New PDF release: Figurate Numbers

By Elena Deza

Figurate numbers have a wealthy heritage with many purposes. the most function of this e-book is to supply a radical and whole presentation of the idea of figurate numbers, giving a lot in their homes, proof and theorems with complete proofs. This booklet is the 1st of this subject written in unified systematic method. It additionally includes many routines with strategies.

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A special summation of the form 1 + 2 + 3 + ··· + n + S3 (n) + S3 (n − 1) = 1 + 2 + · · · + (n − 1) gives S3 (n) + S3 (n − 1) = 1 + 3 + 5 + · · · + (2n − 1) = S4 (n). 2. Similarly, we can construct triangular numbers, using as inner blocks some triangular numbers of smaller size. For example, a triangular number with even index can be constructed using the following formula: S3 (2n) = 3S3 (n) + S3 (n − 1). + (n−1)n = n2 (4n+2) = In fact, one has 3S3 (n)+S3 (n−1) = 3· n(n+1) 2 2 2n(2n+1) = S3 (2n).

Any hexagonal number is triangular, and the situation is trivial. 5. An hexagonal square number is a number which is both hexagonal and square. Such numbers correspond to the positive integer solutions of the Diophantine equation u(2u − 1) = v 2 . Completing the square and rearranging gives (4u − 1)2 − 8v 2 = 1. Substituting x = 4u − 1 and y = 2v gives the Pell’s equation x2 − 2y 2 = 1. It has positive integer solutions (x, y) = (3, 2), (17, 12), (99, 70), (577, 408), . . In terms of (u, v), these give (u, v) = (1, 1), ( 92 , 6), (25, 35), ( 289 2 , 204), .

9. , every hexagonal number is a triangular number. The simplest way to get a proof of this fact is to compare two formulas: (2n − 1)2n n(4n − 2) S3 (2n − 1) = , and = S6 (n). 2 2 The geometrical illustration of this property for n = 3 is given below. ∗ ∗ ∗ ∗ ∗ ∗ • • • = ∗ • • ∗ ∗ • ∗ ∗ ∗ It is easy to see that we just rearranged the four small triangles on the previous picture into one big triangle. com by KAINAN UNIVERSITY on 02/08/15. For personal use only. S6 (n + 1) = S6 (n) + (4n + 1) = S3 (2n − 1) + (4n + 1) 4n2 + 6n + 2 (2n − 1)2n + (4n + 1) = = 2 2 (2n + 1)(2n + 2) = = S3 (2n + 1) = S3 (2(n + 1) − 1).

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