# Tag Info

## New answers tagged binomial-coefficients

3

I have obtained that 3603: 184939, -110384, -170771 4482: 125663, 56483, -129355 Now $$\binom{x}{3}+\binom{y}{3}+\binom{z}{3}$$ can represent all integers among $0,\ldots,5000$.

1

It follows from so-called global central limit theorems (see e.g. Corollary 1, part 1), on page 2 of "Download preview PDF.") that $$\bigg|E\Big|\frac Xn-p\Big|-\sqrt\frac2\pi\,\sqrt{\frac{pq}n}\bigg|<\frac Cn$$ for some absolute real constant $C$ and all natural $n$.

2

This is the mean absolute deviation (MAD) for a binomial distribution, divided by $n$. The expectation is hence $$2 \, (1-p)^{n+1-\lceil np \rceil} \, p^{\lceil np \rceil} \, \binom{n-1}{\lceil np \rceil-1} \;.$$ See this paper (Berend & Kontorovich 2013, doi: 10.1016/j.spl.2013.01.023) for bounds and a reference for the above expression.

1

Here is a confident guess, without a proof. The normal approximation of the binomial distribution gives the estimate $$f(n,p) = \frac{\sqrt{2pq}}{\sqrt{\pi n}}.$$ Now, experimentally, for any fixed $n$ the maximum of $$\frac{\mathbb{E}\, |X/n-p\,|}{f(n,p)}$$ occurs at $p=\frac1{2n}$, where it equals $$c(n) = 2^{-n+1/2} (2n-1)^{n-1/2} n^{-n+1/2}\sqrt{\... 10 As a supplement to @dario2994's result: I have obtained 2613: 27874,-17441,-25379 3337: 60083,-20882,-59229 3362: 20543,19711,-25367 4447: 105313,-35617,-103935 and any solution to \binom{x}{3}+\binom{y}{3}+\binom{z}{3}=n where n=1523,3603,4482 must have |x|,|y|,|z|>50000. The searching method is pretty simple: we have$$ \binom{x}{3}+\binom{y}{...

6

Here are the representations (in a succinct form) of all the numbers in your list but $1523$. 522: -3132 -2827 3766 523: -1296 -576 1335 622: -3401 679 3394 633: -2446 -1646 2675 642: -2710 -2127 3093 843: -2851 -1596 3011 863: -6866 -1313 6884 918: -6033 479 6034 1013: -6891 1087 6884 1458: -1001 230 999 1878: -1961 -225 1964 1983: -3601 2244 3286 A row ...

13

By Cauchy–Bunyakovsky–Schwarz inequality we have $$\left(\sum \prod_i \frac{\binom{a_i}{b_i}}{\binom{A-a_i}{B-b_i}}\right)\left(\sum\prod_i \binom{a_i}{b_i}\binom{A-a_i}{B-b_i}\right)\geqslant \left(\sum \prod_i\binom{a_i}{b_i}\right)^2=\binom{A}{B}^2 .$$ Thus it suffices to prove that $$\sum\prod_i \binom{a_i}{b_i}\binom{A-a_i}{B-b_i}\leqslant \binom{A}... 1 Here is a simple proof of your divisibility relation. It suffices to show that$$\frac{\mathrm{lcm}(1,2,\dots,n+j)}{\mathrm{lcm}(1,2,\dots,n)}\quad\text{divides}\quad\binom{n+j}{n}\quad\text{for}\quad 0\leq j\leq n.$$That is, for any prime p and for 0\leq j\leq n, we have that$$\lfloor\log_p(n+j)\rfloor-\lfloor\log_p(n)\rfloor\leq\sum_{k=1}^\infty \...

9

2-4-6-8, this we don't appreciate! There is a long tradition of exploring additive representations of integers by polynomial sequences. The gold standard for measuring such progress is Waring's problem, but the circle method of Hardy and Littlewood to understand such problems works in many related situations. In particular, one could use the circle ...

0

The Mathematica command Sum[k*Binomial[4 g + 2, 2 k], {k, 0, g}]//FullSimplify performs $$16^g g-\frac{\Gamma (4 g+3) \, _3F_2\left(2,1-g,\frac{3}{2}-g;g+\frac{5}{2},g+3;1\right)}{\Gamma (2 g-1) \Gamma (2 g+5)}.$$

1

According to Maple, $$S_g = \left( g + \frac12\right) \left(16^g - {4 g \choose 2g}\right)$$

1

one sum can be carried out: exchanging the order of summation, $$T(N,K)=\sum_{j=2}^K\sum_{i=j}^K(-1)^{i-j}\binom{i}{j}\frac{j^{N+1}-1}{j-1}=$$ $$=\sum_{j=2}^K\frac{j^{N+1}-1}{j-1}\left[\frac{1}{2^{j+1}}+(-1)^{K-j} \binom{K+1}{j} \, _2F_1(1,K+2;K+2-j;-1)\right]$$

8

Not an answer - but I decided to delete a prior comment and repost as an answer, because I think it puts the 2-4-6-8 conjecture in a different light than considered so far, hopefully leading to some of the experts providing deeper insights and perhaps even a solution. So the stunning fact (to me) is this: 10 numbers up to $5\times 10^{10}$, besides $0$, ...

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