For a combinatorial proof of a symmetric identity

In my paper Supercongruences involving dual sequences [Finite Fields Appl. 46(2017), 179-216], I gave a new symmetric identity which states that if x+y=−1 then \sum_{k=0}^n(-1)^k\binom xk^2\binom{y+z}{n-k}=\sum_{k=0}^n(-1)^k\binom yk^2\binom{x+z}{n-k}. (See (1.17) of the paper.) QUESTION: Is there a combinatorial proof of the above symmetric identity? Answer AttributionSource : Link , Question Author : Zhi-Wei Sun , Answer … Read more

sum calculation

I would like to calculate, or bound from above, the following sum \sum_{i=0}^n(n-2i)^p{p \choose i}, here p\geq 2. Any references are very welcome. Thank you. Answer On of methods for finding an approximation or bound for a finite sum is using the following formula \sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(a) + f(b)}{2} + \sum_{k=1}^\infty … Read more

Approximation a sum involving log and binomial coefficient

I am wondering about the asymptotic approximation of the following expression: S=\sum^{N}_{i=0}\log\Bigg[\binom{\binom{N+1}{i}}{t_i}\Bigg] where t_i=\binom{N}{i}-\binom{N-k}{i-k}+\binom{N-k}{i-1} where k is a positive integer. Also we have that k\ll N. Also for those binomials that have i<k (namely negative) we count them as zero. I am trying to work out the approximation for N \rightarrow \infty. Answer Let B … Read more

Does anyone have ideas about how to simplify this combinatorial expression (mod 2)?

Fix k>0 for ⌈2k+13⌉≤j≤k, 2j−k−1∑i=0(ji)+⌊2k−j2⌋∑b=⌈k+12⌉(2b−k−1∑l=0(bl))(j−b−12j−(2k+1)+b) I think that this expression (mod 2) should be 1 when j=k and 0 otherwise. I can show that it is 1 when j=k and have checked the other cases for relatively small values of k, but have been unable to show in general that when j≠k the expression is … Read more

Partial Sum of the Binomial Theorem [duplicate]

This question already has answers here: Estimating a partial sum of weighted binomial coefficients (3 answers) Closed 7 years ago. The binomial theorem states ∑nk=0Cknrk=(1+r)n. I am interested in the function m∑k=0Cknrk,m<n for fixed n and r, and both m and n are integers. Are there any notable properties for this function? Any literature references? … Read more

How to prove that ∑ni=0(a;q)i(q;q)i(b;q)n−i(q;q)n−ian−i=(ab;q)n(q;q)n\sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}?

By Cauchy identity, {}_1\phi_0(a;—;q,z)=\sum_{n\geq0}\frac{(a;q)_n}{(q;q)_n}z^n=\frac{(az;q)_{\infty}}{(z;q)_\infty},\quad|z|<1,|q|<1, we can obtain the q-analogue of (1-z)^{-a}(1-z)^{-b}=(1-z)^{-a-b}, {}_1\phi_0(a;—;q,z){}_1\phi_0(b;—;q,az)={}_1\phi_0(ab;—;q,z), which is \sum_{n\geq0}\frac{(a;q)_n}{(q;q)_n}z^n\sum_{m\geq0}\frac{(b;q)_m}{(q;q)_m}(az)^m=\sum_{n\geq0}\frac{(ab;q)_n}{(q;q)_n}z^n. Comparing the coefficients of z in the both side of equation above, the identity in the title can be established as \sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}. However, what I really concern is that there is any possible for us to prove this identity directly … Read more

Showing this formula counts these things

I’m writing an article, and I got stuck trying to prove that some numbers are positive. I have a relatively good intuition for guessing what an expression is counting, but in this case I’m not being able to show the formula is right. First of all, let’s introduce this numbers c(ℓ,n,m) which are defined for … Read more

Estimate on sum of squares of multinomial coefficients

I am interested in approximating the sum of the squares of the multinomial coefficients, i.e. apℓ:=∑k0+…+kp=ℓ(ℓ!k0!…kp!)2 or more general, aα0,…,αpℓ:=∑k0+…+kp=ℓ(∏pi=0αkii)2(ℓ!k0!…kp!)2 Here p is prime, and ℓ is an integer smaller then p, ki are non-negative integers. I would like to obtain some simple expression in terms of ℓ and p, a good approximation for large … Read more

Asymptotics of an alternating sum involving the prefix sum of binomial coefficients

Let c>1. Question. What is the asymptotic behaviour of the sum Sn=n∑k=0(−12)k(nk)k∑j=0(cn+kj) as n goes to infinity? I don’t need strong bounds; lim in terms of c will do. Experimentally the limit exists for each c. I tried generating functions, but the prefix-sum-of-binomials term is hard to handle. Also, the dominant term in the alternating … Read more