On sums related to the numerator of generating functions for the k-th power of Fibonacci numbers Pavel Praµzák, Pavel Trojovský University Hradec Králové Rokitanského 62 500 03 Hradec Králové, Czech Republic e-mail:
[email protected] Abstract New results about some sums sn (k; l) of products of the Lucas numbers, which are of similar type as the sums in [9], and sums (k) = Pk2 1 k l=0 l Fk 2l sn (k; l) are derived. These sums are related to the numerator of generating function for the k-th powers of the Fibonacci numbers. sn (k; l) and (k) are expressed as the sum of the binomial and the Fibonomial coe¢ cients. Proofs of these formulas are based on a special inverse formulas. AMS Subj. Classi…cation: 11B39, 05A15, 05A19. Key Words and Phrases: Generating function, Riordan’s theorem, Generalized Fibonacci numbers, Fibonacci numbers, Lucas numbers.
1
1
Introduction
It is well–known that manipulations with generating function of sequences of integers are very helpful in …nding some relations for terms of these sequences. In this paper we will use the Fibonacci numbers, which are de…ned by the recurrence Fn+2 = Fn + Fn+1 with F0 = 0; F1 = 1;and the Lucas numbers, which are de…ned by the same recurrence as the Fibonacci numbers but they have the initial conditions L0 = 2; L1 = 1: In our previous papers [8, 9] we used manipulations with the denominator of generating functions for the k-th power of the Fibonacci numbers to …nd identities for some sums of products of the Lucas numbers, which are a certain generalization of the well–known formula for the Fibonacci and Lucas numbers (see for example [14], p. 179) n X
( 1)i Ln
2i
= 2Fn+1 :
i=0
In this paper we will close this line of papers by …nding relations for sums which are created from the numerator of generating functions for the k-th power of Fibonacci numbers. Generating function for the Fibonacci sequence fFn g was already known to DeMoivre in 1718 and in 1957 S. W. Golomb [2] found generating function …nd a recurrence or a of fFn2 g. Golomb’s e¤ective proof opened the P1e¤ortk to n closed form of generating function fk (x) = n=0 Fn x for the k-th powers of the Fibonacci numbers. Riordan [6] found the general recurrence for fk (x). Carlitz [1], Horadam [3] and Mansour [5] presented some generalizations of Riordan’s result and found similar recurrences for the generating functions for powers of any second–order recurrence sequences. Horadam found some closed forms for the numerator and the denominator of these generating functions, from his general results in [3] it is possible to get specially the following relation for the generating function of the integer powers of the Fibonacci numbers Pk Pi j(j+1) k+1 2 Fik j xi i=0 j=0 ( 1) j ; (1) fk (x) = Pk+1 i(i+1) k+1 i 2 ( 1) x i=0 j where nk are the so-called Fibonomial coe¢ cients de…ned for any nonnegative integers n and k by n Fn Fn 1 Fn k+1 = ; k F1 F2 Fk
where n0 = 1 and nk = 0 for n < k. In [10] Shannon found by Carlitz’ method an expression of the generating function fk (x), which can be written
for any odd integer k as k 1
k 1 2
fk (x) = 5
2 X k j 1 j=0
Fk 2j x ( 1)j Lk 2j x
(2)
x2
and for any even integer k as 0k 2 1 k 2 j X 2 k k 2 ( 1) Lk 2j x k ( 1) A : fk (x) = 5 2 @ ( 1)j + k k j 2 j 1 ( 1) L x + x 2x k 2j 1 ( 1) 2 j=0
(3) In 2003 Stanica [11] gave similar results to Horadam but he obtained some new cases and used them to …nd some combinatorial sums of the type Pn weighted n r i=0 i Fi . Strazdins [12] considered the polynomials from the denominator of the fraction (1) and he found some factorization of them with the help of the Lucas factor of the type 1 Lk x + ( 1)k x2 . Throughout the paper we adopt the conventions that the sum and the product over an empty set is 0 and 1, respectively, and that bxc represents the greatest integer less than or equal to x. In [9] we studied the sequence fSn (k)g1 n=0 where k is a positive integer. The terms of this sequence are de…ned by bk2 1c
S0 (k) = 1 ;
S1 (k) =
X
( 1)i1 Lk
2 i1
i1 =0
and for any positive integer n > 1 bk2 1c
Sn (k) =
X
in =0 in
If we denote
(i; k; n) =
bk2 1c
bk2 1c
X
X
( 1)
i1 =i2 +1
1 =in +1
c b k+1 2
i1 +i2 + +in
n+i i
n Y
Lk
2 ij
:
(4)
j=1
+
b k+1 c 2
n+i i 1
1
(5)
for any positive integers i, k and any nonnegative integer n; we found that (4) can be written as n
Sn (k) =
b2c X i=0
for an odd integer k and
n
( 1)b 2 c
i
(i; k; n)
k+1 n 2i
n
b 2 c n 2i X X j k Sn (k) = ( 1)i+n( 2 +1)+ 2 (j+k+1) (i; k; n) i=0 j=0
for an even integer k:
k+1 j
2
The main results
Let us de…ne the sequence fsn (k; l)g1 n=0 , where k is any positive integer and k 1 l b 2 c is any nonnegative integer, by the following way: bk2 1c
s0 (k; l) = 1 ;
X
s1 (k; l) =
( 1)i1 Lk
2i1
i1 =0 i1 6=l
and for n > 1 bk2 1c
sn (k; l) =
X
bk2 1c
bk2 1c
X
X
in =0 in 1 =in +1 in 6=l in 1 6=l
by
Let the sequence f
1 n (k)gn=0 ,
i1 +i2 + +in
( 1)
i1 =i2 +1 i1 6=l
n Y
Lk
(6)
2 ij
j=1
where k is any odd positive integer, be given
k 1
2 X k Fk n (k) = l l=0
2l
(7)
sn (k; l) :
b k 2 1 c be any nonnegative
Theorem 1. Let k be any positive integer, n and l integers. Then n
b 2 c n 2i X X j ( 1)n(l+1)+ 2 (2l+j+1) (i; k sn (k; l) =
k + 1 F(n 2i j+1)(k Fk 2l j
2; n)
i=0 j=0
2l)
(8)
for an odd positive integer k and n
sn (k; l) =
b 2 c n 2m j X X X
m=0 j=0
k
( 1) 2 (n+j)+n+m+l(j
i)+ 2i (i+1)
i=0
k 2 2
n+m + m
k 4 2
n+m m 1 k + 1 F(j i+1)(k (m; k 2; n) i Fk 2l
2l)
(9)
for an even positive integer k. Theorem 2. Let k be any odd positive integer and n be any positive integer. Then n
n (k)
=5
k 1 2
b 2 c n 2i X X j ( 1)n+ 2 (j+1) (i; k i=0 j=0
2; n)
k+1 Fnk j
2i+1 j
:
(10)
k 2 2
Theorem 3. Let k be any even positive integer, n and l negative integers. Then 0k 1 i 2 X X k @ ( 1)l(i+j+1) L(i j)(k 2l) l j=0 l=0 k
+
k 2
k
k
5 2 Fik j
( 1) 2 (j+i+1)
be any non-
k+1 =0: j
Corollary 4 (Corollary of Theorems 1 and 2). Let k be any positive integer, n > b k 2 1 c and l b k 2 1 c be any nonnegative integers. If k is odd positive integer then n
b 2 c n 2i X X j ( 1) 2 (2l+j+1) (i; k
2; n)
i=0 j=0
k+1 F(n j
2i j+1)(k 2l)
= 0;
n
b 2 c n 2i X X j(j+1) ( 1) 2
2; n)
k+1 j
( 1)( 2 +l)j+m+ 2 (2l+i+1) (i; k
2; n)
(i; k
i=0 j=0
Fnk
2i+1 j
=0
and if k is even positive integer then n
b 2 c n 2m j X X X
m=0 j=0
k
i
i=0
k + 1 F(j i+1)(k i Fk 2l
2l)
= 0:
Corollary 5 (Corollary of Theorems 1 and 2). Let k be any odd positive integer. Then (i) Setting n = 1; 2 in (8) we have k 1
2 X
( 1)i1 Lk
2i1
= Fk+1 + ( 1)l+1 Lk
2l
;
i1 =0 i1 6=l k 1
k 1
2 2 X X
( 1)i1 +i2 Lk
i2 =0 i1 =i2 +1 i2 6=l i1 6=l
=
2i1
Lk
k
1 2
2i2
Fk Fk+1 + ( 1)l+1 Fk+1 Lk
(ii) Setting n = 1; 2 in (9) we have
2l
+
F3(k Fk
2l) 2l
:
k 2
2 X
( 1)i1 Lk
2i1
= Fk+1 + ( 1)l+1 Lk
k
2l
+ ( 1) 2 +1 ;
i1 =0 i1 6=l k 2
k 2
2 2 X X
( 1)i1 +i2 Lk
i2 =0 i1 =i2 +1 i2 6=l i1 6=l
=
4
k
2i1
Lk
2i2
k
+ ( 1) 2 +l Lk
2
k
Fk+1 (( 1) 2 + ( 1)l Lk
2l
2l
+ Fk ) +
F3(k Fk
2l)
:
2l
(iii) Setting n = 1; 2; 3 in (10) we have k 1
2 X k Fk l l=0
k 1
k 1
k 1
2l
=5
k 1 2
k 1
2 2 X X
;
l=0 i1 =0 i1 6=l
2l
Lk
=5
2i1
k 1 2
(Fk+1
1) ;
k 1
2 2 2 X X X
k Fk l
( 1)i1 +i2
l=0 i2 =0 i1 =i2 +1 i2 6=l i1 6=l
2l
Lk
2i1
=5
3
k Fk l
( 1)i1
Lk
k 1 2
2i2
2k +
k
1
Fk+1
2
Fk Fk+1
The preliminary results
As in [8] we denote the denominator of identity (1) by D(x) and its coe¢ cients i by di . The numerator in (1) we denote N (x). Using q–analog ( 1) 2 (i+1) k+1 i of the terminating binomial theorem (see for example [4]) k Y (1
i
q x) =
i=0
for i k+1 0
i
( 1)i q 2 (i
1)
i=0
where q–binomial coe¢ cients k+1 i
k+1 X
=
k+1 i
k+1 i
xi ;
(11)
are de…ned as
(q k+1 1)(q k (q 1)(q 2
1) 1)
(q k (q i
i+2
1)
1)
1 and any complex numbers q, x and any positive integer k, where = 1.
k
Replacing q by = and x by D(x) =
k+1 X
( 1)
x in (11) we have k Y = (1
k+1 i
i(i+1) 2
i=0
k i i
x) :
i=0
Thus for any even integer k we gain k Y D(x) = (1
j
k j
x) = (1
k Y
k 2
(
) x)
j=0
k j
(1
j
x)
j=0;j6= k2 k
k 2
( 1) x)
= (1
1 2 Y
( 1)j (
k 2j
( 1)j Lk
2j x
1
k 2j
+
)x + (
)k
2j 2
(12)
x
j=0
k 2
k 2
( 1) x)
= (1
Y1
(1
+ x2 ) ;
j=0
and for any odd integer k k 1
k Y D(x) = (1
k j
j
x) =
j=0
=
( 1)j
k 2j
)x + (
)k
1
x
1
( 1)j
k 2j
x
j=0
k 1
2 Y
2 Y
( 1)j (
k 2j
( 1)j Lk
2j x
1
k 2j
+
2j 2
(13)
x
j=0 k 1
=
2 Y
(1
x2 ) :
j=0
Further with respect to (2) and (3) we have for any odd integer k 5 fk (x) =
1 k 2
Pk2 1
j=0
and for any even integer k
k j
Fk
2j x
Qk2 1
j=0 (1
fk (x) = where
Qk2 1 i=0 i6=j
(1
( 1)j Lk N (x) ; D(x)
( 1)i Lk
2i x
x2 ) (14)
2j x
x2 )
N (x) = 5
k 2
k 2 2
X
0
k 2
@( 1)
k (1 l
( 1)l
l=0
k 2
2 k Y
k 2
( 1)j Lk
(1
2j x
+ x2 ) +
j=0
k 2 2
k
( 1)l Lk
( 1) 2 x)(2
Y
2l x)
( 1)j Lk
(1
2j x
j=0 j6=l
1
C + x2 )C A :
(15)
b k 2 1 c be any
Lemma 6. Let k be any positive integer and let n and l nonnegative integers. Then sn (k; l) = 0 for each b k 2 1 c < n. Proof. Rewriting relation (6) in the form X
sn (k; l) =
i1 +i2 + +in
( 1)
n Y
Lk
2 ij
j=1
bk2 1c
0 in 0:
r=0
Further using Lemma 11 and the formula Lp Fpq = Fp(q+1) + ( 1)p Fp(q cf. [14], p. 177, we gain pi (k; l) = = =
i 1 X
r=0 i 1 X
r=0 i X r=0
xr+1 di
r
+ ( 1)l Lk
xr+1 di
r +
xr+1 di
=
r
2l
( 1)li Lk Fk 2l i X
xi
( 1)l(i+1) 2l Fi(k 2l)
r+1 dr =
r=0
i X
Fi(k Fk
2l)
( 1)li
2l
F(i
( 1)l (i
1) ,
F(i 1)(k 2l) Fk 2l
1)(k 2l)
r)
F(i
r=0
r+1)(k 2l)
Fk
dr
2l
(32)
hence (28) P follows. Let us deal with relation (29). We de…ne polynomials Qk 2 (x) = ki=02 qi (k; l) xi by k 2
Qk 2 (x) =
2 Y
(1
( 1)j Lk
2j x
+ x2 ) :
(33)
j=0 j6=l
k
With respect to (31) we obtain (1 ( 1) 2 x) Qk 2 (x) = Pk 1 (x) and after multipling of the left–hand side we get
q0 (k; l) = p0 (k; l) = 1 ; qi (k; l) + ( 1)
k +1 2
( 1)
qi 1 (k; l) = pi (k; l) ; i = 1; 2; : : : ; k ;
k +1 2
qk (k; l) = pk+1 (k; l) :
Putting qi (k; l) = 0 for i < 0 or i > k
2 we obtain the general recurrence
k
qi (k; l) + ( 1) 2 +1 qi 1 (k; l) = pi (k; l) ; for any integer i, between coe¢ cients of polynomials Pk Lemma 14 we obtain qi (k; l) =
i X
1
and Qk 2 . Using
k
( 1) 2 (i+j) pj (k; l) :
j=0
and using (32) we obtain (29). Finally we will consider relation (28). The polynomial k 2
(2
( 1)l Lk
k 2
( 1) x)
2l x)(1
2 Y
( 1)j Lk
(1
2j x
+ x2 )
j=0 j6=l
P we denote by Rk (x) = ki=0 ri (k; l) xi . Hence with respect to (29) the identity Rk (x) = (2 ( 1)l Lk 2l x) Pk 1 (x) is valid and after multipling of the right– hand side we get ri = 2pi + ( 1)l+1 Lk 2l pi 1 , for i = 0; 1; : : : ; k. Therefore using (32) we have ri = 2
i X
( 1)l(i
j) F(i j+1)(k 2l)
Fk
j=0
l+1
+ ( 1)
Lk
2l
i 1 X
( 1)l(i
j 1) F(i j)(k 2l)
Fk
j=0
= 2 di +
i 1 X
( 1)l(i+j)
dj
2l
2F(i
Lk
j+1)(k 2l)
Fk
j=0
dj =
2l 2l
F(i
j)(k 2l)
dj :
2l
With respect to well–known identity (see [14], p. 177) Fm Fn +Lm Fn = 2 Fm+n we get i 1 X ri = 2 di + ( 1)l(i+j) L(k 2l)(i j) dj and …nally ri =
Pi
j=0 (
j=0
1)l(i+j) L(k
2l)(i j)
dj , which leads to (30).
4
Proofs of main theorems
Proof of Theorem 1: First we prove relation (8). Analogously as in the proof can de…ne, with respect to identity (13), polynomials Dl (x) = Pk 1 of (28) we i p (k; l) x , where l k 2 1 is any nonnegative integer, by i=0 i k 1
Dl (x) =
1
D(x) ( 1)l Lk 2l x
x2
=
2 Y
( 1)j Lk
(1
2j x
x2 ) :
(34)
j=0 j6=l
P Expanding (1 ( 1)l Lk 2l x x2 ) ki=01 pi (k; l) xi and comparing with D(x) we get some relations, which could be rewritten to the recurrence pn (k; l)
( 1)l Lk
2l
pn 1 (k; l)
pn 2 (k; l) = dn
for any nonnegative integer n, when it is set pn (k; l) = 0 for n < 0 or n > k Hence we obtain as in (32) that for any integer n pn (k; l) =
n X
( 1)l (n
F(n
j)
j+1)(k 2l)
Fk
j=0
1.
(35)
dj :
2l
By the direct multiplication of the product in identity (34) we get for n = 0; 1; 2; : : : ; k 2 1 , respectively for n = 0; 1; 2; : : : ; k 2 3 , n X n+i p2n (k; l) = n i=0
k+1 2
n X n+i n i=0
p2n+1 (k; l) =
s2i (k; l) ;
i k 1 2
i
s2i+1 (k; l) :
When indexes of the sums are shifted similarly as in the proof of Lemma 8 and with respect to Lemma 7, then these two identities can be joined into the following formula n
pn (k; l) =
b2c X
( 1)n
n
i=0
k+1 2
i i
sn
(36)
2i (k; l)
for any nonnegative integer n. Now we have to invert (36) to obtain explicit formula for the sums sn (k; l). To get the assertion we set an = pn (k; l), bn = sn (k; l) and p = k+1 in Lemma 8. Thus we have 2 n
sn (k; l) =
b2c X i=0
( 1)n
k 1 2
n+i i
+
k 3 2
n+i i 1
pn
2i (k; l)
:
From this relation and (35) we get n
b 2 c n 2i X X sn (k; l) = ( 1)n+l (n
j)
(i; k
2; n)
F(n
2i j+1)(k 2l)
Fk
i=0 j=0
dj :
2l
j
Finally putting dj = ( 1) 2 (j+1) k+1 and after simpli…cations we have the j assertion. Now we prove relation (9). By the direct multiplication of the factors in (33) we get the identities n X
q2n+1 (k; l) =
k 2 2
i=0
(2i + 1) s2i+1 (k; l) n i
for n = 0; 1; 2; : : : ; k 2 4 , and q2n (k; l) =
n X
k 2 2
2i s2i (k; l) i
n
i=0
for n = 0; 1; 2; : : : ; k 2 2 . By shifting indexes of summation it is possible to join the previous two identities into the relation n
qn (k; l) =
b2c X
( 1)n
k 2 2
n + 2m sn m
m=0
2m (k; l)
for n = 0; 1; 2; : : : ; k 2. To complete the proof of (9) we have to invert the previous identity. Setting an = qn (k; l), bn = sn (k; l) and t = k 2 2 in the inverse formula in Lemma 9 we obtain n
sn (k; l) =
b2c X
k 2 2
( 1)n+m
m=0
n+m + m
k 4 2
n+m m 1
qn
2m (k; l)
:
From the last identity and identities (32) and (33) we get n
sn (k; l) =
b 2 c n 2m j X X X
( 1)n+m (i; k
2; n)
m=0 j=0 r=0
k
( 1) 2 (n r
Putting dr = ( 1) 2 (r+1) assertion.
k+1 r
2m+j)
( 1)l(j
r) F(j r+1)(k 2l)
Fk
dr :
2l
and after simpli…cations we …nally gain the
Proof of Theorem 2: The numerator of (14) is a polynomial N (x) of odd degree k. If we use the notation from the beginning of the proof of Theorem 1 particularly identity (34) then the following relation holds for it k 1
k 1
2 2 X X 1 k k k 2 Fk 2j x Dj (x) = 5 Fk N (x) = 5 j j j=0 j=0 0 1 k 1 k 2 X k X @5 1 2 k Fk 2j pi 1 (k; j)A xi : = j j=0 i=1 1 k 2
2j
k 1 X
x
pi (k; j) xi =
i=0
Therefore, with respect to (1), for i = 1; 2; : : : ; k we get k 1
5
1 k 2
2 X k Fk j j=0
2j
pi 1 (k; j) =
i 1 X
dj Fik j :
(37)
j=0
Using the following identity for the Fibonomial coe¢ cients, see for example [15], k+1 X j k+1 ( 1) 2 (j+1) Fnk j = 0 ; j j=0
where n, k are any positive integers such that n k + 1, it is possible to generalize (37) for any positive integer i. Hence from (37), using (36), we have b i 21 c
k 1
5
1 k 2
2 X k Fk j j=0
2j
X
i 1
( 1)
i
k+3 2
m m
m=0
si
1 2m (k; j)
=
i 1 X
dj Fik j
j=0
and, with respect to (7), b i 21 c
X
i 1
( 1)
i
m=0
m m
k+3 2
i 1 2m (k)
=5
k 1 2
i 1 X
dj Fik j :
j=0
Setting n = i 1 and replacing m by i the latter identity can be rewritten for any nonnegative integer n as n
hn (k) =
b2c X
n
( 1)n
i
i=0
where we denoted hn (k) = 5
k 1 2
k+1 2
i
Pn
j=0
n 2i (k)
;
(38)
k dj Fn+1 j . To obtain the assertion we
invert (38) using Lemma 8 (similarly as in Theorem 1): n
n (k)
=
b2c X
k 1 2
n
( 1)
n+i i
i=0
=5
k 1 2
+
k 3 2
n
n+i i 1
hn
2i (k)
n
b 2 c n 2i X X ( 1)n (i; k
k 2; n) dj Fn+1
2i j
;
i=0 j=0 j
where we put dj = ( 1) 2 (j+1)
k+1 j
j.
Proof of Theorem 3: As polynomial N (x) in (15) is numerator in (1) we have k
1 2 X
k i k XX ( 1) ( 1)l(i+j) L(i l i=0 j=0 l=0 l
k i k XX
k
+ ( 1) 2
k 2
j)(k 2l)
k
k
( 1) 2 (j+i) dj xi = 5 2
i=0 j=0
k k 2
k X i X
Fik j dj xi
i=0 j=0
for i = 0; 1; : : : ; k. Using Lemma 15 we get 0k 1 i 2 X X k @ ( 1)l(i+j+1) L(i l j=0 l=0 +
dj xi
k
( 1) 2 (j+i+1)
j)(k 2l)
k
5 2 Fik j dj = 0
for i = 0; 1; : : : ; k. Proof of Corollary 4: Both identities can be obtained directly from the identities in Theorem 1 and in Theorem 2 with respect to sn (k; l) = 0 for each even k 2n 1 (see Lemma 6).
5
Concluding remark
It is interesting to compare the e¤ectiveness of formulas (8), (9) and (10) in contrast to de…ning formulas (6) and (7) for computation of sn (k; l) and n (k). Therefore, we found the CPU time (in seconds) required for computation of sums s3 (k; l) for some values of k using the system Mathematica on a standard PC. The measured time is given in Table 1.
Table 1. CPU time in seconds for s3 (k; k 4 1 ) k (6) (8)
101 0.55 0.02
201 4.45 0.02
301 15.28 0.05
401 36.83 0.10
501 73.42 0.16
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