Creating a summatory list without iteration Let list $S_k$ be an arbitrary list of numbers (may not necessarily be ordered).
List $S_{k+1}$ is created via the cumulative sum of elements from list $S_k$.
For example if $S_k$ = [2,5,7,9] then $S_{k+1}$ = [2,7,14,23]
Is there a way to tell what numbers will be in list $S_n$ with $n>k$ without needing to create all the intermediate lists?
 A: Using $S_k(i)$ to indicate the $i^{th}$ term of $S_k$, then 
$$S_n(j) = \sum_{i \le j} {n-k-1+j-i \choose j-i} S_k(i)$$ so you only need to do weighted sums over the original sequence. 
A: You get the next list by multiplication from the left with a lower triangular matrix $L$ with all $1$'s.
$$
L = \begin{pmatrix}
1 & 0 & \dotsc & 0 \\
\vdots & \ddots & \ddots & \vdots \\
\vdots &  & \ddots & 0 \\
1 & \dotsc & \dotsc & 1
\end{pmatrix}
$$
Then you can quickly find sequence $n$ by computing $L^n$ which can be done fairly quickly, e.g by the square-and-multiply method.
A: This is an answer to the request in some comments to show how to arrive at the formula for the entries of $L^h$ 
Here is an example how we can find the entries of $L^h$ symbolically. I do with the $4 \times 4 $ triangular matrix $L$ 
$$ M(h) = \exp( h \cdot \log (L)) = L^h $$ 
and get for $M$
$$ M(h) = \left[ \begin{array} {rrrr}
 1 & . & . & . \\
 1h & 1 & . & . \\
 \frac 12( 1 h^2+ 1 h) & 1h & 1 & . \\
 \frac 16( 1h^3+ 3 h^2+ 2 h) & \frac 12 (1h^2+ 1h) & 1h & 1
 \end{array} \right]$$
Here a trained eye recognizes the Stirling numbers 1st kind as coefficients at the powers of $h$ and because the structure of the matrix has this constant diagonals it is easy to make the formula for the transfer:
$$ M(h)\cdot S_k = S_{k+h} $$
One more step shows, that the evaluation of the polynomials in the entries leads to binomial numbers, which is a well known property of the Stirling numbers first kind (the vandermonde-matrix LDU-decomposes into the matrices of Stirling-numbers 2st kind and of binomial coefficients and thus reduces by the multiplication with the matrix of Stirling numbers 1'st kind (which is the inverse of the 2nd-kind matrix) to binomials)

I had fun to proceed a bit. Factorizing the smbolic entries, assuming the hypothese that we have always the Stirling numbers 1st kind and the fractional cofactors the reciprocal fatorials give
$$ M(h) = \left[ \begin{array} {llll}
 1 & . & . & . \\
 1(h) & 1 & . & . \\
 \frac 12( h(h+1)) & 1(h) & 1 & . \\
 \frac 16( h(h+1)(h+2)) & \frac 12 ( h(h+1)) & 1(h) & 1
 \end{array} \right]$$
and this gives immediately the binomial coefficients
$$ M(h) = \left[ \begin{array} {cccc}
 1 & . & . & . \\
 \binom{h}{1} & 1 & . & . \\
 \binom{h+1}{2} & \binom{h}{1}  & 1 & . \\
 \binom{h+2}{3} & \binom{h+1}{2} & \binom{h}{1}  & 1
 \end{array} \right]$$

and a routine could solve the problem given the vector S of dimension n in the following way:      
   T = 0*S ;  \\ initialize an empty array of size of S
   b  = 1;    \\ contains the current binomial   
   for(j=1,n,
       for(k=j,n, T[k]+=S[k+1-j]*b);
       b *= (h+j-1)/j;
       );
   return(T);

So we have $n^2/2$ operations by the looping.
