Concrete mathematics : a foundation for computer science

534 ANSWERS TO EXERCISES
6.62 (a) A,, = &?A,-, -- A,-2 and B, = &B, 1 - B,~~2. Incidentally,
we also have &A, + B, = 2A,+, and fig,, - A,, = 2B, 1. (b) A table of
small values reveals that
\/5F,, n even;
Ln, n odd.
(cl WA,+1 - B,-l/A, = l/(Frn+l + 1) because B,A, - B, rA,+l = &
and A,A,+l = &(Fz,+l + 1). Notice that B,/A,+l = (F,/F,+r)[n even] +
(L,/L,+li[n odd]. (d) Similarly, xi=, 1/(F2kmkl - 1) = (Ao/BI - Al/B2) +
.. + (Anm~l/B, - A,/B,+j ) = 2 - A,/B,+, This quantity can also be
expressed as (5F,/L,+l) [n even] + (L,/F,+, ) [n odd].
6.63 (a) [z]. There are [;‘-:I with n, = n and (n - l)[nk’] with n, < n.
(b) (i). Each permutation pr . . on -1 of 11,. . , n- 1) leads to n permutations
n17t2.. .n, = p1 pj 1 n pj+l . . . on-l pi. If p1 . pn 1 has k excedances,
there are k+ 1 values of j that yield k excedances in 7~17~2 . . . n,; the remaining
n- 1~ k values yield k+ 1. I-lence the total number of ways to get k excedances
innln2...nnis(k+l)(“,‘)+((n-l)-(k-l))(z:;)=(t).
6.64 The denominator of (l’,‘) is 24nPvJini, by the proof in exercise 5.72.
The denominator of [ ,I:‘,] is the same, by (6.44), because ((i)) = 1 and
KS)
is even for k > 0.
6.65 This is equivalent to saying that (L)/n! is the probability that we
have 1x1 + + xnJ = k, when xl, , x, are independent random numbers
uniformly distributed between 0 and 1. Let yj = (x1 + . . . + Xj ) mod 1. Then
Yl, f.., y,, are independently and uniformly distributed, and 1x1 +. . . + x,J
is the number of descents in the y’s. The permutation of the y’s is random,
and the probability of k descents is the same as the probability of k ascents.
6.66 We have the general formula
((;)) = E (2nr’){n~~:!iik}(-ll”. for n > m > 0,
analogous to (6.38). When m = 2 this equals
((I)) = {n:3}~-(2n+l){n~2}+ (‘“:‘){“:‘}
= ;3n+2 - (2n + 3)2n+’ +i(4nz+6n+3).
6.67 ~n(n+~)(n+l)(2H2n-H,)-&n(10n2+9n-1). (It wouldbenice
to automate the derivation of formulas such as this.)