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A COMBINATORIAL PROBLEM INVOLVING FIBONACCI NUMBERS J . L. Brown, J r . The Pennsylvania State University, University Park, Pa. In Advanced Problem H-70 (this Quarterly, Vol. 3, No. 4, p8 299)f Cm A9 Churchy Jr 0 proposed the following combinatorial results "For n = 2m, first show that the total number of k-combinations of the n natural numbers such that no two elements i and i + 2 appear together in the same selection i s F 2m+2 . and if n = 2m + 1,' & t! F F + . (Solution appears in £l J ) the total is The purpose of this note is to consider by a different method a more general combinatorial problem which includes Church 1 s problem as a special ease* As in the latter problem, the explicit solution will be seen to be expressible entirely in terms of Fibonacci numbers 9 PROBLEMS Given the set S consisting of the first n positive integers and a fixed integer v satisfying 0 < v < n , how many different subsets A of S (including the empty subset) can be formed with the property that af - a" fi v for any two elements a f , a" of A (that i s , subsets A such that integers i and i + v do not both appear in A for any i = 1, 2, • • • , n - p) ? Church 1 s problem is then recovered from the above formulation on taking v = 2. For the solution of the general problem, we let n = m + r with m an integer and 0 =£ r ^ p9 so that n = r (mod v)m Each subset A of S can be made to correspond to an ordered binary sequence of n t e r m s , («i,«2» * * * » a )• by the rule that a. = 1 if i G A and a. = 0 if i (£ A9 A and its corresponding binary sequence (aisa2§*«»9a For a given subset ), we define v ordered binary sequences A lf A 2 ,»*»,Aj / as follows § For l < j < r, and for r < j ^ v 34 1988] A COMBINATORIAL PROB'LEM INVOLVING FIBONACCI NUMBERS e Note that each of the t e r m s al9a29 * ° 9a 35 is included In one and only one of these sequences^ since for j = 1, 2, • • • , v - 1, the sequence A. contains all a.. f s with i = j (modi/) while Ap contains all a . f s with 1 = 0 (mod v). Now If the subset A under consideration satisfies the problem constraint^ then clearly none of the sequences JA.f1' can contain two consecutive ones; conversely, If A contains both i and I + v for some i 0 satisfying 1 < I 0 ^ n - v9 then the sequence A , , cessive ones e where k = i 0 (mod v) will contain two suc- Thus the subset A under consideration will satisfy the given constraint If and only If each A. (j = 1, 29 • • • , v) is a binary sequence without J consecutive ones. But it is well known ([2], Problem 1(b), pe 14; £ 3 j , pp@ 166167) that the total number of binary sequences of length t without consecutive ones is ~P..9 Since each of the r sequences A ls A 2 » @e %A 1 and each of the remaining v - r sequences A has length m + , • • *, Av has length m, it follows that the total number of subsets of A with the desired property is r V—T F F m+3 m+2 To obtain Church 1 s results we take v = 2 and let n = 2m + r where r = 0 or r = 1, so that n = r (mod 2). Then the total number of k-combinatlons of the first n Integers such that no elements 1 and I + 2 appear together is F° ^ F 2 . = F^ if m+3 m+2 m+2 r = o (n even) v and F m+3 , F m+2 .if r = 1 v(n odd)' . Additional references dealing with the case v = 2 may be found In [ 1]|. REFERENCES 1. Ce Ae Church, J r e , Problem H-70, Solution and Comments g The Fibonacci Quarterly, Vol.' 5,' No.' 39 October, 1967, ppe" 253-255," 2@ J e Mordan, An Introduction to Combinatorial Analysis, John Wiley and Sons, Inc.,' N. Ye C s , 1958e Problem 1(b), p8 14. 3ffl J 0 L. Brown, J r e , n Zeckendorf ! s Theorem and Some Applications, t! The Fibonacci Quarterly, Vol.' 2, No. 3, O c t 1964, pps' 163-168. EDITORIAL-NOTE: The restraint that. 0 •-<!/.<-. n can be removed. 0S so that number of subsets becomes F. , F - ,n -= FQF?"~ Set .m.= = 2 .as Is well known for the numbers of subsets of 1»2,.3,., ,,,-n without constraints^. V.E.H, * • • • • ' ~ ~~