Aug 14, 2008 - generalized right$adjusted Pascal matrix that they have the same ... between the generalized Pascal matrix and the Fibonomial coeffi cients ...
The Generalized Fibonomial Matrix Emrah Kilic TOBB University of Economics and Technology, Mathematics Department, 06560 Ankara Turkey
Abstract The Fibonomial coe¢ cients are known as interesting generalization of binomial coe¢ cients. In this paper, we derive a (k + 1)th recurrence relation and generating matrix for the Fibonomial coe¢ cients, which we call generalized Fibonomial matrix. We …nd a nice relationship between the eigenvalues of the Fibonomial matrix and the generalized right-adjusted Pascal matrix that they have the same eigenvalues. We obtain generating functions, combinatorial representations, many new interesting identities and properties of the Fibonomial coe¢ cients. Several cases of our results are given as examples. Key words: The Fibonomial coe¢ cients, generating matrix, recurrence relation, generating function
1
Introduction
The well known Fibonacci numbers are de…ned by Fn = Fn
1
+ Fn
2
with initial conditions F0 = 0 and F1 = 1; for n > 1: The Fibonomial coe¢ cient is de…ned by the relation for n " h i
n m
#
= F
m
1
F1 F2 : : : Fn (F1 F2 : : : Fn m ) (F1 F2 : : : Fm )
h i
with n0 = nn = 1 where Fn is the nth Fibonacci number. These coe¢ cients F F satisfy the relation : "
n m
#
= Fm+1 F
"
n
#
1 m
Preprint submitted to Elsevier Science
+ Fn F
m 1
"
n m
#
1 1
: F
14 August 2008
Let p be a nonzero integer. De…ne the generalized Fibonacci and Lucas sequences by the recurrences: un = pun vn = pvn
+ un 1 + vn 1
2 2
where u0 = 0; u1 = 1 and v0 = 2; v1 = p; respectively, for all n
2:
When p = 1 and p = 2, then un = Fn (nth Fibonacci number) and un = Pn (nth Pell number), respectively. The authors [14] were the …rst to study generalized Fibonomial coe¢ cients formed by terms of sequence fun g as follows: for n m 1 (
)
u 1 u2 : : : un n = (u1 u2 : : : un m ) (u1 u2 : : : um ) m
with
n o n 0
=
n o n n
= 1: When p = 1; the generalized Fibonomial coe¢ cient
is reduced to the Fibonomial coe¢ cient The n
h i
n : m F
n o n m
n generalized Pascal matrix Vn whose (i; j) entry is given by j (Vn )ij = j+i
For example,
1 n
!
1
pi+j
n 1
:
3
2
60 0 0 1 7 7 6 7 6 6 0 0 1 3p 7 7: V4 = 6 7 6 6 0 1 2p 3p2 7 7 6 4 5
1 p p2 p 3
Recently there have been increasing interest in both the Fibonomial coe¢ cients and certain generalized matrix of binomial coe¢ cients, which we call them as generalized Pascal matrices. Regarding left or right adjustments, and certain coe¢ cients generalizations, several authors give various names to Pascal matrices. For example Carlitz [1] considered the right adjusted Pascal matrix and he called it as "matrix of binomial coe¢ cients". In [5], Edelman and Gilbert considered left adjusted matrix of binomial coe¢ cients and he called it "Pascal matrix". In [23], the author considered right adjusted and coe¢ cient generalized matrix of binomial coe¢ cient and he gave a name "Netted Matrix" to this matrix. Regarding generalization of binomial coe¢ cients, several authors have studied 2
the generalized Fibonomial coe¢ cients and their properties (see for more details [7,10,14,21,26,27]). Meanwhile, some authors consider the spectral properties of the generalized Pascal matrix ([1,3,11,22]). Since some relationships between the generalized Pascal matrix and the Fibonomial coe¢ cients have been constructed, the Fibonomial coe¢ cients are considered by some author. In this paper, we give more powerful relationships between the Fibonomial coe¢ cients and certain Pascal matrix. Matrix methods and generating matrices are very useful for solving some problems stemming from number theory. In this paper, we de…ne the generalized Fibonomial matrix and derive an (k + 1)th order linear recurrence relation for the generalized Fibonacci coe¢ cients. Also we show that the generalized Fibonomial and Pascal matrices have the same characteristic polynomials and therefore same eigenvalues. We obtain some explicit and closed formulas for the coe¢ cients and their sums by matrix methods. We give generating functions, properties and combinatorial representations for them. Further we present some relationships between determinants of certain matrices and the generalized Fibonacci coe¢ cients.
2
Generalized Fibonomial Coe¢ cients
This section is mainly devoted to derive a recurrence relation and generating matrix for the generalized Fibonomial coe¢ cients: For the sake of compactness, we shall use the following notations, for …xed k such that 1 i k + 1 : (i 1)(i 2)=2
an;i = ( 1) where n o n m
n o n m
(
)(
n+k k i+1
)
n+i 2 i 1
stands for the generalized Fibonomial coe¢ cients and it is supposed
= 1 when either m > n and n
0:
For later use, we give the following useful result. Lemma 1 For n > 0 and 1
i
k
a1;i an;1 + an;i+1 = an+1;i where an;i be as before.
PROOF. For the …rst case i = 1; the proof can be found in [10]. For the other cases, that is, i > 1; if we simplify the equality a1;i an;1 + an;i+1 = an+1;i , 3
it is reduced to the form: uk+1 un+i + ( 1)i
1
u n uk
i+1
= ui un+k+1 :
The last equality can be easily obtained from the Binet formula of fun g : Thus the proof is complete.
De…ne the (k + 1) follows:
(k + 1) companion matrix Gk and the matrix Hn;k as 3
2
6 a1;1 a1;2 : : : a1;k+1 7
6 6 6 1 Gk = 6 6 6 6 4
0
7 7 7 7 and Hn;k 7 7 7 5
.. .
...
1
0
2
an;2 6 an;1 6 6 6 an 1;1 an 1;2 =6 6 .. .. 6 . . 6 4
an
k;1
an
k;2
3
: : : an;k+1 7 7 : : : an 2;k+1 7 7 7: 7 .. ... 7 .
: : : an
k;k+1
(1)
7 5
The matrix Gk is said to be generalized Fibonomial matrix. Now we give one of the our main results by the following Theorem. Theorem 2 For all n > 0; Gnk = Hn;k : PROOF. By the de…nitions of matrix Hn;k and Fibonomial coe¢ cients, the proof can be easily seen for the case n = 1: Suppose that the equation holds for n 1: Now we show that the equation holds for n + 1: Thus we write Gn+1 = Gk Gnk = Gk Hn;k : k From Lemma 1 and a property of matrix multiplication, we get Gn+1 = Gk Hn;k = Hn+1;k : k Thus the theorem is proven.
It is valuable to note that when A = 1 and k = 1; we obtain well-known fact : 2
3
2
3
61 17 6 Fn+1 Fn 7 5 and H1;n = 4 5:
G1 = 4
10
Fn Fn
1
Now we give a linear recurrence relation for the generalized Fibonomials coe¢ cient. 4
Corollary 3 For n; k > 0; the generalized Fibonomial coe¢ cients satisfy the following order-(k + 1) linear recurrence relation an+1;1 =
k+1 X
a1;i an
i+1;1
(
)(
i=1
or clearly (
)
(
)(
)
n+k+1 k+1 = k k
)
n+k k+1 n+k 1 + k k 1 k ( )( ) n+1 (k 1)(k 2)=2 k + 1 + : : : + ( 1) + ( 1)k(k 1 k n
1)=2
( )
n k
o
PROOF. Since an;1 = n+k and from matrix multiplication, by equating k (1; 1) entries in the equation H1;k Hn;k = Hn+1;k ; the proof is easily seen. Considering the generalized Fibonomial matrix, we obtain following identities. Corollary 4 For n > 0; the following identities hold an
1;1
= an;k+1
am+n+1 an+t+1
i;j i;j
=
=
Pk+1 t=1
Pk+1
m=1
an+1
i;t am+1 t;j
an+r+1
for all m > 0
i;m at r+1 i;j
for t > 0 and t > r
an+1;1 = a1;1 an;1 + an;2 an+1;k+1 = an;1 a1;k+1 an+1;i = a1;i an;1 + an;i+1 for 2
i
k
PROOF. The all identities above can be proved by considering a property of matrix multiplication in Hn+1 = Hn H1 ; Hn+m = Hn Hm and Hn+t = Hn+r Ht r for n; m > 0 and t > r:
3
The eigenvalues of matrix Gk
In this section we determine the eigenvalues of matrix Gk : Since the matrix Gk is a companion matrix, its characteristic polynomial can be easily derived. Let fn;k (x) denote the characteristic polynomial of the matrix Hn;k , then we have the following Corollary. 5
Corollary 5 For n; k > 0, fn;k (x) =
k+1 X
t(t+1)=2
( 1)
t=0
(
)(
n+k k t+1
)
n + t 2 k+1 t x : t 1
Clearly the characteristic polynomial of Gk is given by f1;k (x) =
k+1 X
i(i+1)=2
( 1)
i=0
(
)
k+1 k x i
i+1
:
Here we should note that in [13,9,10,4], the authors gave the characteristic equation of generalized Fibonomial coe¢ cients as shown: n X
Cn (x) =
h(h+1)=2
( 1)
h=0
where
n o n h
( )
n n x h
h
is de…ned as before.
Moreover in [4], the authors proved the conjecture of Horadam and Mahon, and they gave a very nice relationship between the characteristic polynomials of the generalized Fibonomial coe¢ cients and the n n generalized Pascal matrix Vn : Let Rn (x) be the characteristic polynomial of matrix Vn : From [4], we have that Cn (x) = Rn (x) Therefore we derive a nice relationship between the characteristic polynomials of matrix Gk and the polynomial of Vn as follows: f1;n
1
(x) = Cn (x) = Rn (x) :
So we have the following Corollary. Corollary 6 ([4]) Let ; Cm+1 (x) = f1;m (x) are: n
( 1)j
m 2j
; ( 1)j
n
( 1)k ; ( 1)j
m 2j
= A
m 2j
p
A2 + 4 =2: The characteristic roots of
o
j=0;1;:::k 1 o j m 2j 1) j=0;1;:::k 1
; (
if m = 2k
1;
if m = 2k:
Equivalently, the roots of f1;m are n
( 1)m=2 ; ( 1)j
n
( 1)j
m 2j
m 2j
; ( 1)j
; ( 1)j
m 2j
o
m 2j
o
0 j 0; k+1 Y
(x
n i)
=
k+1 X
t(t+1)=2
( 1)
t=0
i=1
(
)(
n+k k t+1
)
n + t 2 k+1 x t 1
and especially for n = 1; k+1 Y i=1
(x
i)
=
k+1 X
i(i+1)=2
( 1)
i=0
7
(
)
k + 1 k+1 i x : i
t
;
=
Considering the results of Corollary 6, we derive the following facts:
f1;4m+4 (x) = (x4 f1;4m+5 (x) = (x4 f1;4m+6 (x) = (x4 f1;4m+7 (x) = (x4
c4m x3 d4m x2 c4m x + 1)f1;4m c4m+1 x3 d4m+1 x2 + c4m+1 x + 1)f1;4m+1 c4m+2 x3 d4m+2 x2 c4m+2 x + 1)f1;4m+2 c4m+3 x3 d4m+3 x2 + c4m+3 x + 1)f1;4m+3 :
In general we obtain the following identity :
f1;t+4 (x) = (x4
where ct = vt+4
ct x3
dt x2 + ( 1)t+1 ct x + 1)f1;t
(2)
vt+2 and dt = vt+4 vt+2 + ( 1)t+1 2:
After arranging the right hand side of (2), equating the corresponding coe¢ cients of xn in the both sides gives us the following new result: Corollary 8 For all t
(
)
(
j;
)
(
)
(
)
(
) (
)
t+5 t+1 t+1 t+1 t+1 t+1 = +( 1)i+1 ct +dt +( 1)i+t ct + i i i 1 i 2 i 3 i 4
where ct is de…ned as before.
PROOF. From (2), we write 8
f1;t+4 (x) = (x4 ct x3 dt x2 + ( 1)t+4 ct x + 1)f1;t = xt+5 ( ) ( )! t+1 t+1 t+4 x + ct 1 0 ( ) ( ) ( )! t+1 t+1 t+1 t+3 ct + dt x 2 1 0 ( ) ( ) ( ) ( )! t+1 t+1 t+1 t+1 t+2 +x + ct + dt ct 3 2 1 0 ( ) ( ) ( ) ( ) ( )! t+1 t+1 t+1 t+1 t+1 t+1 +x ct + dt + ct + 4 3 2 1 0 ( ) ( ) ( ) ( ) ( )! t+1 t+1 t+1 t+1 t+1 t + ct + dt ct + x 5 4 3 2 1 ::: ( ) ( ) ( ) t+1 t+1 t+1 i(i+1)=2 t+5 i i+1 + ( 1) x + ( 1) ct + dt i i 1 i 2 ( ) ( )! t+1 t+1 + ( 1)i+t ct + i 3 i 4 ::: +1 Comparing the coe¢ cient of xi for 1 i polynomial f1;t+4 , the proof is complete.
n in the above equation and the
In [3], the authors show that tr (Vn ) =
u(k+1)n uk
where Vn is the generalized Pascal matrix. Since the matrices Hn;k and Vn have the same eigenvalues, alternatively we have also that u(k+1)n : tr (Hn;k ) = un By Corollary 6, we can give the following result for both the generalized Fibonomial and Pascal matrices. Theorem 9 For n > 0;
tr (Hn;k ) =
8 bm=2c > P > > ( 1)in v(m < i=0 > (m P2)=2
> > :
i=0
2i)n
( 1)in v(n
9
2i)n
if m
1; 3 (mod 4) ;
if m
0; 2 (mod 4) :
4
Diagonalization of the matrix Gk and the Generalized Binet formula
In this section we consider diagonalization of the matrix Gk and then give the generalized Binet formula for the generalized Fibonomial coe¢ cients. From Corollary 6, we know that if 1 ; 2 ; : : : ; k+1 be the eigenvalues of matrix Gk ; then they are all distinct. Thus we can diagonalize the matrix Gk : De…ne the (k + 1) (k + 1) Vandermonde matrix V and diagonal matrix D = diag ( 1 ; 2 ; : : : ; k+1 ) as shown: 2
k 2
k 1
:::
6 6 . . 6 . . 6 . . 6 6 2 2 V =6 6 1 2 ::: 6 6 6 1 2 ::: 6 4
1 1 1
Since
i
6=
j
all i and j for 1
k k+1
3
2 7 7 .. 7 6 6 . 7 6 7 6 7 2 7 and D = 6 6 k+1 7 6 7 6 7 7 4 k+1 7 5
3
1 2
..
. k+1
1
7 7 7 7 7 7 7 7 5
k + 1; det V 6= 0:
i; j
(i)
Let Vj is the (k + 1) (k + 1) matrix obtained from V T by replacing the jth column of V by wi where T
wi =
n i+k+1 1
n i+k+1 2
n
on
:::
n i+k+1 k+1
:
o
n+i 2 Recalling an;i = ( 1)(i 1)(i 2)=2 kn+k ; we give the generalized Binet i+1 i 1 formulas for the generalized Fibonomial coe¢ cients by the following Theorem.
Theorem 10 For n,k > 0; (i)
an
i+1;j
=
det Vj
:
det (V )
PROOF. One can check that Gk V = V D: Since V is the invertible matrix and Gnk = Hn;k , we write Gnk V = Hn;k V = V Dn : Clearly we get the following linear equation system: k 1 hi1 k2
hi1
k 1 1 hi2 k2 1
2 2 1 hi;k 2 22
+ hi2
+ : : : + hi;k
+ hi;k
1
+ hi;k+1 =
+
+ ::: +
+ hi;k
2
+ hi;k+1 =
n i+k+1 1 n i+k+1 2
.. . hi1
k k+1
+ hi2
k 1 k+1
+ : : : + hi;k
2 2 k+1
10
+ hi;k
k+1
+ hi;k+1 =
n i+k+1 : k+1
Thus by the Cramer solution of the above system, we have the conclusion. After some calculations, we present some identities as examples of Theorem 10. Case I When k = 3; det (V ) = (
u22 u3 D3 and for n > 0 we have that
)
n+2 = u1 u2 u3 u3n+3 + ( 1)n+1 u3 un+1 =D; 3 ( )( ) n+3 n = u3n+5 + ( 1)n+1 (u2 un + un+5 ) =u2 D; 2 1 ( )( ) i h n+3 n+1 = u3 u3n+4 + ( 1)n+1 u22 (un+4 + un ) un 4 =u2 u3 D 1 2 where D = A2 + 4: Especially when A = 1; un = Fn (nth Fibonacci number) and so h
Fn Fn+1 Fn+2 = 4 F3n+3 + 2 ( 1)n+1 Fn+1 h
i
=5; i
Fn Fn+2 Fn+3 = F3n+5 + ( 1)n+1 (Fn+5 + Fn+1 ) =5; Fn Fn+1 Fn+3 = [F3n+4 + ( 1)n (Fn+1 + Ln )] =5:
Case II When k = 4; det (V ) = u42 v2 u23 D5 and for n (
0
)
n+3 v4n+6 v4 + v12 + ( 1)n+1 v1 v2 v2n+3 = ; 4 u22 v2 u3 D2 ( )( ) v4n+9 + v5 v3 + v1 + ( 1)n (v1 v2n+2 v2n+5 n+4 n = u1 u2 u 3 D 2 3 1 ( )( ) n+4 n+1 v4n+8 v6 v2 + v0 + ( 1)n (v1 v2n+1 + v2n+6 = 2 2 u22 D2 ( )( ) n+4 n+2 v4n+7 v5 + v3 v1 + ( 1)n (v1 v2n+4 v2n+7 = 1 3 u 2 u3 D 2
v2n+9 ) v2n+8 ) v2n 1 )
From these results, we get Fn Fn+1 Fn+2 Fn+3 = L4n+6
6 + 3 ( 1)n+1 L2n+3 =25;
Fn Fn+2 Fn+3 Fn+4 = (L4n+9 + 8 + ( 1)n (L2n+2 3L2n+7 )) =25; Fn Fn+1 Fn+3 Fn+4 = (L4n+8 19 + ( 1)n (L2n+1 L2n+7 )) =25; Fn Fn+1 Fn+2 Fn+4 = L4n+7
8 + ( 1)n+1 (L2n 15
1
+ L2n+3 + L2n+6 ) =25:
Case III When k = 5; det (V ) = u42 u33 u24 u5 D 2 and for n 11
0
; ; :
(
)
n+4 u5n+10 +( 1)n+1 (u2 u3n+9 +u3 u3n+4 ) (u3 (un+6 +un 2 )+u22 un+2 ) = ; u2 u3 u4 u5 D 2 5 ( )( ) n+1 n+5 n u3 u3n+6 )+u4 un+7 u3 (un+2 +un 2 )) ; = (u5n+14 +( 1) (u3n+14 +u3n+10 u2 u3 u4 D 2 4 1 ( )( ) n+1 n+5 n+1 ) un+11 u2 u3 un+4 u3 un 3 ) = (u5n+13 +( 1) (u2 u3n+12 uu32uu3n+5 ; 2 2 3D 3 2 ( )( ) n+1 n+5 n+2 ) u4 un+7 2u2 un+1 un 2 un 6 ) = (u5n+12 +( 1) (u3 u3n+10 +u2 uu3n+3 ; 2 u D2 2 3 2 3 ( )( ) n+1 n+5 n+3 ) u3n+1 ) u3 un+7 u3 un+3 u4 un 2 ) = (u5n+11 +( 1) (u2 (u3n+10 +u3n+6 u2 u3 u4 D 2 1 4
As applications to the Fibonacci sequence, we obtain
Fn Fn+1 Fn+2 Fn+3 Fn+4 = (F5n+10 + ( 1)n+1 (F3n+9 + 2F3n+4 ) 2 (Fn+6 + Fn 2 ) + Fn+2 )=25 Fn Fn+2 Fn+3 Fn+4 Fn+5 = (F5n+14 + ( 1)n+1 (F3n+14 + F3n+9 + F3n+5 ) +3 (Fn+7 2Fn ))=25 Fn Fn+1 Fn+3 Fn+4 Fn+5 = (F5n+13 + ( 1)n+1 (F3n+12 2F3n+5 ) (Fn+11 + 2Fn+4 + 2Fn 3 ))=25 Fn Fn+1 Fn+2 Fn+4 Fn+5 = (F5n+12 + ( 1)n+1 (2F3n+10 + F3n+3 ) (3Fn+7 + 2Fn+1 + 3Fn 4 ))=25 Fn Fn+1 Fn+2 Fn+3 Fn+5 = (F5n+11 + ( 1)n+1 (F3n+10 + F3n+6 F3n+1 ) (2Fn+7 + 2Fn+3 + 3Fn 2 ))=25
(e )
Let Vj i be a k k matrix obtained from the Vandermonde matrix V by replacing the jth column of V by ei where V is de…ned as before and ei is the ith element of the natural basis for Rn ; that is,
ei = (0; : : : ; 0; 1 ; 0; : : : 0)T # ith
12
and
2
(e ) Vj i
Let
(i) qj
=
(e ) Vj i jV j
6 6 6 6 6 6 6 6 6 6 6 6 6 =6 6 6 6 6 6 6 6 6 6 6 6 6 4
k 1 k 1 1
:::
k j 1
:::
k 1 j 1
.. .
.. .
0
k j+1
0 .. .
k 1 j+1
::: :::
.. .
k i+1 1
:::
k i+1 j 1
0
k i+1 j+1
:::
k i 1
:::
k i j 1
1
k i j+1
:::
k i 1 1
:::
k i 1 j 1
0 .. .
k i 1 j+1
:::
.. .
.. .
.. .
1
:::
j 1
0
j+1
:::
1
:::
1
0
1
:::
# ei
where the (k
(ei )
k) matrices Vj
k k+1
3
7 7 7 7 7 .. 7 . 7 7 7 k i+1 7 7 k+1 7 7 k i 7: k+1 7 7 k i 17 7 k+1 7 .. 7 . 7 7 7 7 7 k+1 7 5 k 1 k+1
1
and V are de…ned as before.
Then we give the following Theorem. Theorem 11 Let 1 ; 2 ; : : : ; k+1 be the distinct roots of xk+1 a1;2 xk 1 : : : a1;k 1 x a1;k+1 = 0: For any integer n and 1 i an;i =
k+1 X
a1;1 xk k + 1;
(i) n+k : j
qj
j=1
PROOF. We consider the following system of k linear equations with k unknowns x1 ; x2 ; : : : ; xk : for 1 i k k 1 x1
+
k 2 x2
+ ::: +
+
k i+1 x2 2
k k+1 xk+1
=0
.. . k i+1 x1 1 k i 1 x1
+
k i 2 x2
+ ::: +
+ ::: +
k i 1 x1 1
+
k i 1 x2 2
1 x1
2 x2
+ ::: +
k i+1 k+1 xk+1
k i k+1 xk+1
+ ::: +
+
k+1 xk+1
x1 + x2 + : : : + xk+1 = 0:
13
=1
k i 1 k+1 xk+1
.. . =0
=0
=0
The system above is equivalent to 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4
k 2
k 1
.. .
k k+1
:::
76 76 76 76 76 76 k i+1 7 6 76 k+1 76 6 k i 7 76 k+1 7 6 76 k i 176 76 k+1 76 6 .. 7 6 . 7 76 76 76 6 k+1 7 76 54
.. .
.. .
k i+1 1
k i+1 2
:::
k i 1
k i 2
:::
k i 1 1
k i 1 2
:::
.. .
.. .
1
2
:::
1
1
:::
32
1
2 3
3
607 6.7 6.7 7 6.7 7 6 7 7 6 7 7 607 7 6 7 7 6 7 7 6 7 7 617 7 6 7 7 7 = 6 7; 6 7 7 607 7 6 7 7 6.7 7 6 .. 7 7 6 7 7 6 7 7 6 7 7 607 7 6 7 7 4 5 5
x1 7 7 x2 .. . .. . xj .. . xk xk+1
0
| {z } ei
by the solution of Vandermonde’s determinants and Cramer rule, we get (ei )
(i) qj
Thus for n; k > 0 and 1
Vj
=
(i = 1; 2; : : : ; k + 1) :
jV j
i
k + 1; fni
=
k+1 X
(i) n+k ; j
qj
j=1
which completes the proof.
As an example of the above result, when k = 2; 1 = 2 ; 2 = 2 ; 3 = 1 are the roots of x3 a1;1 x2 a1;2 x a1;3 = 0: After some computations, we get
(1)
q1 = ( (2)
q1 =
2)
; q2 =
1
2 )( 1
3)
3 )( 1
2
3 )( 2
; q2 =
(
2
3 )( 1
2)
(
1
2 )( 2
; q2 =
1)
; q3 = 2)
14
3 )( 1
2
3)
; q3 =
3)
;
1+ 2
; q3 =
(
(3)
1 3
Therefore, by Theorem 11, we get
1 (
(2)
1+ 3
(3)
2 3 1
(1)
1 (
(2)
2+ 3
(
(3)
q1 = (
(1)
1 3 )( 1
1
2
3 )( 1
3)
1 2
(
2
3 )( 1
3)
:
;
an;1 = = an;2 =
n
n+2 2
(
and since
=
(
(
n+2 1
3)
=
n+1 2
(
3) ( 2
2
+
3)
( (
3) ( 1 n+1 1 1
+
(
3) ( 1
2
n+2 2
3)
3) ( 1
2
o
1
1
+
1)
n+2 3
( 2)
(
1
+
2)
3)
n+2 3
3) ( 1
2
;
3)
1;
n+2 2 3 1
(
n+2 2
+
2)
2) ( 1
1
n
(
+
2
1 2 3
an;3 = =
n+2 1
)( 1 on 3o 1 n+2 n 1 1
n
=
o
2)
3) ( 1
+
n+2 1 3 2
(
2)
1) ( 2 n+1 2
2
+
(
3)
+
1) ( 2
2
n+2 1 2 3
(
3)
3) ( 1 n+1 3
2
+
(
3) ( 1
2
3)
3)
= an 1 ; 1: Since the de…nition of fan;i g for k = 2, the fact an;3 = seen.
5
an
1;1
can also be
On sums of generalized Fibonomial coe¢ cients
In this section, we consider the sum of the generalized Fibonomial coe¢ cients. To compute sum, we de…ne a new generating matrix by extending the matrix Gk given in (1). De…ne the (k + 2)
(k + 2) matrices Tk and Wn as follows: 2
3
61 0 ::: 07 6 7
2
7 7 5
4
6 61 6 6 Tk = 6 60 6 6 .. 6. 6 4
Gk
0
3
7 6 1 0 ::: 07 7 6 7 7 6 7 7 6 Sn 7 7 and Wn = 6 7 7 6 . 7 7 6 .. Hn;k 7 7 6 7
Sn
k
5
where the matrices Gk and Hn;k be as before and also Sn is given by Sn =
n X1
ai;1 =
i=0
n X1 i=0
15
(
)
k+i : k
!
Then we have the following result. Theorem 12 For n,k > 0; Tkn = Wn;k :
PROOF. Since Sn+1 = an;1 + Sn and by Theorem 2, we write the matrix recurrence relation Wn;k = Wn 1;k Tk : By the induction method, we write Wn;k = W1;k Tkn 1 : From the de…nition of Wn;k ; we obtain W1;k = Tk1 and so Wn;k = Tkn : Thus we have the conclusion.
Here we should note that from Corollary 6, we know that the polynomial f1;m has the root 1 for m 0 (mod 4) : Expanding the det ( Ik+2 Tk ) with respect to the …rst row, it is easily seen that the matrix T has also the eigenvalue 1: Thus we see that the matrix T has double eigenvalue for m 0 (mod 4) : For m 6 0 (mod 4) ; we can diagonalize the matrix Tk and so we derive an explicit formula for this sum. De…ne the (k + 2)
(k + 2) matrix M as shown: 2
3
61 0 ::: 07
6 6 6 M =6 6. 6 .. 6 4
where = 1 before.
Pk+1 i=1
a1;i
1
V
7 7 7 7 7 7 7 5
and the Vandermonde matrix V is de…ned as
We can check that Tk M = M D1 where Tk is as before and D1 is a diagonal matrix such that D1 = diag (1; 1 ; 2 ; : : : ; k+1 ). Considering the matrix V; we compute det M with respect to the …rst row and then we …nd det M = det V: Then we give the following Theorem. Theorem 13 For n; k > 0 and k 6 Sn =
0 (mod 4) ;
an;1 + an;2 + : : : + an;k+1 Pk+1 1 i=1 a1;i
1
:
PROOF. Since the matrix M is invertible, we write M 1 Tk M = D1 : Thus the matrix Tk is similar to the matrix D1 : Then we write Tkn M = M D1n : By 16
Theorem 12, Wn;k M = M D1n : Equating the (2; 1)th elements of Wn;k M = M D1n and from a matrix multiplication, we obtain Sn + (an;1 + an;2 + : : : + an;k+1 ) = : Thus Sn =
(an;1 + an;2 + : : : + an;k+1 ) (an;1 + an;2 + : : : + an;k+1 1) = : Pk+1 1 i=1 a1;i
The proof is complete.
As examples of Theorem 13, we give the following cases without required computations. Case I When k = 3; after some simpli…cations, we obtain that for n o n n P 3+i
= u2 u3 (un un+2 un+4 + un+1 un+3 un+5
3
i=0
0
u3 ) =v3 :
Case II When k = 4; the polynomial f1;4 has the roots 1 = 4 ; 2 = 4 ; 3 2 2 = ; 4 = ; 5 = 1: Since f1;4 has the root 1 and so the matrix T4 has double eigenvalue 1 and linear dependent eigenvector, we can not diagonalize it. Case III When k = 5; we get o n n P i+5
i=0
6
5
=
hn
n+5 5
n
o
n+5 2
+
on
n
n+5 5
n+1 3
o
on o n 1
+
n
n
n+5 3
n+5 1
on
n+3 4
on
o
n+1 2
+
n
o
n+4 5
o
i
1 =v1 v3 v5
Generating functions
In this section, we give generating functions of the generalized Fibonomial coe¢ cients. In [17], the author considers k sequences of general higher order linear recurrence relation with arbitrary initial conditions. Then the author give the generating functions for each sequence. Now we use the special results of [17] to obtain generating functions of generalized Fibonomial coe¢ cients. De…ne k sequences of k-th order linear recurrence relation ffni g as shown, for n > 0 and 1 i k fni = c1 fni
1
+ c2 fni 17
2
+ : : : + ck fni
k
(3)
with initial conditions fni =
8 > < 1 if i = 1
n;
for 1
> : 0 otherwise,
k
n
0
k; are constant coe¢ cients, and fni is the nth term of the
where cj ; 1 j ith sequence.
Using the approach of Kalman [15], Er [6] showed that G n = An where k
k companion matrix A and the matrix Gn are as follows: 2
6 c1 6 6 61 6 6 A=6 60 6 6 .. 6 . 6 4
c 2 : : : ck
3
3 2 k 2 1 7 : : : f f f n n n 7 6 7 7 6 7 6 1 7 fn2 1 : : : fnk 1 7 7 6 fn 1 7 7 7 and Gn = 6 6 7 . . . .. .. .. 7 7 6 7 7 6 7 7 5 4 7 k 2 1 5 fn k+1 fn k+1 : : : fn k+1
ck 7
1
0 ::: 0
0
1 ::: 0 .. . . . . . ..
0 .. .
0 0 ::: 1
0
Let G (i; x) = f0i x0 + f1i x1 + f2i x2 + : : : + fni xn + : : : : For the reader convenience, we give the following result with proof. Theorem 14 For n > 0 and 1 G (i; x) =
f0i 1
Pk
i
k;
i v=m+1 cv fm v
c1 x
c2 x2
:::
xm ck xk
for m = 0; 1; 2; : : : ; k
PROOF. Let G (i; x) = f0i x0 + f1i x1 + f2i x2 + : : : + fni xn + : : : : Consider
1
c1 x
c2 x2
:::
ck xk G (i; x)
= f0i + f1i x + f2i x2 + : : : + fki xk + : : : + fni xn + : : : c1 f0i x c1 f1i x2 c1 f2i x3 : : : c1 fki 1 xk : : : ck f0i xk
ck f1i xk+1
ck f2i xk+2
After some arrangements, we write 18
:::
ck fni k xn
c1 fni 1 xn ::::
:::
(4)
1
c2 x2
c1 x
= f0i + f1i fki
ck xk G (i; x)
:::
c1 f0i x + f2i
c1 f1i
c2 f0i x2 + : : : +
1
c1 fki
2
c2 fki
3
:::
+ fki
c1 fki
1
c2 fki
2
: : : ck 1 f0i
fni
c1 fni
c2 fni
1
ck 1 f0i xk
: : : ck fni
2
k
1
ck f1i xk + : : : + xn + : : : :
Now we compute the coe¢ cients of xn of the equation above. Before this, from the de…nition of ffni g ; we get f1i = c1 f0i + c2 f i 1 + : : : + ck f1i
fki
c2 f0i
k
ck f2i k
= c1 f1i
+
::: = c1 fki 1
2
+ c2 fki
3
+ : : : + ck 1 f0i + ck f i 1
1
+ c2 fki
2
+ : : : + ck 1 f1i + ck f0i
1
+ c2 fni
2
+ : : : + ck fni k :
f2i
fki = c1 fki ::: i fn = c1 fni
+ ::: +
and so
f2i
f1i
c1 f0i =
c2 f i 1 + : : : + ck f1i
k
c1 f1i
c2 f0i =
c3 f i 1 + : : : + ck f2i
k
.. . fki
1
c1 fki
2
c2 fki
3
ck 1 f0i = ck f i 1 :
:::
For n k and from the de…nition of ffni g ; the coe¢ cients of xn are all 0: Thus the proof is complete.
Let g (i; x) = a0;i + a1;i x + a2;i x2 + a3;i x3 + : : : + an;i xi + : : : : Then we have the following Corollary. Corollary 15 For 1 g (i; x) =
a0;i 1
a1;1 x
i
k + 1; Pk+2
v=m+1
a1;2 x2
a1;v am :::
where an;i is de…ned as before. 19
v;i
xm
a1;k+1 xk+1
for m = 0; 1; 2; : : : ; k
Especially when i = 1 in Corollary 15, we get 1 X
n=0
(
)
n+k = k 1
n
k+1 k
o
n
k+1 k 1
x
o
x2 +
1
n
k+1 k 2
o
x3 + : : : + ( 1)k(k
1)=2
xk+1
:
Now we consider …rst three cases: Case I When k = 3; we obtain (
1 X 1
1 X 1
and
1 X 1
)
n+3 n x = 3 1
( )(
n 1
4 3
)
n+3 n x = 2 1
(
)(
n+1 2
1
n o
n o 4 2
x
u3 u 4 x
n o 4 3
)
n+3 n x = 1 1
4 3
4 1
v1 v3 x2
n o 4 2
x
n o
n o
x2 +
x2 +
x3 + x4
;
v1 x3
n o 4 1
x3 + x4
u2 u4 x + v 1 x 2 n o 4 2
x
x2 +
n o 4 1
x3 + x4
:
For example, the following generating function for the triple product of consecutive Fibonacci numbers can be found [18]: 1 X
Fn Fn+1 Fn+2 xn =
0
1
2x : 6x2 + 3x3 + x4
3x
Indeed the above n o generating function can also be rewritten via the Fibonomial coe¢ cients n3 as shown : F
1 X 0
(
)
n+2 3
xn =
F
1
x : 6x2 + 3x3 + x4
3x
For the generating function of the powers of Fibonacci numbers, we can refer to [27] and [12]. Case II When k = 4; we get (
)
( )(
)
1 X 0
1 X
1 X 0
1 X 0
(
0
n 1
n+4 n x = 3 1
)(
)
)(
)
n+1 2
(
n+4 n x = 4 1
n+2 3
n+4 n x = 2 1 n+4 n x = 1 1
1
n o
n o
n o
n o
n o
n o
n o
5 4
5 3
x
v2 u5 x
5 4
x
n o 5 4
x
n o 5 4
20
x
x2 +
n o 5 2
x3 +
v 2 u5 x 2
5 3
x2 +
5 2
x2 +
n o 5 3
u5 x
x2 +
5 2
x3 + x3
x3 +
x2
n o 5 2
x5
n o
x5
n o
x5
n o
x5
5 1
x4
u5 x 3 + x 4
v 2 u5 x + u 5 x 2 5 3
n o
x3 +
5 1
5 1
5 1
x4 x4 x4
; ; ; :
Case III When k = 5; we get (
)
( )(
)
)(
)
(
)(
)
(
)(
)
1 X 0
1 X
1 X 0
1 X 0
1 X 0
7
(
0
n 1
n+1 2
n+2 3 n+3 4
n+5 n x = 5 1
n o 6 5
x
1
n o 6 4
x2 +
n o 6 3 2
x3 +
n o 6 2
x4
n o 6 1
x5
n+5 n x = 3 1
n+5 n x = 2 1 n+5 n x = 1 1
v2 v3 u5 x + (u5 u6 =u2 ) x2
n o
n o
n o
n o
n o
n o
n o
n o
6 5
6 5
6 5
x x x
6 4
x2 +
6 3
x3 +
(u5 u6 =u2 ) x 6 4
6 4
x2 +
6 3
u6 x
x3 +
u6 x 3
n o 6 2 2
x4
n o 6 2
x3
x4
(u5 u6 =u4 ) x + x2
x2 +
n o 6 3
x3 +
n o 6 2
x4
x4
n o
x6
n o
x6
n o
x6
6 1
6 1
6 1
x5 x5 x5
In this section, we give combinatorial representations for the generalized Fibonomial coe¢ cients. In [2], by considering a k k companion matrix A given in (4) and its nth power, the authors derive an explicit formula for the elements in the nth power of the matrix A. Let us recall their result by the following Theorem. Theorem 16 ([2]) Let the matrix A = (aij ) be as in (4). The (i; j) entry (n) aij in the matrix Ank is given by the following formula: (n)
aij (c1 ; c2 ; : : : ; ck ) =
P
(t1 ;t2 ;:::tk )
tj +tj+1 +:::+tk t1 +t2 +:::+tk
t1 +t2 +:::+tk t1 ;t2 ;:::;tk
ct11 : : : ctkk
(5)
where the summation is over nonnegative integers satisfying t1 +2t2 +: : :+ktk = n i + j; and the coe¢ cients in (5) is de…ned to be 1 if n = i j: Thus we give the following results. Corollary 17 Let an;i denote the generalized Fibonomial coe¢ cient. Then i+1;j
=
;
n + 5 n (u5 u6 =u2 ) x v2 v3 u5 x (u5 u6 =u2 ) x3 + u6 x4 + x5 n o n o n o n o n o x = ; 6 6 6 2 + 6 x3 + 6 x4 4 1 x x x5 x6 5 4 3 2 1
Combinatorial Representations of the generalized Fibonomial Coe¢ cients
an
x6
P
(t1 ;t2 ;:::tk+1 )
tj +tj+1 +:::+tk+1 t1 +t2 +:::+tk+1
t1 +t2 +:::+tk+1 t1 ;t2 ;:::;tk+1
t
k+1 1 at1;1 : : : a1;k+1
where the summation is over nonnegative integers satisfying t1 + 2t2 + : : : + (k + 1) tk+1 = n i + j 21
; ; :
PROOF. Considering matrices Gk and A; the proof is seen from the result of Theorem 16. Corollary 18 For n (
0;
)
!
X n+k t1 + t2 + : : : + tk+1 t1 tk+1 = a1;1 : : : a1;k+1 k t1 ; t2 ; : : : ; tk+1 (t1 ;t2 ;:::tk+1 )
where the summation is over nonnegative integers satisfying t1 + 2t2 + : : : + (k + 1) tk+1 = n: n
n+k k
PROOF. In Corollary 17, if we take i = j = 1; then an;1 = have the conclusion from (1).
o
and we
For example, when k = 3; we obtain the following representations from Corollary 17: (
)
n+3 P = 3 (r1 ;r2 ;;r3 ;r4 )
r1 +r2 +r3 +r4 r1 ;r2 ;r3 ;r4
( 1)r3 +r4
n or1 +r3 n or2 4 3
4 2
where the summation is over nonnegative integers satisfying r1 + 2r2 + 3r3 + 4r4 = n 1: ( )(
n 1
)
n+3 P r2 +r3 +r4 = r1 +r2 +r3 +r4 2 (r1 ;r2 ;r3 ;r4 )
r1 +r2 +r3 +r4 r1 ;r2 ;r3 ;r4
( 1)r3 +r4
n or1 +r3 n or2 4 3
4 2
where the summation is over nonnegative integers satisfying r1 + 2r2 + 3r3 + 4r4 = n + 1: (
)(
n+1 2
)
n+3 P r3 +r4 = r1 +r2 +r3 +r4 1 (r1 ;r2 ;r3 ;r4 )
r1 +r2 +r3 +r4 r1 ;r2 ;r3 ;r4
( 1)r3 +r4
n or1 +r3 n or2 4 3
4 2
where the summation is over nonnegative integers satisfying r1 + 2r2 + 3r3 + 4r4 = n + 2:
8
Determinantal Representations for the coe¢ cients
In this section, we determine some relationships between determinants of certain matrices and the generalized Fibonomial coe¢ cients. Some similar relationships have been derived by some authors (see for more detail [19,20,24]). 22
Especially Lind [19] gave the …rst result for the relationship between determinant of certain Hessenberg matrices and the generalized Fibonomial coef…cients. For convenience, we give the result of Lind [19]: Let Dn;k denote the n o re(s+r+1)(s r+2)=2 k+1 current n n determinant jars j, where ars = ( 1) ; r; s = s r+1 n
o
n o
F
1; 2; : : : ; n: Then the author showed that Dn;k = n+k where ni is the Fik F F bonomial coe¢ cient. The analogous result holds when the Fibonacci sequence is replaced by an ordinary second-order recurring sequence. Now, by constructing super-diago¬nal matrices, we give some new general results as given by Lind above. De…nition 19 For n > k > 0 ; let Mn = [mij ] denote the k-superdiagonal matrix of order n with mii = ai;1 for 1 i n; mi;i+1 = ai;2 for 1 i n 1; : : : ; mi;i+k = ai;k+1 for 1 i n k. Clearly the matrix Mn takes the form 2
6 a1;1 a1;2 : : : a1;k+1
6 6 6 6 6 6 6 6 6 Mn = 6 6 6 6 6 6 6 6 6 4
0
1 a1;1 a1;2 : : : a1;k+1 ... ... ... ::: a1;1
a1;2
1
a1;1 1
0
3
7 7 7 7 7 7 ... 7 7 7 7 : : : : a1;k+1 7 7 7 .. 7 ::: . 7 7 7 a1;1 a1;2 7 7 5
(6)
1 a1;1
Theorem 20 Then for n > 0; jMn j = an;1 : where jM1 j = a1 : PROOF. (Induction on n) If n = 2; then we have jM2 j =
a1;1 a1;2
= a1;1 a1;1 + a1;2 = a2;1 :
1 a1;1
Suppose that the equation holds for n: Then we show that the equation is true for n + 1: Expanding jMn+1 j by the Laplace expansion of determinant according to the last column and by the de…nition of Mn ; we get jMn+1 j = a1;1 jMn j + a1;2 jMn 1 j + a1;3 jMn 2 j + : : : + a1;k+1 jMn k j : 23
By our assumption and the recurrence relation of fan;1 g, we write jMn+1 j = a1;1 an;1 + a1;2 an = an+1;1 :
1;1
+ a1;3 an
2;1
+ : : : + a1;k+1 an
k;1
Thus the theorem is proven.
For example, we take A = 1; then un = Fn (nth Fibonacci number) and by Theorem 20, we have 3
6
1 3
3
1
0
. 3 .. . 6 ..
6
1 3 0
"
1
1 3 .. .. .. . . .
#
n+3 = 3
3
: F
6
1 3
n n
Let Mn (k) denote the matrix obtained from the matrix Mn = [mij ] by taking m1;j = 0 from 1 j k: For example M3 (1; 2) has the form: 3
2
6 0 0 a1;3 a1;4 7 7 6 7 6 6 1 a1;1 a1;2 a1;3 7
M4 (2) = 6 6
6 0 6 4
0
1 a1;1
0
7: 7 a1;2 7 7 5
1 a1;1
Now we determine some relationships between the sequences fan;i g for 1 < i k and the consecutive determinants of matrix Mn (1; k) for some certain k: Theorem 21 For n > k
i
1; jMn (i)j = an
i;i+1 :
PROOF. Expanding jMn (i)j according to the …rst row by the de…nitions of Mn (i) and Mn ; jMn (i)j = a1;i+1 jMn
i 1j
+ a1;i+2 jMn 24
i 2j
+ : : : + a1;k+1 jMn
k 1j :
Expanding the value of det Mn with respect to the …rst row by considering the de…nition of Mn and the value of det Mn (i) ; we may write after some simpli…cations jMn (i)j = jMn j
a1;1 jMn 1 j
a1;2 jMn 2 j
a1;3 jMn 3 j
:::
a1;i jMn i j
which, by our assumption and Lemma 1, satis…es jMn (i)j = an;1 a1;1 an 1;1 a1;2 an 2;1 a1;3 an 3;1 : : : a1;i an = an 1;2 a1;2 an 2;1 a1;3 an 3;1 : : : a1;i an i;1 = an 2;3 a1;3 an 3;1 : : : a1;i an i;1 .. . = an i+1;i a1;i an i;1 = an i;i+1 :
i;1
Thus the proof is complete. We now de…ne an n compact form:
n upper Hessenberg matrix Dn as in the following 3
2
6 1 1 ::: 17
6 6 6 1 6 6 Dn = 6 6 0 6 6 6 6 4
Mn
1
0
where Mn is as before.
7 7 7 7 7 7 7 7 7 7 7 5
(7)
Then we give the following result without proof. Theorem 22 Then for n > 1 jDn j = Sn where Sn is as before. PROOF. (Induction on n) If n = 2; then jD2 j =
1 1
= a1;1 + 1 = a1;1 + a0;1 = S2 :
1 a1
Suppose that the equation holds for n and n 2: Then we show that the equation holds for n + 1: Expanding jDn+1 j according to the …rst column and 25
by our assumption, we obtain jDn+1 j = jMn j + jDn j which, by Theorem 20, satis…es jDn+1 j = an;1 + Sn = Sn+1 : Thus the proof is complete. To derive other similar some relationships between determinants of certain matrices and the sums of the other products, we de…ne n n matrix Tn;i for 1 i k as follows: 2
Tn;i
3
6 6 6 6 6 6 6 6 =6 6 6 6 6 6 6 4
07 .. 7 .7 7
Mn (i)
0
1 .. . 0 :::
0
11
7 7 7 7 7 7 7 ith row 7 7 7 7 5
where the matrix Mn (i) is as before. Expanding jTn;i j according to the last row, we have the following Theorems without proof. Theorem 23 For n
k
i
1,
jTn;i j = 9
Xn
i 1
i=0
ai;1 :
Conclusion
Throughout this paper we consider the recurrence fun g and its generalized Fibonomial coe¢ cients. Using given results in this paper, one can obtain many applications to the recurrence fun g or its special cases, Fibonacci or Pell sequences. We give some special cases of our results by taking k = 3; 4 and 5: For the case k = 2; the results can be found in [16]. Moreover one can obtain many analogous results for the recurrence fUn g de…ned by Un = AUn 1 BUn 2 with U0 = 0 and U1 = 1: However one should be aware of that in case of recurrence fUn g, a generator matrix can not be found by itself.
26
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