Vol. 10 No 6 - Pi Mu Epsilon

496 PI MU EPSILON JOURNAL
PROBLEMS AND SOLUTIONS 497
already noted. Since the sum of all its digits is equal to 3 times the diagonal
sum, any double triangle number is divisible by 3.
Also solved by PaulS. Bruckman, James Campbell, Mark Evans, Victor
G. Feser, S. Gendler, Richard I. Hess, David E. Manes, Kenneth M.
Wilke, Monte J. Zerger, and the Proposer.
877. [Spring 1996] Proposed by the late John M. Howell, Littlerock,
California.
For given constants a, b, c, d, let Do = a, a 1 = b, and, for n > 1, let
an = can-1 + dan-2·
a) Find an in terms of a, b, c, and d.
b) Find limn--- 011 (an lan-1>·
c) Find integers a, b, c, d so that the limit of part (b) is 3.
Solution by H. -J. Seiffert, Berlin, Germany.
a) Case 1: + 4d = 0. Here, an easy mathematical induction
argument shows that
a,. ( c)"- 1 = l ~ ~( b ac) ac]
- 2 n + 2 , n ~ 1. (1)
If c ¢ 0, this equation remains valid for n = 0.
Case 2: + 4d ¢ 0. Assuming that an is of the form X', then the
recursion formula becomes X'+ 1 = eX' + dx!'- 1 , which reduces to the
quadratic equation x - ex - d = 0. The roots are
and we have
c + .lc2 + 4d c - .fc2 + 4d
u = v and v = -=-----~v'------
2 2
u 2 =cu+d Vl=cv+d u+v=c
' , '
u - v = Jc 2 + 4d, and uv = -d.
Since x = u and x = v each satisfy the recursion formula, it is easily proved
by induction on n that
(b - av)u" - (b - au)v"
a = , n ~ 0.
II U- V
b) We must ensure that an ¢ 0 for all sufficiently large n. This is not
fulfilled if, for example, a = b = 0 or c = d = 0.
Case 1: + 4d = 0, c ¢ 0, and a ¢ 0 orb ¢ 0. From (1) it follows
that
lim a,. = E.
,._.., a,._ 1
2
Case 2: + 4d ¢ 0. Using (2) we find that
if b ~ av and lvl < lui,
if b ~au and lui < I vi,
if b = av, b ~ au and v ~ 0,
if b = au, b ~ av and u ~ 0.
There are many nontrivial cases in which the limit does not exist. For
example, if abd ¢ 0 and c = 0, then a2k = adk and a2k+ 1 = bdk. Here
at'at_ 1 is either adlb or bla, according as k is even or odd, and the limit
does not exist if a 2 d ¢ b 2 •
c) Let r > 1, take c = 1 and d = r(r- 1). Then u =rand v = 1- r,
so that we have I vI < I u I . From part (b) it follows that
a
lim-" = r if b ~ a(l - r).
,._ .. a,._.
Therefore, a possible choice is a = b = c = 1 and d = 6.
Also solved by Paul S. Bruckman, Russell Euler, S. Gendler, Richard
I. Hess, David E. Manes, Rex H. Wu, and the Proposer.
878. [Spring 1996] Proposed by Andrew Cusumano, Great Neck, New
York.
If x is a solution to the equation x - ax + 1 = 0, where a is an
in~er greater than 2, then show that~ can be written in the form p + ·~
qyr , where p, q, and r are integers.
(2)