### On N-Sequences

T. C. Brown, Max L. Weiss
1971 Mathematics Magazine
Citation data: T.C. Brown, On N-sequences, Math. Magazine 44 (1971), 89-92. In  , it is shown that the Fibonacci sequence 1; 1; 2; 3; 5; : : : has the property that if any one term is removed from the sequence then every positive integer can be expressed as the sum of some of the terms that remain, and that if any two terms are removed, then there is a positive integer that cannot be expressed as the sum of some of the remaining terms. We describe this situation by saying that removal of any
more » ... that removal of any two terms from the Fibonacci sequence renders it incomplete, while removal of any one term does not. A sequence which is not incomplete will be called complete. In this note we consider, for each n = 0; 1; 2; : : : ; nondecreasing sequences which are rendered incomplete by the removal of any n + 1 terms, but not by the removal of any n terms. We call such a sequence an n-sequence. Thus the Fibonacci sequence is a 1-sequence.We characterize in a simple way the set of all 1-sequences and show that there are no n-sequences for any n ! 2. A simple description of the set of all 0-sequences seems more difficult, and is left open. Theorem 1. The sequence fa k g ∞ k=1 is a 1-sequence if and only if it has the following three properties: (1) a 1 = a 2 = 1. (2) a k+1 = a k or a k+1 = ∑ k 1 j=1 a j + 1; k ! 2. (3) a k+1 = ∑ k 1 j=1 a j + 1 for infinitely many k. Proof of sufficiency. The proof that the Fibonacci sequence is a 1-sequence given in  translates more or less immediately into a proof of the sufficiency of the given conditions, and is given here, slightly altered, for the sake of completeness. Let fb k g ∞ k=1 be the sequence obtained from fa k g ∞ k=1 by deleting one term from fa k g. Suppose that N is the smallest number which is not the sum of terms from fb k g and that a n N < a n+1 . Then ∑ n 1 j=1 b j ! ∑ n 1 j=1 a j = ∑ n 1 j=1 a j + 1 1 = a n+1 1 ! N. Now let L be the smallest number greater than N which is the sum of terms from fb k g and suppose that L = ∑ r m=1 b j m , where j 1 < j 2 < ¡¡¡ < j r n 1. and hence, since N is not the sum of terms from fb k g, L 1 > N, contradicting the choice of L. If b j 1 > 1 then N ! a n ! b n 1 ! b j 1 , hence N > b j 1 1 ! 1, hence by the definition of N, b j 1 1 = ∑ s p=1 b H j p , and moreover b H j p < b j m for all m ! 2 and all p. Thus ∑ s p=1 b H j p + ∑ r m=2 b j m = L 1 ! N, hence > N, again contradicting the choice of L. Thus fa k g remains complete if one term is deleted.