Lacsap’s Triangle

1 Introduction. let us moot a triangle of fractions Obviously, the burdenmates atomic government issue 18 following some exemplar. In this probe we will try to explain the theory behind this arrangement and to attain a everyday notification surrounded by the comp whizznts tote up and its value. The pattern above is called a Lacsaps Triangle, which inevitably hints at its relation to another arrangement Pascals Triangle (as Lacsap appears to be an anagram of Pascal). The algorithm behind it is very simple each element is the sum of the two elements above it.However, if we represent a triangle as a instrument panel (below), we will be able to notice a pattern between an index emergence of an element and its value tower chromatography column column column column column column 2 0 1 2 3 4 5 rowing 0 1 row 1 1 1 row 2 1 2 1 row 3 1 3 3 1 row 4 1 4 6 4 1 row 5 1 5 10 10 5 1 row 6 1 6 15 20 15 6 6 1 It seems authoritative to us to stress several points that this table mak es obvious ? the issue forth of elements in a row is n + 1 (where n is an index takings of a row) ? the element in column 1 is always bear on to the element in column n 1 ? herefore, the element in column 1 in every row is equal to the number of a given row. flat when we have established the main sequences of a Pascals triangle let us see how they are going to be expressed in a Lacsaps arrangement. We as well suggest looking at numerators and denominators separately, because it seems obvious that the fractions themselves cant be derived from earlier values utilize the progressions of the sort that Pascal uses. Finding Numerators. Lets begin with presenting given numerators in a similar table, where n is a number of a row. n=1 1 1 n=2 1 3 1 n=3 1 6 6 1 n= 4 1 0 10 10 1 n=5 1 15 15 15 15 1 3 Although the triangles appeared similar, the table demonstrates a significant diversity between them. We can see, that all numerators in a row (except 1s) have the same value. Therefore, they do not depend on other elements, and can be obtained from a number of row itself. like a shot a relationship we have to explore is between these numbers 1 1 2 3 3 6 4 10 5 15 If we consider a number of row to be n, hence n=1 1=n 0. 5 2 n 0. 5 (n +1) n n=2 3 = 1. 5n 0. 5 3 n 0. 5 (n +1) n n=3 6 = 2n 0. 5 4 n 0. 5 (n +1) n n=4 10 = 2. 5 n 0. 5 5 n 0. 5 (n +1) n n=5 15 = 3n 0. 6 n 0. 5 (n +1) n Moving from left to right in each row of the table above, we can clearly see the pattern. Dividing an element by a row number we get a serial publication of numbers each one of them is 0. 5 greater than the previous one. If 0. 5 is factored out, the next sequence is 2 3 4 5 6, where each element corresponds to a row number. Using a cyclic method, we have found a general expression for the numerator in the sea captain triangle If Nn is a numerator in a row n, so Nn = 0. 5(n + 1)n = 0. 5n2 + 0. 5n Now we can plot the relation between the row number and the numerator in each row.The graph of a parabolic influence begins at (0 0) and continues to enhance to infinity. It represents a continuous function for which D(f) = E(f) = (0 ) 4 Using a statute for the numerator we can now find the numerators of further rows. For example, if n = 6, hence Nn = 0. 5 62 + 0. 5 6 = 18 + 3 = 21 if n = 7, then Nn = 0. 5 72 + 0. 5 7 = 24. 5 + 3. 5 = 28 and so forth. Another way of representing numerators would be through using factorial notation, for obviously Numeratorn = n Now lets cut down of finding another part of the fraction in the triangle. Finding Denominators.There are two main variables, that a denominator is likely to depend on ? number of row ? numerator To find out which of those is connected with the denominator, let us consider a following table column 1 column 2 column 3 column 4 column 5 column 6 5 row 1 1 1 row 2 1 2 1 row 3 1 4 4 1 row 4 1 7 6 7 1 row 5 1 11 9 9 11 1 It is now evident, that a difference between the successive denominators in a second column incr eases by one with each iteration 1 2 4 7 11, the difference between elements being 1 2 3 4. So if the number of row is n, and the denominator of the second column is D, then D1 = 1D2 = 2 D3 = 4 etc then Dn = Dn-1 + (n 1) = (n-1) + 1 If we now look at the third column with a regard to a factorial sequence, a pattern emerges In the series 1 1 2 3 4 5 6 7 , if d is the denominator of the third column, then d3 = 1 + 1 + 2 = 4 d4 = 1 + 2 + 3 = 6 d5 = 2 + 3 + 4 = 9 dn = (n 2) + 3 To memorize the consistency of this succession, we will continue with the study of the fourth column. By analogy, the allow is as follows Denominatorn = (n 3) + 6 (where n is a number of row) Therefore, it can be represented as followsColumn 2 (n-1) +1 Column 3 (n-2) +3 Column 4 (n-3) +6 It is now clear, that numbers at bottom the brackets follow the (c 1) (where c is the number of column), and the numbers outside are in fact the numerators of the row of the previous index number (comparing to the column) . Therefore, a general expression for the denominator would be Dn = (n (c 1)) + (c 1) 6 where Dn is a general denominator of the triangle n is a number of row c is the number of column Now we can use a formula above to calculate the denominators of the rows 6 and 7. column 2 column 3 olumn 4 column 5 column 6 row 6 (6 1) + 1 = 16 (6 2) + 3 = 13 (6 3) + 6 = 12 (6 4) + 10 = 13 (6 5) +15 =16 row 7 (7 1) + 1 = 22 (7 2) + 3 = 18 (7 3) + 6 = 16 (7 4) + 10 = 16 (7 5) +15 =18 column 7 (7 6) + 21 = 22 Fusing these value with the numerators from the calculations above, we get the 6th and the 7th rows of the Lacsaps triangle Row 6 1 1 Row 7 1 1 If we now let En(r) be the (r + 1)th element in the nth row, starting with r = 0 then the general tilt for this element would be En(r) =Conclusion. To check the validity and limitations of this general statement let us consider the unusual circumstances first of all, will it elaborate for the columns of ones (1st and last column of each row)? if n = 4 r = 0, then En(r) = =1 if n = 5 r = 5, then En(r) = =1 7 therefore, the statement is valid for any element of any row, including the first one En(r) = =1 However, obviously, the denominator of this formula can not equal zero. But as long as r and n are both always positive integers (being index numbers), this limitation appears to be irrelevant.If the numeration of columns was to start from 1 (the 1st column of ones), then the general statement would take the form of En(r) = 8 Bibliography 1) Weisstein, Eric W. Pascals Triangle. From MathWorldA Wolfram Web Resource. http// mathworld. wolfram. com/PascalsTriangle. html 2) Pascals Triangle and Its Patterns an article from All you ever wanted to know http// ptri1. tripod. com/ 3) Lando, Sergei K.. 7. 4 Multiplicative sequences. Lectures on generating functions. AMS. ISBN 0-8218-3481-9