My conjecture is that for any k and n the second sigma is a divisor of the first one, i.e.
Let k be a nonnegative integer, n be a positive integer and be the sum of the kth powers of the infinitary divisors of n. Let be the sum of the kth powers of the odd infinitary divisors of n.
My conjecture is that for any k and n the second sigma is a divisor of the first one, i.e.
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The Happy Ending problem (named so by Paul Erdős as it led to the marriage of his friends Esther Klein and George Szekeres) was originally stated by Esther Klein in the following manner: There are five points on a flat surface, no two of which are coinciding and no three of which are on a straight line. Prove that four of these points are vertices of a convex quadrilateral. This simple theorem has a simple proof but, as the unheard of and complex concept of the so called convex hull is involved, the proof could be unintelligible to nonmathematicians, high school students inclusive. So, here is the simplest proof of mine, in which you can meet the familiar points and lines only. Let us think about the five points as of two lines and an extra point. Line l1 is defined by P1 and P2, line l2 is defined by P3 and P4, and P5 is the extra point. There are only 3 cases to consider: Case 1. Neither line intersects the segment between the points on the other line. One can build a convex quadrilateral by connecting the points on l1 and l2. Case 2. Each line intersects the segment between the points of the other line. One can build a convex quadrilateral by connecting the points on l1 and l2. Case 3. Only one of the lines intersects the segment on the other line. Let us assume without loss of generality that l1 intersects the P3P4 segment of l2. P5 can be in any of the 18 parts of the plane, but fortunately those parts are of only 2 types, O (coming from Other) and S (coming from Same). O means there exists a line dividing the plane so that P5 is in one half of the plane and a triangle consisting of three of the other points exists in the other half such that when P5 is connected to two of its vertices a convex quadrilateral is built (see below the line P2P4 allowing us to build the convex quadrilaterals P2P3P4P5 and P1P2P5P4). S means there exists a line dividing the plane so that P5 is in one half of the plane and a triangle consisting of three of the other points exists in the same half such that when P5 is connected to two of its vertices a convex quadrilateral is built (see below the line P1P2 allowing us to build the convex quadrilateral P1P2P5P4). We have just exhausted all possible arrangements of two lines and a point in a plane and proved that a convex quadrilateral can be built in each and every one of them. Q.E.D.
Find a prime number p, such that all digits of 13*p are the same.
As far as the laws of mathematics refer to reality, they are not certain, and as far as they are certain, they do not refer to reality. Albert Einstein
World maps are not mathematical maps. There is one additional country on the former, the country of mermaids. The color of the sea is the fifth color we need when we want to color the world map. There were many people who traveled the country of mermaids. Sailors sailed to other countries and conquered them. Throw a glance at the map, what do you see? Is Hawaii a part of the USA, Columbia, Japan or India? The History can tell you that but Geography (i.e. the map) cannot. Today one does not need to have sailors in order to invade another country; one only needs soldiers on airplanes or even less (soulless drones or computer viruses). Therefore my version of the Four color theorem is this: Let n be the number of countries. The number of colors needed to color the world map is n+1. Color the map accordingly and you do not need any additional assumptions (i.e. History) to answer the question “Which country does Hawaii belong to?” The Happy Ending Problem is the following simple statement: Given any five points on a flat surface that are in general position, i.e. no two of them coinciding and no three of them on a straight line, prove that four of these points will always form a convex quadrilateral. This is the latest proof of mine: 1. Let us start with the observation that a line can intersect three edges of a triangle at most (see below). The number of intersected edges can be zero, two or three, but the number of intersection points can be zero or two (point of entry and point of exit). 2. An edge (or its extension) of a triangle can intersect three edges of another triangle at most (see 1 above). In the extreme case where every edge (or its extension) of the intersecting triangle intersects three edges of the intersected triangle (see below): a) The number of intersection points is six, i.e. the maximum, and b) Any vertex of any of the triangles is collinear with two other vertices. 3. As any three points in general position form a triangle, any five points in general position form two triangles sharing a vertex. At least one of the edges (or their extensions) of the intersecting triangle (let us call it a "useful edge") intersects two edges of the intersected triangle at most (otherwise we have a contradiction with the general position requirement, see 2b above). In other words, a useful edge (or its extension) does not intersect at least one edge of the intersected triangle (let us call it a “useless edge”). On the other hand, the useless edge (or its extension) does not intersect the useful edge (otherwise the number of intersection points would be more than six, a contradiction with the general position requirement, see 2a above). Therefore, by connecting this pair of (useful and useless) edges we can form a convex quadrilateral (see below). Q.E.D.
Let k be a nonnegative integer, n be a positive integer and be the sum of the kth powers of the positive integer divisors of n. Let be the sum of the kth powers of the odd positive integer divisors of n. Is it true that for every k and n the latter sigma divides the former sigma? In other words, is it true that for every k and n I believe it is true.
I found today that
The sum of odd divisors of any natural number divides the sum of all its divisors. In the language of the OEIS For every n, A000593(n)A000203(n) As one can imagine the margin is too small to contain the proof. Nine is the only natural number n such that the sum of the digits of the first n natural numbers and the sum of the digits of the next n natural numbers are equal. Except for 1, all Ore numbers are Zumkeller numbers.
This is what Leonhard Euler found about the triangular numbers This is what I found about the tetrahedral numbers Can you find x=f(m), y=g(n) and z=h(m, n) such that there is a similar formula for the pentachoronal numbers
It was Yuval Noah Harari who started this train of thought (there) and I’ll finish it differently here.
God helps those who help themselves and does not help those who do not help themselves. Antibiotics are more powerful than God as they help: a) those who help themselves, i.e. those people X who take antibiotics when in need, and b) those who do not help themselves, i.e. those people Y who do not need to take antibiotics, i.e. those Y that have not been infected by X because X had been cured by the antibiotics before they met Y. With my very eyes I saw how this Prometheus lighted his fire and can honestly tell you, gentlemen, it was nothing special. Any slacker or goatherd could do it …..
Apocryphal Tales, Karel Čapek G. H. Hardy once said that any fool could have guessed Goldbach's conjecture. He was wrong since there had lived so many fools before Golldbach's time but none of them had guessed the conjecture (that is why Goldbach's name is attached to it). One might think that any fool could have made G. H. Hardy's remark. But the reasoning in the first paragraph can be applied to the last sentence. Therefore, though G. H. Hardy's remark sounds foolish it is not true that any fool could have made it. There is a vast area of foolishness reserved for the clever people. Blessed are the fools, for they are unable to explore this "special" area. Introduction
Let n be a natural number, D be the set of the divisors of n and sigma(n) be the sum of the elements of D. A number is a Zumkeller number if and only if D can be divided into two disjoint subsets D1 and D2 such that both sums of the elements of D1 and D2 equal sigma(n)/2. Challenge Prove (or disprove) that out of every four consecutive Zumkeller numbers there exists at least one number k such that sigma(k)/2 is also a Zumkeller number. To ask questions like "Where good ideas come from?" is not a good idea. Those with the good ideas do not ask themselves where their ideas come from but what to do with them.
Да питаш откъде идват добрите идеи не е добра идея. Хората с идеите не се питат откъде идват идеите им, а какво могат да направят с тях. Thank God that Just do it is the philosophy of a company producing sneakers, not nuclear power plants.
Благодаря ти, Боже, че Just do it е философията на производител на маратонки, а не на атомни електроцентрали. Barbarians, ancient and contemporary, look alike in their belief they worship the best possible God and inhabit the best possible world.
Древните и съвременните варвари си приличат във вярата си че се кланят на найдобрия от всички възможни богове и живеят в найдобрия от всички възможни светове. There exist natural numbers N, whose names share at least one letter with the name of every other natural number. Find some N.
There is an interesting integer sequence recursively defined in the following manner:
a(0) = 0, a(1) = 1, a(n) = 34*a(n1)  a(n2) + 2. Let sigma(n) be the sum of the divisors of n, k be a nonnegative integer and m = 2*k + 1. Prove that for every m, sigma(a(m)) < 2*a(m). Euclid was able to see the infinitude of primes but was blind to the existence of the number zero. Gregor Mendel left his fingerprint on genetics but did not leave a genetic fingerprint.
Евклид можа да види безкрайността на простите числа, но остана сляп за съществуването на числото нула. Грегор Мендел можа да остави отпечатък в генетиката, но не можа да остави генетичен отпечатък. The bold thing above is a theorem of mine, which I call "The Fiveness Theorem". "The Fiveness Challenge" is to find a number n such that k >= 17 (and for every
j < k, f composed with itself j times does not equal 5 ). Alice and Bob are friends. Alice is an optimist and always sees similarities. Bob is a pessimist and always sees differences. Yesterday they learned that tobacco is more addictive than Facebook.
"It's obvious. There are so many differences between tobacco and Facebook," said Bob. "You are wrong. They are similar, as they both shorten one's life," countered Alice. "On the one hand, you are right. But there's a material difference. Tobacco shortens one's life by shortening its end, i.e. life's worst part. Facebook shortens one's life by shortening its middle, i.e. life's best part. That is why it's easier to give up on Facebook," closed the discussion Bob. Ever the optimist, Amos Tversky used to say that pessimists are unhappy twice: when they predict bad things will happen and when the bad things actually happen. Some pessimist once replied that the pessimists are always happy: when bad things happen, since they are proven right, and when bad things do not happen, since they avoid the bad things.
It seems that Tversky, the optimist, was too pessimistic about pessimism and the pessimist was too optimistic about it. Who was the pessimist and who was the optimist then? Albert SzentGyörgyi, a Nobel prize winner in physiology, said that discovery consists of seeing what everybody has seen, and thinking what nobody has thought. This was independently discovered by another Nobelist  Erwin Schrödinger  who said that the task is...not so much to see what no one has yet seen; but to think what nobody has yet thought, about that which everybody sees.
A true discovery is hard to make. Even if you are a Nobelist. Even if what is discovered is only an aphorism. No rule applies to those who write it. That is why fair and democratic elections put into office unfair and undemocratic people. That is why in the real world the Barber paradox is a nobrainer.

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