Lacsap’s Triangle

1 Introduction. Let us flying field a triplicity of constituents Obviously, the totals argon following some pattern. In this investigation we leave behind try to explain the possibility behind this arrangement and to find a oecumenic relation between the members sum up and its encourage. The pattern above is called a Lacsaps Triangle, which inevitably hints at its relation to another arrangement dadas Triangle (as Lacsap appears to be an anagram of Pascal). The algorithm behind it is very simple from each one atom is the sum of the two elements above it.However, if we represent a triangle as a plank (below), we will be able to notice a pattern between an index yield of an element and its value newspaper chromatography column column column column column column column 2 0 1 2 3 4 5 form 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 important to us to emphasis several points that this accede makes obviou s ? the tot of elements in a row is n + 1 (where n is an index number of a row) ? the element in column 1 is always equal 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. like a shot when we have established the main sequences of a Pascals triangle let us see how they are going to be verbalized in a Lacsaps arrangement. We also suggest understanding at numerators and denominators separately, because it seems obvious that the fractions themselves set upt be derived from earlier values using the progressions of the sort that Pascal uses. Finding Numerators. Lets engender 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 difference between them. We can see, that all numerators in a row (except 1s) have the same value. Therefore, they d o not depend on other elements, and can be obtained from a number of row itself. Now 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 decent in each row of the table above, we can clearly see the pattern. Dividing an element by a row number we get a series 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 original triangle If Nn is a numerator in a row n, then 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 form begins a t (0 0) and continues to rise to infinity. It represents a continuous function for which D(f) = E(f) = (0 ) 4 Using a formula for the numerator we can now find the numerators of further rows. For example, if n = 6, then 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 concentrate 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 increases 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 check the consistency of this succession, we will continue with the study of the fourth column. By analogy, the result 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 inside 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 denomi nator 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 seventh 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 statement 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 work 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 twain 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. hypertext markup language 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