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Transcript 
ELEMENTARY PROBLEMS AND SOLUTIONS
Since n k is the smallest n for which D(n) = k» it follows that
n,
is the smallest n for which R(n) - nk.
Now taking the case where n
= 3 (mod 4 ) , this leads to
"fc+i =i<"fe + 3 ) 2 - 2
and we have also n k+1 = 3 (mod 4). Hence, starting with n3 = 3, we can use the
above recursive algorithm for k ^ 3.
Also solved
Sahib Singh,
by Paul S. Bruckman,
and the
proposer.
Hans Kappus,
L. Kuipers,
Jerry
M.
Metzger,
Generalized Fibonacci Numbers
B-5^9
Proposed
by George N. Philippou,
Nicosia,
Cyprus
Let # 0 , Hl9 ... be defined by H0 = q - p, H1 = p, and Hn+2 = # n+1 + # n for
n = 0, 1,
. Prove that, for n > m > 0,
Solution
by L. A. G. Dresel,
University
Define D(n, m) = J?„+1flm - Hm+1Hn.
of Reading,
England
Then
Z>(n, m) = #n(ff„ + /?„_!> - #„(ffm + 5m_i)
" 3 A - 1 - f l A - i - - ^ ( » - 1. « - 1).
Repeating this reduction step a further m - 2 times, we obtain
D(n, m) = (-l)m_1D(rc - m + 1, 1)
= (-Dm +1(pffn.m+2 - ^ .
m +1 ).
Also solved by Paul 5. Bruckman, Piero Filipponi,
dam, L. Kuipers,
Bob Prielipp,
A. G. Shannon,
J. Suck, and the
proposer.
184
Herta T. Freitag,
P. D. Siafarikas,
A. F. HoraSahib
Singh,
[May
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