Combination of solutionsby Gerry Leversha, St. Paul's School, London, England, and Kenneth M. Wilke, Topeka, KS, USA. Let n, b and c be the sides of the triangle. Since the perimeter n+ b+ c = 4n, if b < n then c > 2n, contradicting the triangle inequality. Hence we can take n < b < c < 2n. Furthermore we can assume that n, b and c are coprime; for, if (n; b) = d > 1 say, then djc since c = 3n ...

The analysis and probability theory on fractals underwent rapid development in last twenty years, see [1, 11, 27, 28] and the references therein. Diffusion processes were constructed for the Sierpiński triangle [4, 15, 21] and more generally for some simple nested fractals [17] and Sierpiński carpets [2, 18, 20, 22]. In [25] Stós introduced a class of subordinate processes on d-sets, called α-s...

Abstract. In this short note the authors give answers to the three open problems formulated by Wu and Srivastava [Appl. Math. Lett. 25 (2012), 1347–1353]. We disprove the Problem 1, by showing that there exists a triangle which does not satisfies the proposed inequality. We prove the inequalities conjectured in Problems 2 and 3. Furthermore, we introduce an optimal refinement of the inequality ...