Lacsap’s Triangle

1 Introduction. Let us consider a triangle of fractions: Obviously, the numbers are following some pattern. In this investigation we will try to explain the theory behind this arrangement and to find a general relation between the element’s number and its value. The pattern above is called a Lacsap’s Triangle, which inevitably hints at its relation to another arrangement – Pascal’s 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 table (below), we will be able to notice a pattern between an index number of an element and its value: column column column column column column column 2 0 1 2 3 4 5 row 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 stress several points that this table makes obvious: ? the number 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. Now when we have established the main sequences of a Pascal’s triangle let us see how they are going to be expressed in a Lacsap’s arrangement. We also suggest looking at numerators and denominators separately, because it seems obvious that the fractions themselves can’t be derived from earlier values using the progressions of the sort that Pascal uses. Finding Numerators. Let’s 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 difference between them. We can see, that all numerators in a row (except 1’s) have the same value. Therefore, they do 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, then 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 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 at (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 let’s 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 = 1

D2 = 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 follows:

Column 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 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 Lacsap’s 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 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. “Pascal’s Triangle. ” From MathWorld–A Wolfram Web Resource. http:// mathworld. wolfram. com/PascalsTriangle. html 2) “Pascal’s 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