MORE ON FIBONACCI NSM
JEREMY C. POND and DONALD F. HOWELLS*
Sussex, England and San Jose State College, San Jose, California
Fibonacci Nim [1] was o r i g i n a l l y stated as follows:
"Consider a game involving two p l a y e r s in which initially t h e r e
is a group of 100 or l e s s o b j e c t s . The f i r s t p l a y e r m a y r e d u c e
the pile by any Fibonacci n u m b e r ( m e m b e r of the s e r i e s 1, 1, 2,
3, 5, 8, 13, 21, . . . ). The second p l a y e r does likewise. The
p l a y e r who m a k e s the last move wins the g a m e . "
Let p e r s o n s A and B be playing the game which A w i n s .
If A is
to win he m u s t be able to r e d u c e the pile to z e r o on his final m o v e .
Thus A m u s t draw from
0+F
(n = 1, 2, 3, . . . )
on his final m o v e .
Looking at the sequence of the n u m b e r of objects from which A
m u s t draw to win on the final move, 1, 2, 3, 5, . . . , we see that 4 is
the f i r s t positive i n t e g e r m i s s i n g . If B is forced to play with 4 objects
r e m a i n i n g , A can c e r t a i n l y win the g a m e .
Now suppose A gets the opportunity to draw from
9, 12, . „ . ).
4+F (5, 6, 7,
A will be able to r e d u c e the pile to 4 objects and can
continue to win.
The s m a l l e s t positive i n t e g e r that is not contained in the union
of the s e t s
(1 0 + F ]) and {4+F ) is 10. If B is forced to d r a w from
n
A • n'
a pile of 10 objects, B cannot reduce the pile to 4 or 0 but B will leave
A in a position to r e d u c e the pile to 4 or 0 and thus A can win.
Now we wish to g e n e r a t e the sequence of positions from which it
is unsafe to draw (0, 4, 10, . . . ).
Let U , = 0 . Then U^ is the s m a l l -
e s t positive i n t e g e r which is not equal to U, +F
(n = 2, 3, . . . ).
is the s m a l l e s t positive i n t e g e r which is not equal to U, +F
or
U~
U?+F
(n= 2, 3, . . . ) .
Therefore
U (r = 2, 3, . „ . ) is the s m a l l e s t positive i n t e g e r
w h i c h i s not equal to U.+F , w h e r e t = 1, 2, . . . , r - 1 and n = 2, 3, . . .
^
t n
Student
61
62
February
MORE ON FIBONACCI NIM
u
n
r
°
410
14
20
24
30
36
40
46
50
56
60
66
72
76
82
86
92
96
1
5
11
15
21
25
31
37
41
47
51
57
61
67
73
77
83
87
93
97
2
5"~
12
16
22
26
32
38
42
48
52
58
62
68
74
78
84
88
94
98
3
7
13
17
23
27
33
39
43
49
53
59
63
69
75
79
85
89
95
99
8
12
18
22
28
32
38
44
48
54
58
64
68
74
80
84
90
94
100
5
9
15
19
25
29
35
41
45
51
55
61
65
71
77
81
87
91
97
13
17
23
27
33
37
43
49
53
59
63
69
73
79
85
89
95
99
21
25
31
35
41
45
51
57
61
67
71
77
81
87
93
97
34
38
44
48
54
58
64
70
74
80
84
90
94
100
55
59
65
69
75
79
85
91
95
89
93
99
The first player can always win if he starts on some position not
equal to U
(r = 1, 2, ...) and always reduces the pile to some U .
Here are all the values of U
0
76
169
254
332
410
488
572
4
82
176
260
338
416
494
576
10
86
186
264
342
420
498
14
92
192
270
348
426
504
20
96
196
274
352
430
510
24
102
202
280
358
436
514
30
108
206
284
364
442
520
36
112
212
290
368
446
524
thus far computed:
40
118
218
296
374
452
530
46
122
222
300
378
456
534
50
128
228
306
384
462
540
56
132
232
310
388
468
552
60
138
238
316
394
472
556
66
150
242
322
400
478
562
72
160
248
326
406
484
566
4.
5.
The following observations can be made:
U r+1
., = Ur + some non-Fibonacci number.
If U ± 1 - U =4, then U ,,-U ,,9^4 since 4+4=8=F/.
r+1 r
r+2 r+1'
o
Thus the average difference of U ,,-U = 5 , r=l, 2, 3, . . .
to
r+1
r
The density of (U } in the positive i n t e g e r s m u s t be ^ l/5„
The p r o b a b i l i t y that the s t a r t i n g p e r s o n can win is Z. 4 / 5 if noth-
10
ing is known about the s t a r t i n g position of the g a m e .
The following questions a r e left u n a n s w e r e d :
Is t h e r e a closed form solution for ] U r | ?
1.
20
3.
1965
M O R E ON F I B O N A C C I NIM
2.
W h a t i s t h e l i m i t i n g d e n s i t y of
Similar
results
a r e found
(U
when
63
} in the positive i n t e g e r s ?
one
considers
"Lucas
Nim"
a n a l o g o u s to F i b o n a c c i N i m .
REFERENCES
1.
B r o t h e r U. A l f r e d ,
onacci Quarterly,
2.
"Research Project:
Fibonacci Nim, " F i b -
1 ( 1 9 6 3 ) , N o . 1, p . 6 3 .
Michael J. Whinihan,
1 ( 1 9 6 3 ) , N o . 4, p p .
"Fibonacci Nim, "
Fibonacci
Quarterly,
9-13.
xxxxxxxxxxxxxxx
ON SUMS F 2 ±
F2
X
y
BENJAMIN SHARPE
State University of New York at Buffalo
F o r m u l a s f o r t h e s u m of t h e s q u a r e s of F i b o n a c c i n u m b e r s a r e :
(1)
F
2
n+2k
2
n =
(2)
F
2
n+2k+l
(3)
F
2
n+2k "
(4)
F
2
n+2k+l '
+ F
F
2
n =
+ F
2
n =
F
F
F
F
n+2k-2
F
2
n =
F
2k+1
2k
F
F
n+2k+l
+ F
2k-1
F
2n+2k-l
2n+2k+l
2n+2k
F
n-1
n+2
+ F
2k
F
2n+2k+2
V a l i d i t y of t h e a b o v e i s e s t a b l i s h e d b y u s i n g :
,_,
F
1 . n
_
0n,
= - ( a
- P ), L
n
^5
For example:
n
= a
n , 0n
1 +J5
1 -J5
,
Q
a
v
v
+ P , a =
, p =
, a P = -1
.
5 v^ * , , , - F 2 ) =
n+2k+l
n'
(a2n+4k+2
L
+ p2n+4k+2)
2n+4k+2 "
5(F
n-l
F
n+2
L
_
(a2n + p2n}
_
2Qnpn (Q2k+lp2k+l
2n ^(-l)n(-2) = L 2 n + 4 k + 2 - ^
+ F
2k
F
2n+2k+2)
_1}
- (-l)11"1
=
^
=
, 2 n + l • a 2 n + l x ^ , 2n+4k+2 __, a 2 n + 4 k + 2 x
i i - l ^ n - l , 3,A3N
(a
+ P
) + (a
+p
)- a
P
(a +P ) - a
L
2n+4k+2
2ka2k, 2n+2,ft2n+2.
p (a
+p
)=
+ (L
2n+l "
L
2n+2- ) " (~l)
L
3
=
XXXXXXXXXXXXXXX
L
2n+4k+2 "
L
2iT
(-1)
L
3e