We next define the arrowhead matrix *N* = [*n*_{i,j}]_{(k+1)×(k+1)} by using the characteristic polynomial of the *k*-step Fibonacci sequence
${P}_{k}^{F}(x)$
as follows:
$$\left[\begin{array}{cccccc}1& -1& -1& \cdots & -1& -1\\ -1& -1& 0& \cdots & 0& 0\\ -1& 0& -1& 0& \cdots & 0\\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ -1& 0& \cdots & 0& -1& 0\\ -1& 0& \cdots & 0& 0& -1\end{array}\right].$$

Now we consider a new (*k*+1)-step sequence which is defined by using the matrix *N* and is called the arrowhead-Fibonacci sequence. The sequence is defined by integer constants *a*_{k+1}(1) = … =*a*_{k+1}(k) =0 and *a*_{k+1}(*k*+1)=1 and the recurrence relation
$${a}_{k+1}(n+k+1)={a}_{k+1}(n+k)-{a}_{k+1}(n+k-1)-\cdots -{a}_{k+1}(n)$$(1)

for *n* ≥ 1, where *k* ≥ 2.

*Let k* = 3, *then we have the sequence*
$$\{{a}_{4}(n)\}=\{0,0,0,1,1,0,-2,-4,-3,3,12,16,4,-27,-59,...\}.$$

By (1), we can write a generating matrix for the arrowhead-Fibonacci numbers as follows:
$${G}_{k+1}\text{\hspace{0.17em}=\hspace{0.17em}}{\left[\begin{array}{cccccc}1& -1& -1& \cdots & -1& -1\\ 1& 0& 0& \cdots & 0& 0\\ 0& 1& 0& \cdots & 0& 0\\ \vdots & & \ddots & & & \vdots \\ 0& 0& \cdots & 1& 0& 0\\ 0& 0& \cdots & 0& 1& 0\end{array}\right]}_{\left(k+1\right)\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}\left(k+1\right)}$$

The matrix *G*_{k+1} is said to be a arrowhead-Fibonacci matrix. It is clear that
$${\left({G}_{k\text{\hspace{0.17em}+ 1}}\right)}^{n}\text{\hspace{0.17em}}\left[\begin{array}{c}{a}_{k+1}\text{\hspace{0.17em}(}k\text{\hspace{0.17em}+ 1)}\\ \text{\hspace{0.17em}}{a}_{k+1}\text{\hspace{0.17em}(}k\text{)}\\ \text{\hspace{0.17em}}\vdots \\ \text{\hspace{0.17em}}{a}_{k+1}\text{\hspace{0.17em}(1)}\end{array}\right]\text{\hspace{0.17em}=\hspace{0.17em}}\left[\begin{array}{c}{a}_{k+1}\text{\hspace{0.17em}(}n\text{\hspace{0.17em}+\hspace{0.17em}}k\text{\hspace{0.17em}+ 1)}\\ \text{\hspace{0.17em}}{a}_{k+1}\text{\hspace{0.17em}(}n\text{\hspace{0.17em}+\hspace{0.17em}}k\text{)}\\ \text{\hspace{0.17em}}\vdots \\ \text{\hspace{0.17em}}{a}_{k+1}\text{\hspace{0.17em}(}n\text{\hspace{0.17em}+ 1)}\end{array}\right]$$(2)

for *n* ≥ 0. Again by an inductive argument, we may write
$$({G}_{k+1}{)}^{n}=\left[\begin{array}{ccc}{a}_{k+1}^{n+k+1}& & -{a}_{k+1}^{n+k}\\ {a}_{k+1}^{n+k}& & -{a}_{k+1}^{n+k-1}\\ {a}_{k+1}^{n+k-1}& {G}^{\prime}& -{a}_{k+1}^{n+k-2}\\ \vdots & & \vdots \\ {a}_{k+1}^{n+1}& & -{a}_{k+1}^{n}\end{array}\right],$$

where *n* ≥ *k*, *a*_{k+1}(*u*) is denoted by ${a}_{k\text{\hspace{0.17em}}+\text{\hspace{0.17em}}1}^{u}$ and ${G}_{k+1}^{\prime}$ is a (*k*+1)×(*k*−1) matrix as follows:
$$({G}_{k+1}^{\mathrm{\prime}})=\left[\begin{array}{cccc}-\left({a}_{k+1}^{n+k}+{a}_{k+1}^{n+k-1}+\cdots +{a}_{k+1}^{n+1}\right)& \cdots & -\left({a}_{k+1}^{n+k}+{a}_{k+1}^{n+k-1}+{a}_{k+1}^{n+k-2}\right)& -\left({a}_{k+1}^{n+k}+{a}_{k+1}^{n+k-1}\right)\\ -\left({a}_{k+1}^{n+k-1}+{a}_{k+1}^{n+k-2}+\cdots +{a}_{k+1}^{n}\right)& \cdots & -\left({a}_{k+1}^{n+k-1}+{a}_{k+1}^{n+k-2}+{a}_{k+1}^{n+k-3}\right)& -\left({a}_{k+1}^{n+k-1}+{a}_{k+1}^{n+k-2}\right)\\ -\left({a}_{k+1}^{n+k-2}+{a}_{k+1}^{n+k-3}+\cdots +{a}_{k+1}^{n-1}\right)& \cdots & -\left({a}_{k+1}^{n+k-2}+{a}_{k+1}^{n+k-3}+{a}_{k+1}^{n+k-4}\right)& -\left({a}_{k+1}^{n+k-2}+{a}_{k+1}^{n+k-3}\right)\\ \vdots & & \vdots & \vdots \\ -\left({a}_{k+1}^{n}+{a}_{k+1}^{n-1}+\cdots +{a}_{k+1}^{n-k+1}\right)& \cdots & -\left({a}_{k+1}^{n}+{a}_{k+1}^{n-1}+{a}_{k+1}^{n-2}\right)& -\left({a}_{k+1}^{n}+{a}_{k+1}^{n-1}\right)\end{array}\right].$$

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 *G*_{3} = −1 *and*
$$({G}_{3}{)}^{n}=\left[\begin{array}{c}{a}_{3}^{n+3}-\left({a}_{3}^{n+2}+{a}_{3}^{n+1}\right)-{a}_{3}^{n+2}\\ {a}_{3}^{n+2}-\left({a}_{3}^{n+1}+{a}_{3}^{n}\right)-{a}_{3}^{n+1}\\ {a}_{3}^{n+1}-\left({a}_{3}^{n+1}+{a}_{3}^{n-1}\right)-{a}_{3}^{n}\end{array}\right]$$

*for n* ≥ 2, *the Simpson formula of sequence* {*a*_{3}(*n*)} *is*
$${a}_{3}^{n+3}({a}_{3}^{n}{)}^{2}-2{a}_{3}^{n+2}{a}_{3}^{n+1}{a}_{3}^{n}+({a}_{3}^{n+2}{)}^{2}{a}_{3}^{n-1}+({a}_{3}^{n+1}{)}^{3}-{a}_{3}^{n+3}{a}_{3}^{n+1}{a}_{3}^{n-1}=1.$$

Let *C*(*c*_{1},*c*_{2},…;*c*_{ν}) be a *ν* × *ν* companion matrix as follows:
$$C\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}({c}_{1},{c}_{2},\cdots ,{c}_{v})\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\left[\begin{array}{cccc}{c}_{1}& {c}_{2}& \cdots & {c}_{\upsilon}\\ 1& 0& & 0\\ \vdots & \ddots & & \vdots \\ 0& \cdots & 1& 0\end{array}\right].$$

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

(Chen and Louck [20]). *The* (*i*,*j*) *entry*
${c}_{{i}_{},j}^{(u)}({c}_{1},{c}_{2},\cdots ,{c}_{v})$
*in the matrix C*^{u} (*c*_{1},*c*_{2},…,*c*_{v}) *is given by the following formula*:
$${C}_{i,j}^{(u)}({c}_{1},{c}_{2},\cdots ,{c}_{v})=\sum _{({t}_{1},{t}_{2},\cdots ,{t}_{v})}\frac{{t}_{j}+{t}_{j+1}+\cdots +{t}_{v}}{{t}_{1}+{t}_{2}+\cdots +{t}_{v}}\times \left(\begin{array}{l}{t}_{1}+\cdots +{t}_{v}\\ \phantom{\rule{1em}{0ex}}{t}_{1},\cdots ,{t}_{v}\end{array}\right){c}_{1}^{{t}_{1}}{\displaystyle \cdots {c}_{v}^{{t}_{v}}}$$

*where the summation is over nonnegative integers satisfying t*_{1} + 2*t*_{2} + … + *vt*_{v} = *u* − *i* + *j*,
$\left(\begin{array}{l}{t}_{1}+\cdots +{t}_{v}\\ \phantom{\rule{1em}{0ex}}{t}_{1},\cdots ,{t}_{v}\end{array}\right)={\displaystyle \frac{({t}_{1}+\cdots +{t}_{v})!}{{t}_{1}!\cdots {t}_{v}!}}$
*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*
$${a}_{{m}_{},n}(n)=\sum _{({t}_{1},{t}_{2\cdots},{t}_{k+1})}\frac{{t}_{k+1}}{{t}_{1}+{t}_{2}+\cdots +{t}_{k+1}}\times \left(\begin{array}{l}{t}_{1}+\cdots +{t}_{k+1}\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{t}_{1},\cdots ,{t}_{k+1}\end{array}\right)(-1{)}^{k+1}$$

*where the summation is over nonnegative integers satisying t*_{1}+2*t*_{2}+…+(*k*+1)*t*_{k+1} = *n*.

In Theorem 2.3, if we choose *v* = *k*+1, *u* = *n*, *i* = *j* = *k*+1, *c*_{1} = 1 and *c*_{2} = … *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
$f(x)={x}^{k+1}-{x}^{k}+{\displaystyle \frac{{x}^{k}-1}{x-1}=\frac{{x}^{k}((x-1{)}^{2}+1)-1}{x-1}.}$
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
${u}_{1}=1+i{\displaystyle \frac{k+1}{k+2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{u}_{1}=1-i\frac{k+1}{k+2}}$
while *f*(*u*) = 0 shows *u*^{k}((*u*−1)^{2} + 1) = 1 so that
$({u}_{1}{)}^{k}={\displaystyle \frac{(k+2{)}^{2}}{2k+3}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}({u}_{2}{)}^{k}=\frac{(k+2{)}^{2}}{(k+1{)}^{2}+(k+2{)}^{2}}}$
which are contradictions since *k* ≥ 2.

If *x*_{1}, *x*_{2}, …, *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 *x*_{1}, *x*_{2}, …, *x*_{k+1} are distinct. Let *V*^{k+1} be (*k*+1) × (*k*+1) Vandermonde matrix as follows:
$${V}^{k+1}=\left[\begin{array}{llll}({x}_{1}{)}^{k}& ({x}_{2}{)}^{k}& \cdots & ({x}_{k+1}{)}^{k}\\ ({x}_{1}{)}^{k-1}& ({x}_{2}{)}^{k-1}& \cdots & ({x}_{k+1}{)}^{k-1}\\ \phantom{\rule{1em}{0ex}}\vdots & \phantom{\rule{1em}{0ex}}\vdots & & \phantom{\rule{1em}{0ex}}\vdots \\ \phantom{\rule{1em}{0ex}}{x}_{1}& \phantom{\rule{1em}{0ex}}{x}_{2}& & {x}_{k+1}\\ \phantom{\rule{1em}{0ex}}1& \phantom{\rule{1em}{0ex}}1& \cdots & \phantom{\rule{1em}{0ex}}1\end{array}\right].$$

Let
$${U}_{i}^{k+1}=\left[\begin{array}{l}({x}_{1}{)}^{n+k+1-i}\\ ({x}_{2}{)}^{n+k+1-i}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\vdots \\ ({x}_{k+1}{)}^{n+k+1-i}\end{array}\right]$$

and supoose that
${V}_{{i}_{},j}^{k+1}$
is a(*k*+1) × (*k*+1) matrix obtained from *V*^{k+1} by replacing the *j* th column of *V*^{k+1} by
${U}_{i}^{k+1}.$

*For n* ≥ *k* ≥ 2,
$${g}_{{i}_{},j}^{(n)}=\frac{det{V}_{i,j}^{k+1}}{det{V}^{k+1}},$$

*where*
$({G}_{k+1}{)}^{n}=[{g}_{{i}_{},j}^{(n)}].$

Since the eigenvalues of the matrix *G*_{k+1}, *x*_{1}, *x*_{2}, …, *x*_{k+1} are distinct, the matrix *G*_{k+1} is diagonalizable. Let *D*_{k+1} = (*x*_{1}, *x*_{2} …, *x*_{k+1}), we easily see that *G*_{k+1} *V*^{k+1} = *V*^{k+1} *D*_{k+1}. Since det *V*^{k+1} ≠ 0, the matrix *V*^{k+1} is invertible. Then it is clear that (*V*^{k+1})^{−1} *G*_{k+1} *V*^{k+1} = *D*_{k+1}. Thus, the matrix *G*_{k+1} is similar to *D*_{k+1}. So we get (*G*_{k+1})^{n} *V*^{k+1} = *V*^{k+1} (*D*_{k+1})^{n} for *n* ≥ *k* ≥ 2. Then we can write the following linear system of equations:
$$\left\{\begin{array}{}\phantom{\rule{2em}{0ex}}{g}_{i,1}^{(n)}({x}_{1}{)}^{k}+{g}_{i,2}^{(n)}({x}_{1}{)}^{k-1}+\cdots +{g}_{i,k+1}^{(n)}=({x}_{1}{)}^{n+k+1-i}\\ \phantom{\rule{2em}{0ex}}{g}_{i,1}^{(n)}({x}_{2}{)}^{k}+{g}_{i,2}^{(n)}({x}_{2}{)}^{k-1}+\cdots +{g}_{i,k+1}^{(n)}=({x}_{2}{)}^{n+k+1-i}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\vdots \\ {g}_{i,1}^{(n)}({x}_{k+1}{)}^{k}+{g}_{i,2}^{(n)}({x}_{k+1}{)}^{k-1}+\cdots +{g}_{i,k+1}^{(n)}=({x}_{k+1}{)}^{n+k+1-i}\end{array}\right.$$

for *n* ≥ *k* ≥ 2. So, for each *i*, *j* = 1,2,…, *k* + 1, we obtain
${g}_{{i}_{},j}^{(n)}$
as follows
$${g}_{{i}_{},j}^{(n)}=\frac{det{V}_{i,j}^{k+1}}{det{V}^{k+1}}.$$

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*
$${a}_{k+1}(n)=-\frac{det{V}_{k+{1}_{},k+1}^{k+1}}{det{V}^{k+1}}.$$

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

*A u* × *v* *real matrix* *M* = [*m*_{i,j}] *is called a contractible matrix in the* *k*^{th} *column* (*resp. row*) *if the* *k*^{th} *column* (*resp. row*) *contains exactly two non-zero entries*.

Suppose that *x*_{1} *x*_{2}, …, *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,k} *x*_{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*) =
$[{m}_{{i}_{},j}^{(n)}]$
be the *n* × *n* super-diagonal matrix, defined by
$${m}_{i,j}^{(n)}=\left\{\begin{array}{l}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\mathrm{i}\mathrm{f}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}i=j=\alpha \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{f}\mathrm{o}\mathrm{r}\phantom{\rule{thinmathspace}{0ex}}1\underset{\_}{<}\phantom{\rule{thinmathspace}{0ex}}\alpha \underset{\_}{<}n\\ 1\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\\ \phantom{\rule{2em}{0ex}}i=\alpha +1\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}j=\alpha \phantom{\rule{thinmathspace}{0ex}}\mathrm{f}\mathrm{o}\mathrm{r}\phantom{\rule{thinmathspace}{0ex}}1\underset{\_}{<}\alpha \underset{\_}{<}n-1,\\ \phantom{\rule{2em}{0ex}}\mathrm{i}\mathrm{f}\phantom{\rule{thinmathspace}{0ex}}i=\alpha -u\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}j=\alpha \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{f}\mathrm{o}\mathrm{r}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}u+1\underset{\_}{<}\alpha \underset{\_}{<}n\\ -1\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\mathrm{s}\mathrm{u}\mathrm{c}\mathrm{h}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}1\underset{\_}{<}u\underset{\_}{<}k,\\ \phantom{\rule{2em}{0ex}}0\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}.\end{array}\right.$$

that is,
$$\begin{array}{c}\begin{array}{ccccccccc}\begin{array}{cccccccccccc}\begin{array}{ccc}& & \end{array}& & & & & & & & & & (k+1)\text{th}& \end{array}& & & & & & & & \end{array}\\ \begin{array}{ccccccccc}\begin{array}{cccccccccc}\begin{array}{ccc}& & \begin{array}{cc}& \end{array}\end{array}& & & & & & & & \downarrow & \end{array}& & & & & & & & \end{array}\\ {M}_{k+1}(n)=\left[\begin{array}{cccccccccc}1& -1& -1& \cdots & -1& 0& \cdots & 0& 0& 0\\ 1& 1& -1& -1& \cdots & -1& 0& \cdots & 0& 0\\ 0& 1& 1& -1& -1& \cdots & -1& 0& \cdots & 0\\ \vdots & \ddots & \ddots & \ddots & & & & \ddots & \ddots & \vdots \\ 0& \cdots & 0& 1& 1& -1& -1& \cdots & -1& 0\\ 0& 0& \cdots & 0& 1& 1& -1& -1& \cdots & -1\\ 0& 0& \cdots & 0& 0& 1& 1& -1& \cdots & -1\\ \vdots & \vdots & & & & \ddots & \ddots & \ddots & \ddots & \vdots \\ 0& 0& & \cdots & & 0& 0& 1& 1& -1\\ 0& 0& & \cdots & & & 0& 0& 1& 1\end{array}\right]\end{array}$$

Then we have the following theorem.

For and *n* ≥ 1 *and k* ≥ 2,
$$per{M}_{k+1}(n)={a}_{k+1}(n+k+1{)}_{},$$

*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:
$$per{M}_{k+1}(2)=\left[\begin{array}{l}1-1\\ 1\phantom{\rule{1em}{0ex}}1\end{array}\right]$$

and
$$per{M}_{k+1}(3)=\left[\begin{array}{l}1-1-1\\ 1\phantom{\rule{1em}{0ex}}1-1\\ 0\phantom{\rule{1em}{0ex}}1\phantom{\rule{1em}{0ex}}1\end{array}\right].$$

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
$$per{M}_{k+1}(n+1)=per{M}_{k+1}(n)-per{M}_{k+1}(n-1)-\cdots -per{M}_{k+1}(n-k).$$

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
${R}_{k+1}(n)=[{r}_{{i}_{},j}^{(n)}]$
be the *n* × *n* matrix, defined by
$${r}_{i,j}^{(n)}=\left\{\begin{array}{l}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\mathrm{i}\mathrm{f}\phantom{\rule{thinmathspace}{0ex}}i=j=\alpha \phantom{\rule{thinmathspace}{0ex}}\mathrm{f}\mathrm{o}\mathrm{r}\phantom{\rule{thinmathspace}{0ex}}1\underset{\_}{<}\alpha \underset{\_}{<}n,\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}i=\alpha +1\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}j=\alpha \phantom{\rule{thinmathspace}{0ex}}\mathrm{f}\mathrm{o}\mathrm{r}\phantom{\rule{thinmathspace}{0ex}}1\underset{\_}{<}\alpha \underset{\_}{<}n-k-1\\ 1\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\\ \phantom{\rule{2em}{0ex}}i=n-k+\alpha \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n-k+\alpha \underset{\_}{<}j\underset{\_}{<}n\phantom{\rule{thinmathspace}{0ex}}\mathrm{f}\mathrm{o}\mathrm{r}\phantom{\rule{thinmathspace}{0ex}}1\underset{\_}{<}\alpha \underset{\_}{<}k-1,\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\mathrm{i}\mathrm{f}\phantom{\rule{thinmathspace}{0ex}}i=\alpha \phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}j=\alpha +u\phantom{\rule{thinmathspace}{0ex}}\mathrm{f}\mathrm{o}\mathrm{r}\phantom{\rule{thinmathspace}{0ex}}1\underset{\_}{<}\alpha \underset{\_}{<}n-k\\ -1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\mathrm{s}\mathrm{u}\mathrm{c}\mathrm{h}\phantom{\rule{thinmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}1\underset{\_}{<}u\underset{\_}{<}k,\\ \phantom{\rule{2em}{0ex}}0\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}.\end{array}\right.$$

that is,
$$\begin{array}{c}\begin{array}{ccccccccc}\begin{array}{ccc}\begin{array}{ccc}& & \end{array}& (k+1)\text{th}& \end{array}& & & & & & & & \end{array}\\ \begin{array}{ccccccccc}\begin{array}{ccc}\begin{array}{ccc}& & \begin{array}{cc}& \end{array}\end{array}& \downarrow & \end{array}& & & & & & & & \end{array}\\ {R}_{k+1}(n)=\left[\begin{array}{cccccccccc}1& -1& \cdots & -1& 0& 0& \cdots & 0& 0& 0\\ 1& 1& -1& \cdots & -1& 0& 0& \cdots & 0& 0\\ 0& 1& 1& -1& \cdots & -1& 0& 0& \cdots & 0\\ \vdots & \ddots & \ddots & \ddots & \ddots & & \ddots & & \ddots & \vdots \\ 0& \cdots & 0& 1& 1& -1& \cdots & -1& -1& 0\\ 0& 0& \cdots & 0& 1& 1& -1& \cdots & -1& -1\\ 0& 0& 0& \cdots & 0& 0& 1& 1& \cdots & 1\\ \vdots & & \ddots & \ddots & & \ddots & \ddots & \ddots & \ddots & \vdots \\ 0& 0& 0& 0& \cdots & 0& 0& 0& 1& 1\\ 0& 0& 0& 0& 0& \cdots & 0& 0& 0& 1\end{array}\right]\leftarrow (n-k)\text{th}\end{array}$$

Assume that the *n* × *n* matrix
${T}_{k+1}(n)=[{t}_{{i}_{},j}^{(n)}]$
is defined by
$$\begin{array}{c}\begin{array}{ccccccccc}\begin{array}{cccccccccccc}\begin{array}{ccc}& & \end{array}& & & & & & & & & & (k+1)\text{th}& \end{array}& & & & & & & & \end{array}\\ \begin{array}{ccccccccc}\begin{array}{cccccccccc}\begin{array}{ccc}& & \begin{array}{cc}& \end{array}\end{array}& & & & & & & & \downarrow & \end{array}& & & & & & & & \end{array}\\ {T}_{k+1}(n)=\left[\begin{array}{ccccc}1& \cdots & 1& 0& \cdots 0\\ 1& & & & \\ 0& & & {R}_{k+1}(n-1)& \\ \vdots & & & & \\ 0& & & & \end{array}\right]\end{array},$$

where *n* > *k* + 2 such that *k* ≥ 2.

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

*Let* *a*_{k+1}(*n*) *be the nth the arrowhead-Fibonacci number for* *k* ≥ 2. *Then*

*For* *n* > *k* + 1,
$$per{R}_{k+1}(n)={a}_{k+1}(n+1).$$

*For* *n* > *k* + 2,
$$per{T}_{k+1}(n)=\sum _{i=1}^{n}{a}_{k+1}n(i).$$

Let the equation hold for *n* > k + 1, then we show that the equation holds for *n* + 1. If we expand the *per***R**_{k+1}(*n*) by the Laplace expansion of permanent with respect to the first row, then we obtain
$$per{R}_{k+1}(n+1)=per{R}_{k+1}(n)-per{R}_{k+1}(n-1)-\cdots -per{R}_{k+1}(n-k).$$

Also, since
$$per{R}_{k+1}(n)={a}_{k+1}(n+1),\phantom{\rule{thinmathspace}{0ex}}per{R}_{k+1}(n-1)={a}_{k+1}(n),\dots ,\phantom{\rule{thinmathspace}{0ex}}per{R}_{k+1}(n-k)={a}_{k+1}(n-k+1),$$

it is clear that
$$per{R}_{k+1}(n+1)={a}_{k+1}(n+2).$$

It is clear that expanding the *perT*_{k+1}(*n*) by the Laplace expansion of permanent with respect to the first row, gives us
$$per{T}_{k+1}(n)=per{T}_{k+1}(n-1)+per{R}_{k+1}(n).$$

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:
$$g(x)=\frac{{x}^{k}}{1-x+{x}^{2}+\cdots +{x}^{k+1}},$$

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*:
$$g(x)={x}^{k}\mathrm{exp}\left(\sum _{i=1}^{\mathrm{\infty}}\frac{{x}^{i}}{i}{\left(1-x-\cdots {x}^{k}\right)}^{i}\right),$$

*where* *k* ≥ 2.

Since
$$\mathrm{ln}g(x)=\mathrm{ln}{x}^{k}-\mathrm{ln}(1-x+{x}^{2}+\cdots +{x}^{k+1})$$

and
$$-\text{ln}(1-x+{x}^{2}+{\displaystyle \cdots +{x}^{k+1})=-\mathrm{ln}(1-(x-{x}^{2}-\cdots -{x}^{k+1}))=\sum _{i=1}^{\mathrm{\infty}}\frac{{x}^{i}}{i}(1-x-\cdots {x}^{k}{)}^{i},}$$

it is clear that
$$\mathrm{ln}g(x)-\mathrm{ln}{x}^{k}=\mathrm{ln}\frac{g(x)}{{x}^{k}}=\sum _{i=1}^{\mathrm{\infty}}\frac{{x}^{i}}{i}(1-x-\cdots {x}^{k}{)}^{i}.$$

Thus we have the conclusion.

Now we consider the sums of arrowhead-Fibonacci numbers.

Let
$${S}_{n}=\sum _{i=1}^{n}{a}_{k+1}(i)$$

for *n* ≥ 1 and *k* ≥ 2, and suppose that *A*_{k+2} is the (*k* + 2)×(*k* + 2) matrix such that
$${A}_{k+2}=\left[\begin{array}{lll}1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0& \cdots & 0\\ 1& & \\ 0& {G}_{k+1}& \\ \vdots & & \\ 0& & \end{array}\right].$$

Then it can be shown by induction that
$$({A}_{k+2}{)}^{n}=\left[\begin{array}{llll}\phantom{\rule{1em}{0ex}}1& 0& \cdots & 0\\ {S}_{n+k}& & & \\ {S}_{n+k-1}& & ({G}_{k+1}{)}^{n}& \\ \phantom{\rule{1em}{0ex}}\vdots & & & \\ \phantom{\rule{1em}{0ex}}{S}_{n}& & & \end{array}\right].$$

## Comments (0)