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Concrete mathematics : a foundation for computer science

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526 ANSWERS TO EXERCISJES<br />

5.94 This is a consequence of Henrici’s “friendly monster” identity,<br />

f(a,z)f(a,wz)f(a,w"z)<br />

=<br />

F (<br />

;a-+, ia++ 42 3<br />

5a,3a+~i,~a+5,3a-~,fa,~a+~,a I( 9 >) '<br />

where f (a, z) = F(; a; z). This identity can be proved by showing that both<br />

sides satisfy the same differential equation. If we replace 3n by 3n + 1 or<br />

3n + 2, the given sum is zero.<br />

5.95 See [78] <strong>for</strong> partial reisults. The <strong>computer</strong> experiments were done by<br />

V. A. Vyssotsky.<br />

5.96 All large n have the property, according to S&k&y [256’]. Paul ErdGs<br />

conjectures that, in fact, ma+, cp ((2,“)) tends to infinity as n + 00.<br />

5.97 The congruence surely holds if 2n + 1 is prime. Steven Skiena has also Ilan Vardi notes<br />

found the example n = 2953, when 2n + 1 = 3.11 .179.<br />

that the condition<br />

holds <strong>for</strong><br />

6.1 2314,2431,3241,1342,3124,4132,4213,1423,2143, 3412,4321. 2n+l =p’,<br />

where p is prime,<br />

6.2 { E}n-&, because every such function partitions its domain into k non- if and only if<br />

empty subsets, and there are rnk ways to assign function values <strong>for</strong> each 2pm’ mod p2 = 1.<br />

partition. (Summing over k gives a combinatorial proof of (6.10).) This yields two<br />

more examples:<br />

6.3 Now dk+’ 6 (center of gravity) --E = 1 -e+(d’ +...+dk)/k. This n= (‘093~-‘)/2;<br />

recurrence is like (6.55) but with 1 - c in place of 1; hence the optimum<br />

solution is dk+’ = (1 - c)Hk. This is unbounded as long as c < 1.<br />

n = (35112-1)/2.<br />

6.4 Hln+’ - :H,,. (Similarly EC”=, (-l)kp’/k = Hz,, - H,.)<br />

6.5 U,(x,y) is equal to<br />

+ ky)n-‘. The<br />

k31 (;)(-l)kp'(~+ky)n-' =<br />

This proves (6.75). Let R,(x,y) =<br />

x~“U,(x,y); then Ro(x,y) =: 0 and R,(x,y) = R,-'(x,y) + l/n+y/x, hence<br />

R,(x,y) = H,+ny/x. (Incidentally, the original sum U, = U,(n, -1) doesn’t<br />

lead to a recurrence such as this; there<strong>for</strong>e the more general sum, which detaches<br />

x from its dependenice on n, is easier to solve inductively than its<br />

special case. This is another instructive example where a strong induction<br />

hypothesis makes the difference between success and failure.) The Fibonacci recurrence<br />

is additive,<br />

6.6 Each pair of babies bb present at the end of a month becomes a pair but the rabbits are<br />

of adults aa at the end of the next month; and each pair aa becomes an multiplying.

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