dimension $n=322$ (so that the highest exponent is $k=321$ in row $322$)
    n=322
    result = Gp^2 \cdot V(123456789)         

    k  | ss_k(123456789)
    ---!----------------------
    0    7.62078943682E15
    1    3.13612736252E23
    2    1.93588108604E31
    3    1.43398598823E39
    4    1.18023538008E47
...
  317  1.53837082910E2576
  318  1.88735309863E2584
  319  2.31554800916E2592
  320  2.84094533467E2600
  321  3.48562264082E2608


Generalization
That use of the $G$-matrix allows immediate generalization to further iterates by further powers-of-$G$. Even more, we have the simple option to let our sums start at another value than $[1,1,1^2,1^3,...]^T$ for instance
$$ G^2 \cdot \left( V(5)-V(2) \right) = SS(5)-SS(2) = S(5)+S(4)+S(3) = \small \begin{bmatrix} 
 12 \\ 
 31 \\ 
 99 \\ 
 361 \\ 
 1431 \\ 
 6001 \\ 
 26199 \\ 
 117841
 \end{bmatrix} \tag 8
$$

Appendix:
If you want to exercise with this it might be easier to create the matrix $G$ from the Faulhaber-matrix $F$ and $F$ itself as inversion or the reduced Pascalmatrix:
$$ P^* = \small \left[  \begin{array} {r}
 1 & . & . & . & . & . & . & . \\ 
 -1 & 2 & . & . & . & . & . & . \\ 
 1 & -3 & 3 & . & . & . & . & . \\ 
 -1 & 4 & -6 & 4 & . & . & . & . \\ 
 1 & -5 & 10 & -10 & 5 & . & . & . \\ 
 -1 & 6 & -15 & 20 & -15 & 6 & . & . \\ 
 1 & -7 & 21 & -35 & 35 & -21 & 7 & . \\ 
 -1 & 8 & -28 & 56 & -70 & 56 & -28 & 8
 \end{array} \right]
$$
which is easier to create programmatically and then 
$$ F = (P^*)^{-1} \tag 9$$
and $G$ is $F$ with one empty column prepended.