On the arrowhead-Fibonacci numbers

Let k = 3, then we have the sequence

By (1), we can write a generating matrix for the arrowhead-Fibonacci numbers as follows:

The matrix G_k+1 is said to be a arrowhead-Fibonacci matrix. It is clear that

for n ≥ 0. Again by an inductive argument, we may write

where n ≥ k, a_k+1(u) is denoted by ak + 1u and Gk+1′ is a (k+1)×(k−1) matrix as follows:

It is important to note that det G_k+1=(−1)^k+1 and the Simpson identity for a recursive sequence can be obtained from the determinant of its generating matrix. From this point of view, we can easily derive the Simpson formulas of the arrowhead-Fibonacci sequences for every k ≥ 2.

Since det G₃ = −1 and

for n ≥ 2, the Simpson formula of sequence {a₃(n)} is

Let C(c₁,c₂,…;c_ν) be a ν × ν companion matrix as follows:

See [18, 19] for more information about the companion matrix.

(Chen and Louck [20]). The (i,j) entryci,j(u)(c1,c2,⋯,cv)in the matrix C^u (c₁,c₂,…,c_v) is given by the following formula:

where the summation is over nonnegative integers satisfying t₁ + 2t₂ + … + vt_v = u − i + j, t1+⋯+tvt1,⋯,tv=(t1+⋯+tv)!t1!⋯tv!is a multinomial coefficient, and the coefficients in (3) are defined to be 1 if u = i − j.

Then we can give a combinatorial representation for the arrowhead-Fibonacci numbers by the following Corollary.

Let a_k+1(n) be the nth the arrowhead-Fibonacci numberfor k ≥ 2. Then

where the summation is over nonnegative integers satisying t₁+2t₂+…+(k+1)t_k+1 = n.

In Theorem 2.3, if we choose v = k+1, u = n, i = j = k+1, c₁ = 1 and c₂ = … c_k+1 = −1, then the proof is immediately seen from (G_k+1)ⁿ.

Now we consider the Binet formulas for the arrowhead-Fibonacci numbers by using the determinantal representation.

The characteristic equation of the arrowhead-Fibonacci sequence x^k+1 − x^k + x^{k − 1} + x^k−2 + …+ 1 = 0 does not have multiple roots.

Let f(x) = x^k+1 − x^k + x^k−1 + x^k−2 + … + 1, then f(x)=xk+1−xk+xk−1x−1=xk((x−1)2+1)−1x−1. It is clear that f(0) = 1 and f(1) = k for all k ≥ 2. Let u be a multiple root of f(x), then u ∉ {0,1}. If possible, u is a multiple root of f(x) in which case f(u) = 0 and f′(u) = 0. Now f′(u) = 0 and u ≠ 0 give u1=1+ik+1k+2andu1=1−ik+1k+2 while f(u) = 0 shows u^k((u−1)² + 1) = 1 so that (u1)k=(k+2)22k+3and(u2)k=(k+2)2(k+1)2+(k+2)2 which are contradictions since k ≥ 2.

If x₁, x₂, …, x_k+1 are roots of the equation x^k+1 − x^k + x^k−1 + x^k−2 + … + 1= 0, then by Lemma 2.5, it is known that x₁, x₂, …, x_k+1 are distinct. Let V^k+1 be (k+1) × (k+1) Vandermonde matrix as follows:

Let

and supoose that Vi,jk+1 is a(k+1) × (k+1) matrix obtained from V^k+1 by replacing the j th column of V^k+1 by Uik+1.

For n ≥ k ≥ 2,

where(Gk+1)n=[gi,j(n)].

Since the eigenvalues of the matrix G_k+1, x₁, x₂, …, x_k+1 are distinct, the matrix G_k+1 is diagonalizable. Let D_k+1 = (x₁, x₂ …, x_k+1), we easily see that G_k+1V^k+1 = V^k+1D_k+1. Since det V^k+1 ≠ 0, the matrix V^k+1 is invertible. Then it is clear that (V^k+1)⁻¹G_k+1V^k+1 = D_k+1. Thus, the matrix G_k+1 is similar to D_k+1. So we get (G_k+1)ⁿV^k+1 = V^k+1 (D_k+1)ⁿ for n ≥ k ≥ 2. Then we can write the following linear system of equations:

for n ≥ k ≥ 2. So, for each i, j = 1,2,…, k + 1, we obtain gi,j(n) as follows

Then we can give the Binet formulas for the arrowhead-Fibonacci numbers by the following corollary.

Let a_k+1(n) be the nth the arrowhead-Fibonacci numberfor k ≥ 2. Then

Now we consider the permanent representations of the arrowhead-Fibonacci numbers.

A u × vreal matrixM = [m_i,j] is called a contractible matrix in thek^thcolumn (resp. row) if thek^thcolumn (resp. row) contains exactly two non-zero entries.

Suppose that x₁x₂, …, x_u are row vectors of the matrix M. If M is contractible in the k^th column such that m_i,k ≠ 0,m_j,k ≠ 0 and i ≠ j, then the (u − 1)×(v − 1) matrix M_ij:k obtained from M by replacing the i^th row with m_i,kx_j + m_j,k x_i and deleting the j^th row. The k^th column is called the contraction in the k^th column relative to the i^th row and the j^th row.

In [21], Brualdi and Gibson obtained that per (M) = per(N) if M is a real matrix of order α >1 and N is a contraction of M.

Let M_k+1(n) = [mi,j(n)] be the n × n super-diagonal matrix, defined by

that is,

Then we have the following theorem.

For and n ≥ 1 and k ≥ 2,

where perM_k+1(1) = 1.

First we start with considering the case n < 4. The matrices M_k+1(2) and M_k+1(3) are reduced to the following forms:

and

It is easy to see that perM_k+1(2) = 0 and perM_k+1(3) = −2. From definition of arrowhead-Fibonacci sequence it is clear that a_k+1 (k + 2) = 1, a_k+1 (k + 3) = 1 and a_k+1 (k + 4) = −2. So we have the conclusion for n < 4. Let the equation hold for n ≥ 4, then we show that the equation holds for n + 1. If we expand the perM_k+1(n) by the Laplace expansion of permanent with respect to the first row, then we obtain

Since perM_k+1(n) = a_k+1(n+k+1), perM_k+1(n−1) = a_k+1(n+k), …, perM_k+1(n − k) = a_k+1(n+1) , we easily obtain that perM_k+1(n+1) = a_k+1(n+k+2).So the proof is complete.

Let n > k + 1 such that k ≥ 2 and let Rk+1(n)=[ri,j(n)] be the n × n matrix, defined by

that is,

Assume that the n × n matrix Tk+1(n)=[ti,j(n)] is defined by

where n > k + 2 such that k ≥ 2.

Then we can give more general results by using other permanent representations than the above.

Leta_k+1(n) be the nth the arrowhead-Fibonacci number fork ≥ 2. Then

1. Forn > k + 1,

   perRk+1(n)=ak+1(n+1).

2. Forn > k + 2,

   perTk+1(n)=∑i=1nak+1n(i).

1. Let the equation hold for n > k + 1, then we show that the equation holds for n + 1. If we expand the perR_k+1(n) by the Laplace expansion of permanent with respect to the first row, then we obtain

   perRk+1(n+1)=perRk+1(n)−perRk+1(n−1)−⋯−perRk+1(n−k).

   Also, since

   perRk+1(n)=ak+1(n+1),perRk+1(n−1)=ak+1(n),…,perRk+1(n−k)=ak+1(n−k+1),

   it is clear that

   perRk+1(n+1)=ak+1(n+2).

2. It is clear that expanding the perT_k+1(n) by the Laplace expansion of permanent with respect to the first row, gives us

   perTk+1(n)=perTk+1(n−1)+perRk+1(n).

   Then, by the result of Theorem 2.10. (i) and an induction on n, the conclusion is easily seen.

   It is easy to show that the generating function of the arrowhead-Fibonacci sequence {a_k+1(n)}is as follows:

   g(x)=xk1−x+x2+⋯+xk+1,

   where k ≥ 2.

   Then we can give an exponential representation for the arrowhead-Fibonacci numbers by the aid of the generating function with the following theorem.

The arrowhead-Fibonacci numbers have the following exponential representation:

wherek ≥ 2.

Since

and

it is clear that

Thus we have the conclusion.

Now we consider the sums of arrowhead-Fibonacci numbers.

Let

for n ≥ 1 and k ≥ 2, and suppose that A_k+2 is the (k + 2)×(k + 2) matrix such that

Then it can be shown by induction that

The sequenc{ak+1m(n)} is simply periodic for k ≥2. That is, the sequence is periodic and repeats by returning to its starting values.

Since{a42(n)}={0,0,0,1,1,0,0,0,1,1,...},P4(2)=5.

Given an integer matrix X=[x_i,j], X(mod m) means that all entries of X are modulo m, that is, X(mod m)= (x_i,j(mod m)) . Let us consider the set 〈X〉m={Xi(modm)|i>_0}. If gcd (m,det X)=1, then 〈X〉_m is a cyclic group. Let |〈X〉m| denote the cardinality of the set 〈X〉_m. Since det G_k+1=(−1)^k+1, 〈G_k+1〉_m is a cyclic group for every positive integer m. By (2), it is easy to show that Pk+1(m)=|〈Gk+1〉m|

Now we give some useful properties for the period P_k+1(m) by the following theorem.

1. Letpbe a prime and suppose thatuis the smallest positive integer withP_k+1(p^u+1) ≠ P_k+1(p^u). ThenP_k+1(p^v) = p^v−u· P_k+1(p) for everyv>u andk≥2.

2. Ifmhas the prime factorization

   m=∏i=1u(pi)αi,(u≥1),thenPk+1(m)

   equals the least common multiple of theP_k+1((p_i)^αi)’ s fork≥2.

3. IfkisAeven integer such thatk≥2, thenP_k+1(m) is even for every positive integerm.

1. If I is the (k+1)×(k+1) identity matrix and t is a positive integer such that

   (Gk+1)Pk+1(pt+1)≡I(modpt+1),thenGk+1)Pk+1(pt+1)≡I(modpt).

   Then, it is clear that P_k+1(p^t) divides P_k+1(p^t+1). On the other hand, if we denote (Gk+1)Pk+1(pt)=I+(ai;j(t)⋅pt), then by the binomial expansion, we may write

   (Gk+1)Pk+1(pt)⋅p=(I+(ai,j(t)⋅pt))p=∑i=0ppi(ai,j(t)⋅pt)i≡I(modpt+1).

   This yields that P_k+1(p^t)· p is divisible by P_k+1(p^t). Then, P_k+1(p^t+1) = P_k+1(p^t) or P_k+1(p^t+1) = P_k+1(p^t)· p, and the latter holds if and only if there is a ai,j(t) which is not divisible by p. Due to fact that we assume u is the smallest positive integer such that P_k+1(p^u+1)≠ P_k+1(p^u), there is an ai,j(u) which is not divisible by p. Since there is a ai,j(u) such that p does not divide ai,j(u) it is easy to see that there is an ai,j(u+1) which is not divisible by p. This shows that P_k+1(p^u+2)≠ P_k+1(p^u+1). Then we get that Pk+1(pu+2)=p⋅Pk+1(pu+1)=p⋅(p⋅Pk+1(pu))=p2⋅Pk+1(pu). So by induction on u we obtain P_k+1(p_v) = p^v−u ·P_k+1(p) for every v>u. In particular, if Pk+1(p2)≠Pk+1(p),thenPk+1(pv)=pv−1⋅Pk+1(p).

2. It is clear that the sequence {ak+1(pi)αi(n)} repeats only after blocks of length λ⋅Pk+1((pi)αi) where λ is a natural number. Since P_k+1(m) is period of the sequence {ak+1m(n)}, the sequence {ak+1(pi)αi(n)} repeats after P_k+1(m) terms for all values i. Thus, we easily see that P_k+1(m) is of the form λ⋅Pk+1((pi)αi) for all values of i, and since any such number gives a period of P_k+1(m). Therefore, we conclude that

   Pk+1(m)=lcm[Pk+1((p1)α1),…,Pk+1((pu)αu)].

3. Since det G_k+1 = −1 when k is a even integer and Pk+1(m)=|Gk+1m|, we have the conclusion.