اشترك بالحزمة الذهبية واحصل على وصول غير محدود شمرا أكاديميا
تسجيل مستخدم جديدNumber cubes with consecutive line sums
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We settle the existence of certain anti-magic cubes using combinatorial block designs and graph decompositions to align a handful of small examples.
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For a real number $t$, let $r_ell(t)$ be the total weight of all $t$-large Schr{o}der paths of length $ell$, and $s_ell(t)$ be the total weight of all $t$-small Schr{o}der paths of length $ell$. For constants $alpha, beta$, in this article we derive
recurrence formulae for the determinats of the Hankel matrices $det_{1le i,jle n} (alpha r_{i+j-2}(t) +beta r_{i+j-1}(t))$, $det_{1le i,jle n} (alpha r_{i+j-1}(t) +beta r_{i+j}(t))$, $det_{1le i,jle n} (alpha s_{i+j-2}(t) +beta s_{i+j-1}(t))$, and $det_{1le i,jle n} (alpha s_{i+j-1}(t) +beta s_{i+j}(t))$ combinatorially via suitable lattice path models.
We give an explicit form of Inghams Theorem on primes in the short intervals, and show that there is at least one prime between every two consecutive cubes $xsp{3}$ and $(x+1)sp{3}$ if $loglog xge 15$.
Let $vec{w} = (w_1,dots, w_n) in mathbb{R}^{n}$. We show that for any $n^{-2}leepsilonle 1$, if [#{vec{xi} in {0,1}^{n}: langle vec{xi}, vec{w} rangle = tau} ge 2^{-epsilon n}cdot 2^{n}] for some $tau in mathbb{R}$, then [#{langle vec{xi}, vec{w} ran
gle : vec{xi} in {0,1}^{n}} le 2^{O(sqrt{epsilon}n)}.] This exponentially improves the $epsilon$ dependence in a recent result of Nederlof, Pawlewicz, Swennenhuis, and Wk{e}grzycki and leads to a similar improvement in the parameterized (by the number of bins) runtime of bin packing.
Let $f(n,r)$ denote the maximum number of colourings of $A subseteq lbrace 1,ldots,nrbrace$ with $r$ colours such that each colour class is sum-free. Here, a sum is a subset $lbrace x,y,zrbrace$ such that $x+y=z$. We show that $f(n,2) = 2^{lceil n/2r
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In this paper, we consider the problem about finding out perfect powers in an alternating sum of consecutive cubes. More precisely, we completely solve the Diophantine equation $(x+1)^3 - (x+2)^3 + cdots - (x + 2d)^3 + (x + 2d + 1)^3 = z^p$, where $p
$ is prime and $x,d,z$ are integers with $1 leq d leq 50$.
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