FIBONACCI EXPRESSIONS ARISING FROM A COIN-TOSSING SCENARIO INVOLVING PAIRS OF CONSECUTIVE HEADS

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1 FIBONACCI EXPRESSIONS ARISING FROM A COIN-TOSSING SCENARIO INVOLVING PAIRS OF CONSECUTIVE HEADS MARTIN GRIFFITHS Abstract. In this article we study a combinatorial scenario which generalizes the wellknown problem of enumerating sequences of coin tosses containing no consecutive heads. It is shown how to derive formulas enumerating, for a fixed value of k, the sequences of length n containing exactly k distinct pairs of consecutive heads. We obtain both exact expressions and asymptotic relations. 1. Introduction Suppose that a coin is tossed 1 times and the resulting sequence of heads and tails is recorded. Two possible outcomes are as follows: a) HTTTHTHTTTTH and b) HTTHHTTHHHHT. Notice that sequence a) contains no consecutive heads whereas b) does. It is in fact wellknown [3, 7] that of the n possible outcomes when a coin is tossed n times, the number containing no consecutive heads is given by F n+. Looking at sequence b) again, we can actually be a little more specific and say that it contains exactly 4 distinct pairs of consecutive heads. Counting from left to right, these pairs are positioned at 4, ), 8, 9), 9, 10) and 10,11). Thus, on using Un,k) to denote the number of sequences of length n containing exactly k HH s, we see that a) contributes 1 to U1,0) while b) contributes 1 to U1,4). Our aim here is to derive formulas that allow us to enumerate, for a fixed value of k, the sequences of length n containing exactly k distinct pairs of consecutive heads. In an interesting though rather incomplete initial foray into this problem [], the following recurrence relation was obtained for Un,k): Un,k) = Un 1,k)+Un 1,k 1)+Un,k) Un,k 1), 1.1) where n, k 0, Um, 1) = 0 for all m 0, and U0,0) = 1 by definition. In order to use 1.1) to calculate Un,k) recursively, we need simply to note that U1,0) = and U0,m) = U1,m) = 0 for all m 1. As is shown in [], it is reasonably straightforward to derive the relation 1.1) by using the intermediate results UTn,k) = UTn 1,k)+UHn 1,k) 1.) and UHn,k) = UTn 1,k)+UHn 1,k 1), 1.3) whereutn,k) and UHn,k) represent the numberof sequences of n tosses containing exactly k HH s such that the last toss is a tail and a head, respectively. Result 1.) is true since any sequence A of length n 1 containing exactly k HH s either ends in a head or a tail, and the sequence of length n that results when a tail is appended to the end of A still has exactly this many pairs of HH s. On the other hand, a sequence of length n with exactly k HH s such that AUGUST

2 THE FIBONACCI QUARTERLY the last toss is a head either has its first n 1 tosses containing exactly k HH s and ending a tail or containing exactly k 1 HH s and ending in a head. In order to obtain explicit formulas for Un,k) from 1.1), we illustrate a nice application of exponential generating functions. It is shown that Un, k) becomes an ever-more complex expression in n involving the Fibonacci numbers as k increases. A general asymptotic relation for Un,k) is also obtained.. Some Results on Exponential Generating Functions The exponential generating function Gx) for the sequence a 0,a 1,a,... is defined as Gx) = a 0 0! + a 1 1! x+ a! x + a 3 3! x3 + a k = k! xk, k=0 where the coefficient of x n / in this series is a n [1, 4]. Since G x) = a 1 0! + a 1! x+ a 3! x + a k+1 = x k,.1) k! k=0 a n+1 is the coefficient of x n / in G x). Also, xgx) = a 0 0! x+ a 1 1! x + a! x3 + a 3 3! x4 + = a 0 1! x+ a 1! x + 3a 3! x3 + 4a 3 4! x4 + ka k 1 = x k,.) k! k=1 so the coefficient of x n / in xgx) is na n 1. The above may be generalized to show that the coefficient of x n / in x m Gx) is nn 1) n m+1)a n m. In particular, the exponential generating function for the Fibonacci numbers is given by Hx) = exp sinh ) see [] or sequence A00004 in [6], for example). This function will play a key role in our quest for formulas enumerating the sequences of length n containing a certain fixed number of pairs of consecutive heads. x ).3) 3. Initial Calculations The exponential generating function of Un,k) for fixed k is given by G k x) = Un,k) x n. 3.1) 0 VOLUME 49, NUMBER 3

3 FIBONACCI EXPRESSIONS ARISING FROM A COIN-TOSSING SCENARIO On using 1.1) and setting k = 1 we obtain Un+,1) x n = Un+1,1) x n + from which it follows, by way of.1), that Un+1,0) x n + Un,1) x n Un,0) x n, 1 x) G 1 x) G 1x) = G 0 x) G 0x). 3.) However, since Un,0) = F n+, it is the case, utilizing.1) once more, that G 0 x) = H x). Thus 3.) gives the linear second-order differential equation 1 x) G 1 x) G 1x) = H x) H x) = 1 exp ) { cosh This has the auxiliary equation λ λ 1 = 0 with solutions λ 1 = 1+ The complementary function is therefore given by 1+ ) 1 ) Aexp x +Bexp x ) { = exp A+B)cosh + } sinh. 3.3) and λ = 1. } +A B)sinh, where A,B R. Then, as exp cosh ) and x exp sinh ) ) are part of the complementary function, we try the particular integral ) Cxexp cosh +Dxexp sinh ) ) x for some C,D R. On using the initial conditions G 1 0) = G 1 0) = 0, the solution to 3.3) is found to be G 1 x) = 1 ) { exp xcosh + } x )sinh. 3.4) In order to obtain a formula for Un,1) it is now a matter of extracting the coefficient of x n / from the right-hand side of 3.4). First, from.3) we obtain ) { H x) = exp cosh + 1 } sinh. 3.) AUGUST x )

4 THE FIBONACCI QUARTERLY Then, on using 3.) and.3) in turn, we have ) exp cosh ) x = H x) 1 exp sinh ) = H x) 1 Hx). 3.6) Therefore, from 3.4), 3.6), and.3), it is the case that G 1 x) = x H x) 1 ) Hx) + x Hx) 10 = x Hx) 1 Hx)+ x H x). Using this result, along with.1),.), and 3.1), we obtain Un,1), the coefficient of x n / in G 1 x): which is valid for any n 0. Un,1) = n F n 1 1 F n + n F n = 1 {nf n+1 +F n 1 ) F n }, 3.7) 4. Further Results This process may now be continued indefinitely. Next, we have x) G x) G x) = G 1 x) G 1x) = ) { exp xcosh The solution of 4.1) is G x) = 1 ) { 1 exp 0xcosh +3 sinh }. 4.1) + x } 8)sinh. 4.) Such differential equations are somewhat tedious to solve by hand, and we performed the calculations with the assistance of Mathematica [8]. In conjunction with 3.1),.1),.), and the generalization of.), 4.) gives, Un,) = 1 0 {nn 1)F n +4n )F n } for n 0, where we adopt the convention that F m = 1) m+1 F m for any m 0. Similarly we have 3x) G 3x) G 3 x) = G x) G x) = 1 ) { 0 exp xx 4)cosh + x +40x+8 ) sinh }. VOLUME 49, NUMBER 3

5 FIBONACCI EXPRESSIONS ARISING FROM A COIN-TOSSING SCENARIO This has the solution 1 ) { 70 exp x x +9x+6)cosh + 3 3x 18x 1 ) sinh }, from which we obtain Un,3) = 1 10 {nn 1)n )F n 3 +F n )+3nn 1)F n 1 +n )F n 3 ) 6F n }, and so on. Each member of the resulting family of linear second-order differential equations has the form k x) G k x) G kx) = G k 1 x) G k 1x) ) { = aexp f k x)cosh +g k x) sinh }, 4.3) for some a Q and polynomials f k x) and g k x) having integer coefficients and, for k 3, degree k 1. The solution to 4.3) is of the form ) { G k x) = bexp r k x)cosh +s k x) } sinh, 4.4) where b Q and r k x) and s k x) are polynomials having integer coefficients and, for k 3, degree k. From thisit is possibletodeducethat, forany particular valueof k, Un,k) canbeexpressed as m h i n)f n i, i=0 for some integer m 0 and family {h i n) : i = 0,1,...,m} of polynomials in n some of which could be the zero polynomial). It needs to be born in mind of course that a particular representation of Un, k) is certainly not unique; indeed, the Fibonacci recurrence relation allows us to express these representations in different ways. First, on using the result. Asymptotic Results F n φn, which can be found in [4], for example, 3.7) gives Un,1) = 1 {nf n+1 +F n 1 ) F n } n F n+1 +F n 1 ) n ) φ n+1 + φn 1 = nφn 1 +φ). AUGUST 011 3

6 THE FIBONACCI QUARTERLY Then similar, if somewhat more involved, calculations yield Un,) n φ n 10 and Un,3) n3 φ n φ). In general, it is a question of picking out the dominant terms in the expression 4.4) for G k x) and hence for Un,k)), which, for k 3, arise as the coefficients of x k in both r k x) and s k x). It may be shown from this that Un,k) nk φ n k F k 1 φf k ) and Un,k) nk φ n k L k 1 φl k ) k! k+1 for k even and k odd, respectively, and thus that Then, since we have the result k! k+ Un,k) nk φ n k 1) k F k 1 φf k ). k! k+1 ) 1) k F k 1 φf k = φ k 1 +φ) k, Un,k) nk φ n k+ k! +φ) k.1) for any fixed k 0. The asymptotic formula.1) does, however, give rather poor approximations for small values of n. For example, we have to wait until n = 186 before it provides us with an approximation for Un,) having a relative error of less than 1%. 6. Acknowledgement The author would like to thank the referee for suggestions that have helped improve the clarity of this article. References [1] P. J. Cameron, Combinatorics: Topics, Techniques, Algorithms, Cambridge University Press, [] M. N. Deshpande and J. P. Shiwalkar, The number of HH s in a coin-tossing experiment and the Fibonacci sequence, Mathematical Gazette, 9 008), [3] M. Griffiths, No consecutive heads, Mathematical Gazette, ), [4] D. E. Knuth, The Art of Computer Programming, Volume 1, Addison-Wesley, [] T. Koshy, Fibonacci and Lucas Numbers with Applications, Wiley-Interscience, 001. [6] N. J. A. Sloane Ed.), The On-Line Encyclopedia of Integer Sequences, 011, njas/sequences/. [7] E. W. Weisstein, Coin Tossing. From MathWorld A Wolfram Web Resource, [8] Wolfram Research Team, Mathematica, Version 6, Wolfram Research, Champaign, Illinois, 007. MSC010: 0A1, 0A16, 11B37, 11B39, 34A0. School of Education, University of Manchester, Oxford Road, Manchester, M13 9PL, United Kingdom address: martin.griffiths@manchester.ac.uk 4 VOLUME 49, NUMBER 3