Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
Fantastic answer, thank you – user51819 Dec 31 '12 at 19:32
@Amzoti: Yes: all the non-zero terms in the powers of his matrix are these binomial coefficients - as Marvis now seems to have noted – Henry Dec 31 '12 at 19:34
@Amzoti As I have indicated in the comment to WimC answer, we can prove that $L^k(i,j) = \dbinom{k-1+i-j}{k-1}$ – user17762 Dec 31 '12 at 19:35
@Amzoti: the entries of $L^m$ are polynomials using the unsigned Stirling numbers first kind. If they are evaluated for some $m$ they result in binomials leading to Henry's formula. – Gottfried Helms Dec 31 '12 at 19:36
@Amzoti I can imagine that another link is: how do you efficiently compute these weights? – WimC Dec 31 '12 at 19:39

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.

share|cite|improve this answer
Is there any way to do it by iterating through the positions of S_n instead of needing a large matrix? – user51819 Dec 31 '12 at 19:24
You don't actually need a large matrix, since all entries along a diagonal in $L^n$ are equal. – WimC Dec 31 '12 at 19:27
In fact $L^k(i,j) = \dbinom{k-1+i-j}{k-1}$ – user17762 Dec 31 '12 at 19:34

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(k=j,n, T[k]+=S[k+1-j]*b);
       b *= (h+j-1)/j;

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

share|cite|improve this answer
Is there a way to do it in better than $n^2/2$ operations? – user51819 Dec 31 '12 at 20:50
@user51819: hmm, I don't think that we can get below that $n^2/2$ scalar operations. If we however measure in terms of array/matrix-operations, the inner loop can be counted as one vector-operation, giving then an order of $n$-operations, and so on, but I think this is only (and even only remotely) interesting in the unlikely case that we deal with huge arrays $S$ and a near-hardware implementation of matrix-operations makes them similar to scalar operations in respect to their time-consumtion. (But I've no expertise for that latter question) – Gottfried Helms Dec 31 '12 at 21:01
@user51819 : to reduce the computing time of pre-multiplication of a vector by a lower triangular matrix below the magical $n^2/2$ scalar operations you might look for something like "karatsuba" multiplication (don't have the name at hand) and discrete fast fourier transformation. I think I remember vaguely that those methods help to break that lower limit. – Gottfried Helms Dec 31 '12 at 21:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.