arXiv:1703.00323v1 [math.CO] 1 Mar 2017 S,B 0MOe mr 60,E arc,AGES ALGERIA. ALGIERS, Harrach, El 16309, Smar, Oued 108M BP 1 ESI, Alia El 32, ALGERIA. BP ALGIERS, Laboratory RECITS Mathematics, of Faculty USTHB, omni nvri´.Uiested oe.LbrtieLTSE LITIS Laboratoire Rouen. Universit´e. Universit´e de Normandie TEN URURYFRANCE. ROUVRAY DU ETIENNE SH,Fclyo ahmtc,RCT aoaoyB 2 lAi 1 Alia El 32, ALGERIA. BP ALGIERS, Laboratory RECITS Mathematics, of Faculty USTHB, eeaie ioac ubr n eeaie ua ubr com (resp. numbers sequences Lucas two generalized to and refer numbers Fibonacci Generalized Introduction 1 nomial ´ 3 1 2 [email protected] [email protected] t [email protected] eea om o h qaiiso hn,ZoWn n Mans giv and we Zao-Wang Zhang, consequence, of t a equalities for As the formula for context. a forms that b general also a in give involving we appearing equation and efficients one function only generating in ordinary s stated involves that nicely which show are We Mansour formulas numbers. by Fibonacci hence generalized and of Wang convolution and Zao by eralized Hac`ene Belbachir ntesl-ovlto fgeneralized of self-convolution the On Q efcso aiyo qaiispoerdb hn n gen- and Zhang by pioneered equalities of family a on focus We = x 2 − px , + ioac numbers Fibonacci q hc a w itntrossc that such roots distinct two has which ofi Djellal Toufik , 1 , ac ,2017 2, March U k and Abstract , 1 V k ihtesm hrceitcpoly- characteristic same the with ) enGbilLuque Jean-Gabriel , 2 18 60 SAINT- 76800 4108. A 11BbEzzouar, Bab 6111 11BbEzzouar, Bab 6111 l these all ivariate our. eco- he the e U elf 0 monly 3 0, = U1 = 1, V0 = 2, and V1 = p,[3, 8]. Under this assumption the sequences have the following Binet form αk βk U = − , V = αk + βk. (1) k α β k − Generalized Fibonacci numbers have several applications in numerical analy- sis for instance for solving nonlinear equations, (see e.g. [10, 11]). One of the authors also investigated closed topics such as sums of product of Fibonacci polynomials [5], q-analogs [4, 6, 7]. We focus on a family of equalities pioneered by Zhang [17] and generalized by Zao and Wang [16] and hence by Mansour [12] which involves self convolu- tion of Fibonacci numbers. The starting point of these equalities is that the discriminant of a quadratic form can be written as a differential operator. In particular, Mansour [12] gave a method to obtained all the equations but did not provide a closed form. We show that all these formulas are nicely stated in only one equation involving a bivariate ordinary generating function and we give also a formula for the coefficients appearing in that context. As a consequence, we give the general forms for the equalities of Zhang, Zao-Wang and Mansour.

In the last section, we explain how to generalize these results for higher level recurrence by the use of classical invariant theory and we detail the example of Tribonacci numbers. Throughout the paper we use extensively materials issued from the theory of generating functions; readers should refers to [9, 15] for a survey of the topic.

2 A polarization formula

Consider a quadratic polynomial Q in x with two distinct roots. Without ∂2 lost of generality we assume Q unitary (i.e. ∂x2 Q = 2). If ∆ denotes the discriminant of Q, one has ∂ 2 ∆= Q 4Q. (2) ∂x −   As a consequence, setting F = 1 , one has − Q 1 ∂ ∂ F 2 = Q F +4F . (3) ∆ ∂x ∂x    2 Our purpose is to describe the polynomials αi,n(x) such that

n 1 − ∂i F n = α (x) F, (4) i,n ∂xi i=0 X which are obtained applying successively many times (2). We consider the generating function yF (y)= F nyn = . (5) F 1 yF n 1 X≥ − Dy We want to find a differential operator x satisfying

(y)= Dy F. (6) F x

1 1 ∂Q Remark first, that if we replace x by x = x + x′ with x′ = 2 4y 1 ∂x √1 ∆ − − in Q, we obtain   Q + y Q (x′)= . (7) 1 4 y − ∆ We translate this equality in terms of differential operators considering the polarization operator PX sending each f(x) to f(x + X). This operator is

∂ P = exp X . (8) X ∂x   We set y y D P ′ x = y X X x , (9) 1 4 | → − ∆ where PX X x′ means that we first act by the operator PX and hence we | → ∂Q apply the substitution X x′. This operation is necessary because ∂x is not a constant. So we obtain→ y y yF Dy F = = Q = = (y). (10) x −Q + y −1+ y 1 yF F Q − Dy Expanding x and comparing to equality (4), we deduce our main result:

3 Theorem 2.1 We have ∂Q i ∂x αi,n = n 1 α˜i,n, (11) ∆ −
where the ordinary generating function of the α˜i,n’s is

y 1 1 A(y, z)= α˜ ziyn = exp z 1 . (12) i,n 1 4y 2 √1 4y − 0 i

2 1 2 3 A(y, z) = y +(z + 4) y + 7 z + 2 z + 16 y + 38 z +5 z2 + 1 z3 + 64 y4 6
+ 187 z + 69 z2 + 13 z3 + 1 z4 + 256 y5 + . 2 6
24 ··· From Theorem 2.1 we deduce

∆F 2 = ∂Q ∂F +4F • ∂x ∂x 2 2 3 1 ∂Q 2 ∂ F ∂Q ∂F ∆ F = 2 +7 + 16F • 2 ∂x ∂x ∂x ∂x 3 2 3 4 1 ∂Q
3 ∂ F ∂Q 2 ∂ F ∂Q ∂F ∆ F = 3 +5 2 + 38 + 64F • 6 ∂x ∂x ∂x ∂x ∂x ∂x 4 3 2 4 5 1 ∂Q
4 ∂ F 13 ∂Q
3 ∂ F 69 ∂Q 2 ∂ F ∂Q ∂F ∆ F = 4 + 3 + 2 + 187 + 256F • 24 ∂x ∂x 6 ∂x ∂x 2 ∂x ∂x ∂x ∂x ...


• Remark 2.3 In our proof of Theorem 2.1, the fact that Equality (12) is obtained by applying successively many times formula 2 is somewhat hidden. In appendix A we give an alternative proof which makes this property more visible. This proof is based on a solving of a bivariate differential equation.

3 Itering self-convolution

Consider two sequences of complex numbers C = (cn)n 0 and D = (dn)n 0. ≥ ≥ The convolution of C and D is the sequence defined by

C ⋆ D = cidj . (13) i+j=n !n N X ∈ 4 In other words, the ordinary generating function of the convolution of two sequences is the product of the ordinary generating function of the sequences,

SC⋆D(z)= SC (z)SD(z). (14)

⋆k For any sequence C =(cn)n 0, we denote by cn = i1 i n ci1 cik the ≥ + + k= k ··· ··· × ⋆k ⋆k P nth term of the sequence C = C⋆ ⋆C = cn n 1. ··· ≥ z }| {
3.1 Central binomial coefficients We consider the generating function, 1 1 B(y)= 1 , (15) 2 √1 4y −  −  appearing in the argument of exp in the right hand side of equality (12). It B(y) is well known that y is the ordinary generating function of the binomial 2n+1 coefficient n+1 . Hence,
k 1 1 B(y) 1 B(y) k k exp 2 z √1 4y 1 = exp zy y = z y − − k! y k 0   n  o n o X⋆k≥ (16) 1 2n +1 = zkyn+k. k! n +1 k,n 0   X≥ Branching this equality in (12), one obtains

1 2n +1 ⋆k A(y, z) = zkyk+1 4nyn yn k! n +1 k 0 n 0 ! n 0   ! X≥ X≥ X≥ (17) 1 2j +1 ⋆k = 4i zkyn. k! j +1 0 k

Hence, for a fixed k the sequence (˜αk,n)n can be written as a convolution. Proposition 3.1

1 2j +1 ⋆k α˜ = 4i . (18) k,n k! j +1 i+j=n k 1   X− − 5 The first values ofα ˜k,n are collected in Table 1. Example 3.2 For instance we have

⋆3 ⋆3 ⋆3 α˜ = 1 5 +4 3 + 16 1 = 1 (57+4 9+16) . 3,6 6! 3 2 1 6! · = 109 . 6 




n 1 1 From Proposition 3.1, we observe thatα ˜0,n =4 − andα ˜n 1,n = . − n! k n 1 2 3 4 5 6 7 8 9 10 \ 0 1 4 16 64 256 1024 4096 16384 65536 262144 1 0 1 7 38 187 874 3958 17548 76627 330818 2 0 0 1 10 69 406 2186 11124 54445 259006 3 0 0 0 1 13 109 748 4570 25879 138917 4 0 0 0 0 1 16 158 1240 8485 52984 5 0 0 0 0 0 1 19 216 1909 14471 6 0 0 0 0 0 0 1 22 283 2782 7 0 0 0 0 0 0 0 1 25 359 8 0 0 0 0 0 0 0 0 1 28 9 0 0 0 0 0 0 0 0 0 1

Table 1: First values of k!˜αk,n.

The following sequences are registred in OEIS encyclopedia :α ˜1,n is A000531 and count the number from area of cyclic polygon of 2n + 1 sides. 2!˜α2,n (n 3) is A038806, i!˜αi,i+3 (n 3) is A081270 and i!˜αi,i+2 (n 2) is A016777.≥ ≥ ≥

3.2 Application to Fibonacci numbers and related se- quences

We consider a sequence A =(an)n 0 whose generating function is of the form ≥ γ S (x)= a xk = − , (19) A k Q k 0 X≥

6 where γ is a complex number and Q is a unitary quadratic polynomial with two distinct roots. So we have

⋆n k n ySA(x) yγF ak x y = = , (20) 1 ySA(x) 1 yγF 1 n 0 k X≤ X≤ − − where F = 1 . Hence from equality (10), one has − Q ⋆n k n Dγy ak x y = x F. (21) 1 n 0 k X≤ X≤ Equaling the coefficient of yn in the left hand side and the right hand side, we find n 1 i γ n 1 − ∂Q ∂iS (x) (S (x))n = − α˜ A (22) A ∆ i,n ∂x ∂xi i=0   X   where ∆ is the discriminant of Q. Now extracting the coefficient of xn in (22), one finds

n 1 n 1 − ⋆n γ − i k1 i+k2 k ak = α˜i,n 2 ( p) − (k2 + i)iak2+i ∆ k1 − i=0 k1+k2=k     Xk+n 1 nX1 n 1 − − γ − i k s+i s k = α˜ 2 − ( p) − (s) a ∆ i,n k s + i − i s s=k i=s k !   −  −  X X (23) 2 where Q = x px+q and (s)i = s(s 1) (s i+1) denotes the Pochham- mer symbol. − − ··· −

Let α and β be two distincts non zero complex numbers. We consider, as αn βn in [16], the generalized Fibonacci numbers defined as Un = α−β and the n n − generalized Lucas numbers defined as Vn = α + β . Setting q = αβ and k Vk 2 Q (x)= q− k x + x . One has k − q

Uk n qk Fk(x) := Uk(n+1)x = . (24) Qk n 0 X≥

7 2 k Vk 4q Since α = β the discriminant ∆ = −2 of Q is not zero. So we can apply 6 k q k k equality (22) to Fk and obtain

n 1 Uk − n 1 i i n −qk − ∂Q ∂ Fk(x) (Fk(x)) = α˜i,n i ∆k ∂x ∂x ! i=0   Xn 1 n 1 − i Uk − n i 1 (n i 1)k k i ∂ Fk = ( 1) − − q − − α˜ (V 2xq ) . V 2 4qk − i,n k − ∂xi  k −  i=0 X (25)

Example 3.3 For instance, for n = 4 one obtains

3 3 2 4 Uk 1 k 3 ∂ Fk k k 2 ∂ Fk (Fk) = V 2 4qk 6 (Vk 2xq ) ∂x3 5q (Vk 2xq ) ∂x2 k − − − − 2k k ∂Fk 3k +38q (Vk 2xq ) 64q F , − ∂x − as expected in [16].
Consider also the bigger example for n = 10 which gives

9 9 8 10 Uk 1 k 9 ∂ Fk 1 k k 8 ∂ Fk (Fk) = 2 k (Vk 2xq ) 9 q (Vk 2xq ) 8 Vk 4q 9! ∂x 1440 ∂x − 7 − − 6 − 359 2k k 7 ∂ Fk 1391 3k k 6 ∂ Fk + q (Vk 2xq ) 7 q (Vk 2xq ) 6 5040 ∂x5 360 ∂x4 14471 4k − k 5 ∂ Fk− 6623 5k − k 4 ∂ Fk + q (Vk 2xq ) 5 q (Vk 2xq ) 4 120 3∂x 3 ∂x 2 138917 6k − k 3 ∂ Fk − 7k − k 2 ∂ Fk q (V 2xq ) 3 129503q (V 2xq ) 2 6 k − ∂x − k − ∂x +330818q8k(V 2xqk) ∂Fk 262144q9kF . k − ∂x − (k)
Set fj = Uk(j+1). From (23), we obtain

n 1 n 1 ⋆n − (k) Uk − k n s 1 s (k) f = ( q ) − − β (j + n)V f , (26) j V 2 4qk − n,s k s+j k s=0    −  X n 1 i s i where βn,s(x) = i=−s 2 − i s α˜i,n(s n + x)i. See in Appendix B for the first values of these polynomials.− Since− we have P
⋆n f (k) = U U , (27) j kj1 ··· kjn j1+ +jn=j+n   ···X we deduce the following result.

8 Corollary 3.4

n 1 n 1 − Uk − k n s 1 s Ukj1 Ukjn = 2 k ( q ) − − βn,s(j)Vk Uk(s+j n+1). ··· Vk 4q − − j1+ +jn=j   s=0 ···X − X Example 3.5 Let us treat the case where n = 6. Applying Corollary 3.4, one finds

n 1 U − U U U U U U = k β (j)V 5U kj1 kj2 kj3 kj4 kj5 kj6 V 2 4qk 6,5 k kj j1+ +j6=j  k −  ···Xk 4 2k 3 3k 2 4k q β6,4(j)Vk Uk(j 1) + q β6,3Vk Uj 2 q β6,2Vk Uj 3 + q β6,1VkUj 4 − 5k − − − − − q β6,0Uj 5 , − −
as in [12] (the values of β6,i(x) are in Appendix B). Also consider the bigger example where n = 10:

n 1 U − 1 U U = k (j 1) V 9U kj1 ··· kj10 V 2 4qk 362880 − 9 k kj j1+ +j10=j  k −   ···X1 k 8 1 2k 2 7 20160 q (j 2)8(j + 4)Vk Uk(j 1) + 5040 q (j 3)7(2j + 14j + 19)Vk Uk(j 2) − 1 3k − 2 − 6 − − 1080 q (j 4)6(j + 3)(2j + 12j + 1)Vk Uk(j 3) − 1 4k − 4 3 2 − 5 + 360 q (j 5)5(2j + 20j + 40j 5j 87)Vk Uk(j 4) 1 5k − 4 3 2 − − 4 − 180 q (j 6)4(2j + 16j 12j 176j + 75)Vk Uk(j 5) − 1 6k − 6 5− −4 3 2 − 3 + 270 q (j 7)3 (4 j + 36 j 50 j 840 j 404 j + 3054 j + 1125) Vk Uk(j 6) 1 7k − 6 − 5 − 4 − 3 − 315 q (j 8)2 (j + 1)(4 j + 24 j 206 j 984 j − 2 − 2 − − +2860 j + 7752 j 11025) Vk Uk(j 7) 2 8k 8 − 7 6 − 5 4 3 + 315 q (j 9)(j +4 j 84 j 266 j + 1974 j + 4396 j 2− − − 12916 j 15159 j + 11025) VkUk(j 8) − 4 9k −2 2 2 2 2 2 −2 2 q (j 2 )(j 4 )(j 6 )(j 8 )Uk(j 9) − 2835 − − − − −
4 Tribonacci, quadrabonacci and beyond

The technical described in the paper can be generalized for higher level recur- rences. We illustrate our purpose with a level 3 recurrence. We assume that the generating series is given by F = 1/Q where Q = x3 +3ax2 +3bx + c is a unitary cubic polynomial with three− distinct roots. We consider the binary cubic P (x, y) = x3 +3ax2y +3bxy2 + cy3. Polynomial invariants of binary

9 forms are studied since the middle of the Nineteenth century. These invari- ants are obtained by applying the Cayley Omega process (see e.g. [14]). A transvection is a bilinear operation defined by

∂ ∂ k ∂x1 ∂y1 (P, Q)k = P (x ,y )Q(x ,y ) . (28) 1 1 2 2 ∂ ∂ ∂x2 ∂y2 x1=x2=x y1=y2=y

For instance, the algebra of polynomial invariants of a binary cubic is gener- ated by 1 j = ((P,P )2, (P,P )2)2 =6abc 4b3 4a3c +3a2b2 c2. (29) 1296 − − − 3 1 From P (x, y)= y Q(xy− ), equality (29) gives ∂Q ∂Q2 ∂Q2 2 ∂Q 2 ∂Q 2 36Q + 16 (30) ∂x ∂x2 ∂x2 − ∂x ∂x         !   ∂Q2 3 2 Q 108Q2 = 108j. − ∂x2 −   ∂F ∂Q 2 Since ∂x = ∂x F , we obtain

2 2 2 1 ∂ Q ∂Q ∂Q ∂F F = 2 16 108j ∂x − ∂x ∂x ∂x   3  (31) ∂2Q  ∂Q
∂2Q
+2 2 18 2 F + 108 . ∂x − ∂x ∂x     
  Now, consider the (shifted) Tribonaci numbers defined by T0 = T1 = 1, T2 = 2 and Tn = Tn 1 + Tn 2 + Tn 3 for n> 2. For this polynomial we have 2 ∂Q − 2 − − ∂ Q 108j = 176, =3x +2x + 1, and 2 =6x + 2. So we obtain, − ∂x ∂x 1 ∂F F 2 = (4(9x4 +12x3 +16x2 +8x +3) +8(27(x3 + x2 + x)+7)F +108). 176 ∂x (32) In other words, by extracting the coefficient of xn in the left and right hand sides of (32), we observe

⋆2 1 Tn = 3(n + 1)Tn+1 + 2(4n + 7)Tn + 2(8n + 19)Tn 1 (33) 44 −  + 2(6n + 15)Tn 2 + (9n + 27)Tn 3 , − −  10 for any n> 0. In principle, the same strategy can be applied for computing linearization formula for self-convoluted higher Fibonacci-like numbers. Furthermore, the algebra of polynomial invariants is richer from binary quartic forms since it has more than one generator. So for any invariant in a Hilbert basis, one can find a family of formulas. The only limit to this process is the computational complexity. The formulas obtained not only have an increasing size but also the algebra of invariants is difficult to describe. For binary forms the last works obtained a description for the binary nonic [1] and the binary decimic [2]. An alternative way should consist to compute rational invariants [14] instead of polynomial invariants. It is well known that the rational invariants are simpler and can be computed by using the so-called associated forms [13] which is obtained from the ground form by applying certain substitutions. But now the difficulties is that we have to recover the differential operators. Anyway, the closed connexion between the classical invariant theory and the formulas involving self-convolute of generalized Fibonacci numbers need to be investigated and should bring a series of very interesting and deep results.

A An alternative proof for Theorem 2.1

Assuming (4) and applying (2), one obtains

∂ ∂ n n 1 ∂ ∂ ∂x Q ∂x F = nF − ∂x F ∂x Q n 1 2 (34) = nF − (∆F 4F ).


−
So, we have 1 ∂ ∂ F n+1 = Q F n +4nF n . (35) n∆ ∂x ∂x      Substituting F n by its expansion (formula (4)) and identifying the coefficient ∂i of ∂xi F in the left and right hand sides of the resulting formula, we find that the family of polynomials (αi,n(x))i,n defined by the recurrence

α =0 for i< 0, n< 1 or i n, i,n ≥ α0,1 =1, 1 ∂ ∂ ∂ αi,n+1 = Q αi,n + Q αi 1,n +4nαi,n in the other cases ∆n ∂x ∂x ∂x −        (36)

11 satisfies (4). Notice that (4) define a unique sequence of polynomials. A straightforward induction yields 4 n α = , (37) 0,n ∆   and 1 1 ∂ n α = Q . (38) n,n+1 n! ∆ ∂x   More generally, one has

Lemma A.1 The polynomial αi,n is a degree i polynomial which is equal, up ∂ i to a multiplicative constant, to ∂x Q . Proof We proceed by induction on
the pair (i, n i) lexicographically or- − dered. If i = 0 then equality (37) implies the result. Suppose now that i> 0. If n i = 1 then the result is a consequence of equality (38). Now suppose n i>− 1. From − 1 ∂ ∂ ∂ αi,n = Q αi,n 1 + Q αi 1,n 1 +4nαi,n 1 , (n 1)∆ ∂x ∂x − ∂x − − − −        ∂2  and ∂x2 Q = 2 we show that induction hypothesis implies the lemma.

For simplicity, we set n 1 ∆ − α˜i,n = αi,n. (39) ∂ i ∂x Q According the Lemma A.1,α ˜i,n is a rational
number satisfying the recurrence α˜ =0, for i< 0, n< 1 or i n, i,n ≥ α˜0,1 =1, (40) 2i 1 α˜i,n+1 = α˜i,n + α˜i 1,n +4˜αi,n, in the other cases . n n − Consider the double generating function

i n A(y, z)= α˜i,nz y , (41) 0 i

12 Equivalently to (40), the series A(y, z) is the unique solution of ∂ y A(t, z) y A(t, z) A(y, z)= y +2yz dt + zy dt +4yA(y, z). (42) ∂z t t Z0 Z0 It is easier to manipulate this equation under the form ∂2 ∂ A(y, z) ∂ ∂2 (1 4y) A(y, z)= z 1+2 +(z+8) A(y, z)+2z A(y, z), − ∂y2 ∂z y ∂y ∂y∂z   (43) ∂ y with the initial conditions A(0, z) = 0, ∂y A (0, z) = 1, and A(y, 0) = 1 4y . − We check that   y 1 1 A(y, z)= exp z 1 . (44) 1 4y 2 √1 4y − −   −  is the unique solution of (43). Hence,

n n n 1 ∂Q i ∂i y n − n 1 F y = ∆ n 1 i=0 α˜i,n(x) ∂x ∂xi ∆ F ≥ y ≥ ∂ = ∆A ,X ∂x X= ∂Q .F  P P∆ P | ∂x

(45) y
1 ∂Q 1 = 1 4 y F x + 2 ∂x y 1 . ∆ √1 4 ∆ − −   − 

B First values of βn,p(x)

In this section we compute the first values for the polynomial βn,p(x) involved in Corollary 3.4. n=1:

β1,0(x)=1. n=2:

β2,0(x)=2x, β (x)= x 1. 2,1 − n=3:

β (x)=2(x + 1)(x 1), 3,0 − β (x)=(2x + 1)(x 2), 3,1 − 1 β (x)= (x 1)(x 2). 3,2 2 − − 13 n=4: 4 β (x)= x(x + 2)(x 2), 4,0 3 − β (x)=2 x2 + x 1 (x 3) , 4,1 − − β (x)=(x +1)(x 2)(x 3) , 4,2 −
− 1 β (x)= (x 1)(x 2)(x 3) . 4,3 6 − − − n=5: 2 β (x)= (x + 3)(x + 1)(x 1)(x 3), 5,0 3 − − 1 β (x)= (2 x + 1) 2 x2 +2 x 9 (x 4) , 5,1 3 − − 1 β (x)= 2 x2 +4x 1 (x 3)(
x 4) , 5,2 2 − − − 1 β (x)= (2 x +3)(x 2)(
x 3)(x 4) , 5,3 6 − − − 1 β (x)= (x 1)(x 2)(x 3)(x 4) . 5,4 24 − − − − n=6: 4 β (x)= x(x + 2)(x + 4)(x 2)(x 4), 6,0 15 − − 2 β (x)= x4 +2 x3 10 x2 11 x +9 (x 5) , 6,1 3 − − − 1 β (x)= (x + 1) 2 x2 +4 x 9 (x 4)(
x 5) , 6,2 3 − − − 1
β (x)= 2 x2 +6 x +1 (x 3)(x 4)(x 5) , 6,3 6 − − − 1 β (x)= (x +2)(x 2)(
x 3)(x 4)(x 5) , 6,4 12 − − − − 1 β (x)= (x 1)(x 2)(x 3)(x 4)(x 5) . 6,5 120 − − − − −

14 n=7: 4 β (x)= (x + 1)(x + 3)(x + 5)(x 1)(x 3)(x 5), 7,0 45 − − − 2 β (x)= (2 x + 1) x4 +2 x3 21 x2 22 x + 75 (x 6) , 7,1 15 − − − 1 β (x)= x4 +4 x3 7 x2 22 x +3 (x 5)(x
6) , 7,2 3 − − − − 1 β (x)= (2 x +3)(x +4)(x 1)(x
4)(x 5)(x 6) , 7,3 9 − − − − 1 β (x)= x2 +4 x +2 (x 3)(x 4)(x 5)(x 6) , 7,4 12 − − − − 1 β (x)= (2 x +5)(x
2)(x 3)(x 4)(x 5)(x 6) , 7,5 120 − − − − − 1 β (x)= (x 1)(x 2)(x 3)(x 4)(x 5)(x 6) . 7,6 720 − − − − − −

It is easy to prove that βn,k is a degree n 1 polynomial admitting − 1 (x n + k) as a factor but the polynomials β (x n + k)− remains − k n,k − k to be properly investigated. For instance, numerical evidence suggests an interesting closed form for βn,0.

Acknowledgment: The paper was partially supported by the project PHC MAGHREB ITHEM 14MDU929M and GRR project MOUSTIC.

15 References

[1] A. E. Brouwer, M. Popoviciu, The invariants of the binary nonic, Jour. Symb. Comp., 45 6 (2010) 709-720.

[2] A. E. Brouwer, M. Popoviciu, The invariants of the binary decimic. Jour. Symb. Comp., 45 8 (2010), 837-843.

[3] A. F. Horadam, Basic properties of a certain generalized sequence of numbers, Fibonacci Quart. 3(1965), 161-176.

[4] H. Belbachir, A. Benmezai, A q-analogue for bisnomial coefficients and generalized Fibonacci sequences. Comptes Rendus Mathematique, 352 3 (2014), 167-171.

[5] H. Belbachir, F. Bencherif, Sums of products of generalized Fibonacci and Lucas numbers. Ars Combinatoria, 110 (2013), 33-43.

[6] H. Belbachir, A. Benmezai, Expansion of Fibonacci and Lucas polyno- mials: an answer to Prodinger’s question. Journal of Integer Sequences, 15 2 (2012).

[7] H. Belbachir, A. Benmezai, An alternative approach to Cigler’s q-Lucas polynomials. Applied Mathematics and Computation, 226 2 (2014), 691-698.

[8] H. Belbachir, , F. Bencherif, Linear recurrent sequences and powers of a square matrix. Integers, 6, (2006).

[9] P. Flajolet, R. Sedgewick, Analytic Combinatorics, Cambridge univer- sity press (2009).

[10] M. J. Jamieson, Fibonacci numbers and Aitken sequences revisited, Amer. Math. Monthly, 97(1990), 829-831.

[11] J. H. McCabe, G. M. Phillips, Aitken sequences and generalized Fi- bonacci numbers, Math. Comp., 45 (1985), 553-558.

[12] T. Mansour, Generalizations of some identities involving the Fibonacci numbers, Fibonacci Quart., 43 4 (2005) 307-315.

16 [13] W. F. Meyer, Allgemeine Formen und Invariantentheorie, Bd. 1, Bin¨are Formen Leipzig: G J Goschensche Verlagshandlung (1909).

[14] P. Olver, Classical invariant theory, London Mathematical Society Stu- dent text, 44, Cambridge university Press 1999.

[15] F. S. Wilf, generatingfunctionology, Academic Press, Inc (1994).

[16] F. Zao, T. Wang, Generalizations of some identities involving the Fibonacci numbers, Fibonacci Quart., 39 (2001) 165-167.

[17] W. Zhang, Some identities involving Fibonacci numbers, Fibonacci Quart., 35 (1997) 225-229.

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