Hello, My name is Mike and this is a secondary level math question, and I am a student. My question is: Are there two formulas that define sequences that agree on their first four terms, but differ on the fifth term and all succeeding terms? Thank you for any help Hi Mike, One classic example: Take n points in general position on the circumference of a circle, and let R(n) be the maximum number of regions the circle can be divided into by joining all pairs of the given points. Then for n = 1, 2, 3, 4, 5, 6 R(n) = 1, 2, 4, 8, 16, 31. The number 31 always comes as a surprise since one's first guess is that R(n) = 2 n-1. (To make things go wrong on the 5th step, you can start the sequence with n = 2 points.) For hundreds of other naturally occurring examples you can look through Neil Sloane's dictionary of sequences: sequences@research.att.com For example, if you send the message lookup 1 2 4 8 16 you get a response that lists dozens of sequences that he is aware of beginning with 1, 2, 4, 8, 16, and explanations of where they come from. (You can also use his web page, which does lookups "online"! Go to: http://www.research.att.com/~njas/sequences/ The whole sequence table is also visible there, as well as an explanation of the symbols used in the table. Alternatively, you can generate your own sequences using Lagrange's interpolation formula. You can devise a polynomial of degree at most n that takes on your prescribed values at 1, 2, 3, ..., n+1. Here's how it works. Suppose you want a 2nd degree polynomial whose value at 1 is 5, at 2 is 7, and at 3 is 11. Add up the three terms of the form "your value at k times the product of terms (x-r)/(k-r) where r takes on all values from 1 to 3 except k": 5(x-2)(x-3)/2 + 7(x-1)(x-3)/(-1) + 11(x-1)(x-2)/2. Simplify to get 15x 2 - 57x + 47. Check by plugging in 1, 2, 3. Plug in 4 and you get 59, so the resulting sequence is 5, 7, 11, 59. Chris Go to Math Central