Concrete mathematics : a foundation for computer science

546 ANSWERS TO EXERCISES
7.40 The egf for (nF, 1- F,) is (z - lip(z) where i(z) = J&0 F,z”/n! =
(e@’ - e&‘)/fi. The egf for (ni) is e -‘/(l - z). The product is
5 l/2 Ii 112 -el@ ‘12) = 5 liqe -42 _~ e 6’).
(e
We have i(z = -?(-z). So the answer is (--l)"F,.
7.41 The number of up-down permutations with the largest element n in
position 2k is (z”k ‘,)A ok ,A,, -2k. Similarly, the number of up-down permutations
with the smallest element 1 in position 2k + 1 is (‘&‘)A2kAn ok 1,
because down-up permutations and up-down permutations are equally numerous.
Summing over all possibilities gives
AkA,~r~~k+2[n=O] + [n=l]
The egf A therefore satisfies 2A’(2) = A(z)’ + 1 and A(0) = 1; the given
function solves this differential equation.
7.42 Let a, be the number of Martian DNA strings that don’t end with c
or e; let b, be the number that do. Then
a n = 3~~1 + 2b,- 1 + [n = 01, b, -= 2a n I + h-1 ;
A(z) = 3zA(z) +2zB(z) + 1 , B(z) = ZzA(z)+zB(z);
and the total number is [z”] ( 1 + z)/( 1 - 42 ~ z2) = F3,, +2.
7.43 By (5.45), g,, = An6(0). The nth difference of a product can be
written
A”A(z)B(z) = t ‘k” (ACED kA(Z))(AR k~(~)) )
k
0
and En k = (1 + A)nPk = xi (“j “)Ai. Therefore we find
hn = G (E) (“yk) fjfkgn-k.
3
This is a sum over all trinomial coefficients; it can be put into the more
symmetric form
fj+kgk+L.