Concrete mathematics : a foundation for computer science

6 RECURRENT PROBLEMS
through any of the existing intersection points (hence it intersects them all
in different places). The recurrence is therefore
Lo = 1;
L, = L,-l +n, for n > 0.
(1.4)
The known values of L1 , Lz, and L3 check perfectly here, so we’ll buy this.
Now we need a closed-form solution. We could play the guessing game
again, but 1, 2, 4, 7, 11, 16, . . . doesn’t look familiar; so let’s try another
tack. We can often understand a recurrence by “unfolding” or “unwinding”
it all the way to the end, as follows:
L, = L,_j + n
= L,-z+(n-l)+n
= LnP3 + (n - 2) + (n - 1) + n
= Lo+1 +2+... + (n - 2) + (n - 1) + n
= 1 + s,, where S, = 1 + 2 + 3 + . . + (n - 1) + n.
In other words, L, is one more than the sum S, of the first n positive integers.
The quantity S, pops up now and again, so it’s worth making a table of
small values. Then we might recognize such numbers more easily when we
see them the next time:
n 1 2 3 4 5 6 7 8 9 10 11 12 13 14
S, 1 3 6 10 15 21 28 36 45 55 66 78 91 105
Unfolding?
I’d call this
“plugging in.”
These values are also called the triangular numbers, because S, is the number
of bowling pins in an n-row triangular array. For example, the usual four-row
array ‘*:::*’ has Sq = 10 pins.
To evaluate S, we can use a trick that Gauss reportedly came up with
in 1786, when he was nine years old [73] (see also Euler [92, part 1, $4151): It seems a lot of
stuff is attributed
to Gauss-
s,= 1 + 2 + 3 +...+ (n-l) + n
either he was really
+Sn= n + (n-l) + (n-2) + ... + 2 + 1 smart or he had a
great press agent.
2S, = (n+l) + (n+l) + (n+l) +...+ (n+1) + (n+l)
Maybe he just
We merely add S, to its reversal, so that each of the n columns on the right ~~~s~n~,!~etic
sums to n + 1. Simplifying,
s _ n(n+l)
n-
2 ’
for n 3 0. (1.5)