MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
0

1

Given a number of length $l$, the sum of the digits range from $0$ to $9l$

For each sum, there are $x$ permutations of digits. find the sequence of length $9l+1$ that solves $x$.

I found a function that solves this problem without expanding the polynomial equation $l$ times:

$f(n \in S,l,b) = \sum\limits_{i=0}^{\lfloor\frac{n}{b}\rfloor}(-1)^{i}{l \choose i}{n+l-1-bi \choose l-1}$

($l=l$, $b=10$,$S={0,1,...,9l}$)

Where $f(n,l,b)$ is the $n^{th}$ coefficient of a polynomial of length $l$ and base $b$ where $S={0,1,...,(b-1)l}$

(thanks to Tom De Medts for giving me the proper terms.)

Are there other functions that give the $n^{th}$ polynomial coefficient? I can't seem to find one.

flag
3 
My first impression is that this question would fit better on math.stackexchange.com, but perhaps I am missing some subtlety or connection to more advanced themes – Yemon Choi Feb 18 2011 at 8:36

1 Answer

7

Your sequences are simply the coefficients of the polynomials $$(1 + x + x^2 + \dots + x^9)^L .$$

(This is a straightforward application of the theory of generating functions.)

link|flag
I am willing to believe you, but does your argument still hold after examining past the first 10 numbers in the sequence, when the summation starts to be used? I added a latex that I think is the closed form. – mna Feb 18 2011 at 16:12
1 
Yes, the argument is always correct. For each digit, the possible outcomes are 0,1,...,9, so the generating function for each digit is $1 + x + x^2 + \dots + x^9$. Since you are asking for the number of possible strings consisting of $L$ digits and summing up to a given number $t$, the answer is precisely the coefficient of $x^t$ in the product of these $L$ generating functions. – Tom De Medts Feb 18 2011 at 16:31
You have convinced me thanks! Now taking a side-step, is there a simpler function than my f(t,L)? – mna Feb 18 2011 at 17:02

Your Answer

Get an OpenID
or

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