Even if you prove a theorem, you cannot prove it is interesting.
Дори и да докажеш една теорема, не можеш да докажеш че тя е интересна.
Дори и да докажеш една теорема, не можеш да докажеш че тя е интересна.
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Thinking is similar to moving in two-dimensional space. Intelligence drives you forward, imagination drives you sideways. When there is no way forward, sideways travel is recommended.
Perfect numbers were discovered about 2500 years ago, but human intelligence has not revealed much about them. It is not known whether their number is infinite or whether there are any odd perfect numbers. In the early 21st century Reinhard Zumkeller pressed the pedal of his imagination and generalized the perfect numbers. Instead of unsuccessfully searching for proofs for the infinitude of the perfect numbers, mathematicians started thinking about the Zumkeller numbers and finding interesting things about them. It is now known for certain that there are infinitely many Zumkeller numbers, that some of them are odd, and that out of every 12 consecutive natural numbers at least one is a Zumkeller number. Elementary, my dear Watson, when you can't go forward and find something about a second thing, you can go sideways and find a third thing about a fourth thing. “Sometimes a cigar is just a cigar,” Freud said from the comfort of his chair. "Oh, Sigmund! How naive you are!" Bill Clinton exclaimed from the couch and Freud felt Bill had a story to tell.
Intelligence is like flatulence. People tolerate it if you keep it to yourself but are annoyed if you let it roam free.
Интелигентността е като газовете - търпят я ако я сдържаш, но се дразнят ако я пуснеш на воля. Why should one care about the DARK matter out there when there is so much GRAY and WHITE matter in here?
Обаждам се на приятел да го поздравя по случай рождения му ден. Разговорът е симетричен: започва с моите пожелания и завършва с неговите - да съм публикувал нещо математическо. Как да му кажа че математическите ми открития са несвързани едно с друго и имат средна дължина от 2 изречения? Такива неща никой не публикува. Затова ... решавам сам да си публикувам нещо старо, отпреди десетина години. When he was a 5-year-old boy Andrey Kolmogorov asked questions like How many distinct patterns can you create with a thread while sewing on a four-hole button? Unfortunately, there is no information on the Web what the answer was and had Kolmogorov found it himself. In a book about Grigori Perelman, Masha Gessen admitted that two professional mathematicians, both former students of Kolmogorov, had given different answers. We found the number of patterns p in several cases where single-color thread was involved. We also found the number of patterns in several cases where threads of different colors c were used, including the case where the buttonholes were located on the vertices of a generic convex n-gon (i.e. a convex n-gon with no more than two diagonals intersecting at any point in its interior). Now, let us try the more complicated case involving different color threads and buttonholes located on the vertices of a “plain vanilla” regular convex n-gon and then generalize our findings with respect to all convex n-gons. Assumptions: A1. It is mandatory to utilize the buttonholes. There are many ways to sew on a button without utilizing the buttonholes but let us forget about them for the time being. A2. All n buttonholes are located on the vertices of a regular convex n-gon. A3. Different-color threads might be used for different segments between the buttonholes. A4. Only buttonhole-to-buttonhole connections are used. No cross-border connections are allowed. Calculation: 1. Some of the regular convex n-gons are generic ones – those, whose number of vertices is an odd number (n = 2*q + 1) plus the square, where n = 4. We have already found (see OEIS A209916) that for the generic convex n-gons the number of patterns p is p = ((c+1)^((n-1)*n/2) * (c*(c-1))^C(n, 4)) - 1 as all possible intersections totaling C(n, 4) must be counted twice when two differently painted diagonals, totaling c*(c-1), intersect (as two different patterns exist for every two colors, which you can see below).  Now we will cover those regular convex n-gons where n is an even number greater than 4 (i.e. when n = 2*q and q > 2). 2. Thanks to Bjorn Poonen and Michael Rubinstein and their article “THE NUMBER OF INTERSECTION POINTS MADE BY THE DIAGONALS OF A REGULAR POLYGON” we know some things about regular polygons that we can use now: a) When n = 2*q the number of diagonals intersecting at a point (except for the center of the n-gon) may be 2, 3, 4, 5, 6 or 7; b) When n = 2*q the number of diagonals intersecting at the center of the n-gon is n/2. 3. The number of intersection points I(n) of the diagonals can be represented in the following manner: I(n) = i(2)+i(3)+i(4)+i(5)+i(6)+i(7)+i(n/2), where i(k) (k = 2, …, 7) is the number of points where k diagonals intersect and i(n/2) is the center of the regular n-gon where n/2 diagonals intersect (in the case of n = 2*q, q > 1). Therefore, i(n/2) must be separately counted only in the cases where n = 2*q and q > 7. 4. In it. 1 above, we saw that two different-color threads (e.g. red and green) look differently when the threads intersect (depending on which one is on top of the other). The same is true when the number of different-color threads is 3, 4 and more (see below how many patterns are there when three differently painted diagonals intersect at a point).  Therefore, when trying to find the number of different patterns, it is not enough to count the number of edges and diagonals. We also have to count twice the cases where two different-color diagonals intersect, three times the cases where three different-color diagonals intersect etc. 5. Based on it. 2-4 above the formula in it. 1 above takes the following form: p = (M1*M2*M3) - 1, where  and where -1 stands for the case when the button is attached to the cloth only by “0-color” threads, i.e. the case where the button is not attached at all.
6. This formula could be used for any convex n-gon, as long as we remember that in the cases of n = 2*q + 1, as well as in the cases of n = 2*q and q ≤ 7, we do not need to multiply by the third term (i.e. by M3) as the central intersection point either does not exist (when n = 2*q + 1) or had already been handled (when n = 2*q and q ≤ 7). 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 non-mathematicians, 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 part of the plane, but fortunately those parts are of only 2 types, O (coming from Other) and S (coming from Same). О 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 P4P1P2P5).  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.
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, and c) No edges of the two triangles coincide.  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 2 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 six or more, a contradiction with the general position requirement, see 2 above). Therefore, by connecting this pair of (useful and useless) edges we can build a convex quadrilateral (see below). Q.E.D.  Let k be a non-negative integer, n be a positive integer and  be the sum of the k-th powers of the positive integer divisors of n. Let  be the sum of the k-th 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  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. I saw how this Prometheus guy 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 най-добрия от всички възможни богове и живеят в най-добрия от всички възможни светове. There is an interesting integer sequence recursively defined in the following manner:
a(0) = 0, a(1) = 1, a(n) = 34*a(n-1) - a(n-2) + 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 (if one wants bad things to happen, then he's a madman, not a pessimist).
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 Szent-Györgyi, 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.
Original discoveries are hard to make. Even if you are a Nobelist. Even if what is discovered is only an aphorism. |
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