I have been trying to find the sum of this series. It is an inclusion-exclusion formula. I have tried representing it as coefficient of [$x^{m}$] in a polynomial, but could not find the polynomial or m. Is there some other way to find it?

The series is:

$${ S }=\sum _{ i=0 }^{ n } { (-1) }^{ i }\binom{n}{i} \binom{N-i*k-1}{n-1}$$

where $${n} = \lceil{N/k}\rceil$$

I think that it needs to be utilized to find the sum, but so far I have not been able to use it.

Update : The original problem is to find the number of restricted compositions of a number N given that each summand is a positive integer less than equal to b and the number of parts is minimal.

The minimum number of parts ${k}=\lceil{N/b}\rceil$. Let us denote the solution as $C(N, k, 1, b)$ where ${ N }=\sum _{ i=0 }^{ k }t_i$ and ${1}\le{t_i}\le{b}$. The formula $${C(N, k, 1, b)}=\sum _{ i=0 }^{ k } { (-1) }^{ i }\binom{k}{i} \binom{N-ib-1}{n-1}$$ can be derived by using inclusion exclusion principle. The first term means that there is no upper bound on the summand, then subtract the number of compositions which have atleast one term greater than b, then add those terms(due to overlapping) which have atleast two terms greater b and so on. The solution assumes that $\binom{n}{r} = 0$, if ${n}<{r}$ even if n is negative. I want to minimize the asymptotic complexity of calculating this formula.

  • $\begingroup$ Suggestion: find out how many terms you have with negative top argument in the binomial coefficient, then use my result to write those with positive argument in terms of those with a negative one (there don't seem to be that many negative ones). $\endgroup$ – Chappers May 15 '17 at 17:34
  • $\begingroup$ @Chappers I get that. Thank you. Can it be improved further? I am new to this community. I apologize if I am asking too much. $\endgroup$ – abc xyz May 15 '17 at 18:35
  • $\begingroup$ I don't think there are general forms for such sums that will be useful to you (sums of binomial coefficients generically involve hypergeometric functions). More helpful is probably working out precisely where $N-ik$ becomes smaller than $n$. $\endgroup$ – Chappers May 16 '17 at 2:38

The answer is zero, for slightly strange reasons. The binomial coefficient $$ \binom{ai+b}{r} $$ is a polynomial in $i$ of degree $r$. One can show that $$ \sum_{i=0}^n (-1)^i \binom{n}{i} P(i) = 0 $$ for any polynomial $P$ of degree strictly less than $n$. Since $\binom{N-ik-1}{n-1}$ is a polynomial in $i$ of degree $n-1<n$, it follows that $S=0$.

  • $\begingroup$ The answer is not 0. I have many counter examples. eg. Take N = 2, k = 2. Therefore n = 1. Now when i = 0, value of the expression is 1 and when i = 1, value of the expression is 0. So the sum = 1 in this case. $\endgroup$ – abc xyz May 14 '17 at 23:51
  • $\begingroup$ When $i=1,n=1$, $ (-1)^i \binom{n}{i} = -n = -1$, and for $N=k=2$ as well, $ \binom{N-ik-1}{n-1} = \binom{2-2-1}{0} = 1 $ since $\binom{a}{0}=1$ for any $a$. So the $i=1$ term is $-1$, not $0$. $\endgroup$ – Chappers May 15 '17 at 0:02
  • $\begingroup$ But isn't $\binom{n}{r}=0, {n} < {r}$ ? As another example take N = 7 and k = 3, for which n = 3. When i = 0, term is $\binom{3}{0}\binom{7-0-1}{3-1} = {15}$, for i = 1, the term is ${-1}*\binom{3}{1}\binom{7-3-1}{3-1} = {-9}$, for i = 2, the term is $\binom{3}{2}\binom{7-6-1}{3-1} = {0}$, for i = 3, the term is ${-1}*\binom{3}{3}\binom{7-9-1}{3-1} = {0}$. Therefore the sum is 6 and not 0. $\endgroup$ – abc xyz May 15 '17 at 7:08
  • $\begingroup$ Ah, I see the problem. While $\binom{n}{r}=0$ for $r>n$ if $n$ is a nonnegative integer, this is not true in general. For example, $\binom{-1}{r} = \frac{(-1)(-2)\dotsm(-1-r+1)}{r!} = (-1)^r$ for any $r \geq 0$. (Recall that the binomial coefficient $\binom{a}{r}$ is defined as $$ \frac{a(a-1)\dotsm (a-r+1)}{r!}, $$ which equals $ \frac{a!}{r!(a-r)!} $ only when $a$ is a nonnegative integer.) $\endgroup$ – Chappers May 15 '17 at 14:54
  • $\begingroup$ I didn't knew that n has to be non-negative to follow that rule. It seems that I didn't consider that while deriving the formula. The above mentioned formula is actually answer to a combinatorics problem. I am stating the original problem and my approach as a update in the question. Please suggest if anything is still wrong. $\endgroup$ – abc xyz May 15 '17 at 15:43

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