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ftDVANCED PROBLEMS AND SOLUTIONS
Edited By
RAYMOND E. WHITNEY
Lock Haven State College, Lock Haven, Pennsylvania
Send all communications concerning Advanced Problems and Solutions to
Raymond Whitney, Mathematics Department, Lock Haven State College, Lock
Haven, Pennsylvania, 17745* This department especially welcomes problems
believed to be new or extending old results.
Proposers should submit solu-
tions or other information that will assist the editor,
To facilitate their con-
sideration, solutions should be submitted on separate signed sheets within three
months after publication of the problems,,
H-119
Proposed by L* C a r i s f z , Duke U n i v e r s i t y , Durham, N o * C a r o l i n a .
Put
H<».n.^|||;-^*(-i)()-)(-"-')(»--_»-i)
/
•
/n - j +p - k \ / p - k + i \
\
p - k
A
•*
;
Show that H(m,n,p) = 0 unless m, n,p are all even and that
xT/o
O O \
H(2m,2n,2p) =
min(m,n*p)
x^
t i\T
£
(-1)
v.
(m + n + p - r )J
r , r , ( m _ r ) , g _ r ) , fe _
t
r),=
(The formula
/m + n\
H(2m, 2n)
{ m )
2
•
where
»«-.-> - s£«-« i ^( i r ) )(""j 1 + J )C-=r , X m "i-"" J )
251
252
ADVANCED PROBLEMS AND SOLUTIONS
[Oct
is p r o v e d in the Fibonacci Q u a r t e r l y , Vol. 4 ( 1 9 6 6 ) , pp. 323-325.)
H-120 Proposed by M . N . S . Swamy, Nova Scotia Technical College, Halifax,,
Canada.
T h e Fibonacci polynomials a r e defined by
f n + 1 (x)
= x . f n (x) + fn_lW
fi(x) = 1,
If z
(i)
f2(x) = x .
= f (x) . f (y) then show that
z
s a t i s f i e s the r e c u r r e n c e r e l a t i o n ,
2
2
z ,. - xv
+ 2)z , - xJ y z , + z
J • z lo - x(x + Jy
n+4
n+3
n+2
n+l
n
(ii)
x
H-121
n
(x + Jy) 2 • ]T z
V
r
=0.
= (z , - z
) - v(xy
- l)(z
_,_, - z ).
J
x
n+2
n-l
n+l
n
Proposed by H, H. Ferns, University of Victoria,, Victoria, B . C . , Canada.
Prove the following identity.
?j° Vk
Fmi+K = F
1=1
where F
i s the n
X
\ nTkj nk A
Fibonacci number,
F
+
"
F
X
< m * k> >
m, A a r e any i n t e g e r s o r z e r o and
k is an even i n t e g e r o r z e r o .
W r i t e t h e f o r m t h e identity t a k e s if k i s an odd i n t e g e r .
Find an analogous identity involving L u c a s n u m b e r s .
H-122
Proposed by R. E. Whitney, Lock Haven State College, Lock Haven, Pa„
Let F
denote t h e n
Fibonacci n u m b e r e x p r e s s e d in b a s e 2. Conn
th
sider the ordered a r r a y F - ^ F s ° ° °.
Let g denote t h e n
digit of t h i s
a r r a y . Find a formula for g . If p o s s i b l e , g e n e r a l i z e for any b a s e .
1967]
H-70
ADVANCED PROBLEMS AND SOLUTIONS
2 53
Proposed by C . A e C h u r c h , J r . , W e V i r g i n i a U n i v 9 / M o r g a n t o w n , W . V i r g i n i a .
For n = 2m5 show that the total number of k-combinations of the first
n natural numbers such that no two elements i and i + 2 appear together in
the same selection is F 2 , and if n = 2m + 1, the total is F
F
,
m+2
m+2 m+35
Solution and comments by the proposer-
For his quick solution of the "probleme desmenages" Kaplanskyf2l gives
two results for combinations with restricted positions. We state them in the
following form:
The number of k-combinations of the first n natural numbers, on a
line, with no two consecutive is
\
rI/ n~- :kk +•Il n
; o< k<^j-i
a)
if arranged on a circle, so that 1 and n are consecutive, the number is
n
(2)
/n - k \
^^x r : ~ l ,
n - k 1 k J
See also [4, p0 198Je
Summed over k,
0<k<|
(1) and (2) give the Fibonacci and
Lucas numbers, respectively,,
For the problem as stated we use (1).
The restriction that i and i + 2 cannot appear in any selection can be
stated as (a) no two consecutive even integers appear and (b) no two consecutive odd integers appear s
If n = 2m, a k-combination with the stated restrictions will be made
up of s integers from among the m even, no two consecutive, and k - s
from among the m odd, no two consecutive
k
(3)
v
k\
Thus there are
/ m - s + 1 \ / m - (k - s) + l \
s
A
k s
-
/
k-combinations of the first 2m natural numbers such that i and i + 2 do not
appear.
254
ADVANCED PROBLEMS AND SOLUTIONS
[Oct,
Summing (3) over k we get the total number
2
F2
m+2
=
~
R?]
V
yk / m - s + l \ / m - ( k - s ) + 1\
s
k
s=o\
A
~S
/
feo
with the usual condition that
0
(>)
for
b > a > 0
For n = 2m + 1 we choose s from among the m even integers, no
two consecutive, and k - s from among the m + 1 odd integers, no two consecutive, to get that there are
k / m - s + l \ / m - (k - s) + 2 \
yx
(4)
s
k\
s=<
)\
k s
-
/
k-combinations of the first 2m + 1 natural numbers such that i and i + 2 do
not appear.
Summed over k,
(4) gives the total number
m+l
F
k
•fisr-.-r-t-n
F
m+2 m+3
It is also of interest to consider the circular analog of this problem by
way of (2)0
For n = 2m, 2 and 2m are taken to be consecutive as are 1 and 2m.
- 1. By the same argument as before we find that there are
k
v-^
^ m m
- s
s=o
(
m - s\
s
m
/m - (k - s)\
J m - (k - s) \
k-s
J
circular k-combinations such that i and i + 2 do not appear and a total of
1967]
ADVANCED PROBLEMS AND SOLUTIONS
2
L
2
m
=
H k
m
V
V
S
s=o
m
-
s
('m s
I
s
\
/
m
ni
k
255
/ m - (k - s ) \
k
s
s
- * "^V
"
/ '
F o r n = 2m + 1, 2 and 2m a r e consecutive as a r e 1 and 2m + 1 and
we have the total
m
L
L
m
^4
m+l
V
fco
k
m
V
s^o
m
"
/ m ~ s\
s
I
s
/
m + l
m
- * -
s)
/ m - (k - s) + l \
+
1
k
\
Mixed r e s u l t s can also be obtained using both (1) and (2).
-
s
/
For example,
one can take l i n e a r combinations on the evens and c i r c u l a r combinations on
the odds.
Remarks.
T h e p r o b l e m posed in H-70 f i r s t appeared in t h e l i t e r a t u r e in
a p a p e r by N. S. Mendelsohn ["3J; an explicit formula was not obtained.
f i r s t explicit formula w a s given by M. A b r a m s o n [l,
solution for t h e n u m b e r of
k-combinations
l e m m a 3j„
such that
i
and
The
AbramsonTs
i + 2 do not
a p p e a r t o g e t h e r is
W
/ n - 2k + s + 2 \
k s
-
k\
/ k -
s
s\
M )
REFERENCES
1.
M. A b r a m s o n ,
"Explicit E x p r e s s i o n s for a C l a s s of P e r m u t a t i o n P r o b -
lems, " C a n a d ^ J M a ^
7(1964), pp. 345-350.
2. I. Kaplansky, Solution of the " P r o b l e m e d e s Menages,
M
Bull. A m e r . Math
Soco , 49(1943), 784-785.
3.
N. S. Mendelsohn, "The Asymptotic S e r i e s for a C e r t a i n C l a s s of P e r m u tation P r o b l e m s , Canad. J. of Math. , 8(1956), pp. 234-244.
4.
J . Riordan, An Introduction to Combinatorial A n a l y s i s , New York,
1958.
H-73 Proposed by V . E . Hoggatt,Jr 8 / San Jose State ColIege / San Jose, Calif.
Let
f0(x) = 0,
f t (x)
= 1
256
ADVANCED PROBLEMS AND SOLUTIONS
and
f n+2 (x) = xf n + i (x) + f n (x) ,
[Oct.
n - 0
and let b (x) and B (x) be the polynomials in H-69; show
f
^ (x) = xB v(x 2 ) ,
2n+2
n
and
f
2n+l
<x) = b (x 2 )
n
.
T h u s t h e r e is an intimate r e l a t i o n s h i p between the Fibonacci polynomials, f (x)
and the Morgan-Voyce polynomials b (x) and B (x)„
Solution by Douglas L i n d , U n i v . o f V i r g i n i a , C h a r l o t t e s v i l l e , V i r g i n i a .
U s i n g t h e explicit r e p r e s e n t a t i o n s of B (x) and b (x) given in H-69, and
of f (x) given in B - 7 4 , we find
x
2n-2r
2n-2r+i
-
c
t .
= f 2n +2( x ) >
,v
T h e s e r e l a t i o n s have been given by Rc Ae Hayes \_ "Fibonacci and Lucas P o l y n o m i a l s , " ( M a s t e r ' s T h e s i s ) ; equations (3.4-1) and ( 3 . 4 - 2 ) ] .
H-77
Proposed by V . E . H o g g a t t , J r 0 /
San Jose State C o l l e g e , San Jose, C a l i f .
Show
2n+i / _
, i \
i
=h )
for all i n t e g e r s k.
F
2k+2j+l
_
5
L
2n+2k+2
Set k = -(n + 1) and d e r i v e
5W
F
-
^n
1967]
ADVANCED PROBLEMS AND SOLUTIONS
257
a r e s u l t of S. G. Guba P r o b l e m No* 174, I s s u e No, 4, J u l y - A u g u s t 1965, p .
' 73 of Matematika V. SkSle.
Solution by L. Carlifz, Duke University, Durham, North Carolina.
Since
Ln = a
+ ^B s,
Fn =
n
0n
a - p— '
]8 = I
(1 - V 5 )
where
a = |
(1 + V 5 ) 5
1 + Q'2 = o^V5 ,
1 + p2 = -pV5
,
it follows that
n
n
/ \
=
k+2
/ \
^
a k ( l + a2)
- £ k ( T + j3 2 ) n
a - j8
=
( a k + n - (-l)n/3k+n) (V5)n
V5
( 5(n-l)/2L,A
_ J
k+n
( 5
Fk+n
thus g e n e r a l i z i n g the s t a t e d result 0
§(")*Note that
(n odd)
(n even) ,
In p a r t i c u l a r , for k = - n ,
2.5M/2
j
J
I
0
(nodd)
.
(n even) .
we get
258
2
ADVANCED PROBLEMS AND SOLUTIONS
f72n+l\F
_f/2n+l\F
+
2
[Oct.
^/2n+l\
-2n-i+2j
-5("; l )w^£(T 1 )',
2n+i-2j
2
S(2"-/)
>. i
„
r i F2J+1
3=l
so that
1
Sfr )
F2j+i
"
5n
Similarly, we have
n / \
';) L k+2 . = « k a + C 2) n + /s k a + /s2)r
SO
fck+n
i
In particular, since
+ (-l)n^k+n)(V5)n
5
5
L, ,
(n even)
k+n
V
k+n
(n odd) .
(n+1)/
gfflw-W •$(?)--*-?(T)
-2n+2j
so that
LATE ACKNOWLEDGEMENT: Problems H-64, H-71, H-72, H-73, and H-77
were also solved by M. N. S. Swamy.
****«-