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Unformatted text preview: This is, of course, the famous Fibonacci sequence. An easy proof that this list will always work was offered by Douglas Vander Griend. Take a number N . If it’s a Fibonacci number, it’s on the list. Otherwise, subtract off the largest Fibonacci number, F less that N . The result will be a number strictly less than the Fibonacci number before F . Induction finishes the argument. Burkart Venzke suggests modifying the third condition so that no n consecutive numbers on the list are allowed to appear in any one sum. With n = 3 the first few terms are 1, 2, 3, 4, 6, 9, 13, 28,. .. where each number now is the sum of the previous one and the one three steps before, that is a k = a k1 + a k3 . The general case leads to a k = a k1 + a kn . Eliminating the condition on consecutive list numbers is the same as setting n = 0, which gives a list of powers of 2....
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This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.
 Spring '10
 AlbertoDelgado
 Combinatorics

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