1 Introduction

Wituła and Słota in their papers [21,22,23] introduced original number systems called quasi-Fibonacci numbers of n-th order. These systems arose naturally on a base of two well-known equivalent relations for Fibonacci numbers, that is

$$\begin{aligned} (1+\eta ^k+\eta ^{4k})^n=F_{n+1}+F_n(\eta ^k+\eta ^{4k}) \end{aligned}$$

and

$$\begin{aligned} (1+\eta ^{2k}+\eta ^{3k})^n=F_{n+1}+F_n(\eta ^{2k}+\eta ^{3k}), \end{aligned}$$

where \(F_n\) is the n-th Fibonacci number and \(n\in {\mathbb {N}}_0,\ k\in {\mathbb {N}}{\setminus } 5{\mathbb {N}},\ \eta \in {\mathbb {C}}{\setminus }\{1\},\) \(\eta ^5=1\), i.e. k is not divisible by 5 and \(\eta \) is a primitive 5-th root of unity (see also [14, 19]).

The simplest system of this type consists of \(\delta \)-Fibonacci numbers discussed in many papers (e.g. [10, 11, 23]). Recall that \(\delta \)-Fibonacci numbers are members of sequences \((a_n(\delta )),\ (b_n(\delta ))\) defined in the following way

$$\begin{aligned} \bigl (1+\delta (\eta ^k+\eta ^{4k})\bigr )^n=a_n(\delta )+b_n(\delta )(\eta ^k+\eta ^{4k}), \end{aligned}$$
(1)

where \(n\in {\mathbb {N}}_0,\ k\in {\mathbb {N}}{\setminus } 5{\mathbb {N}},\ \eta \in {\mathbb {C}}{\setminus }\{1\},\ \eta ^5=1\).

They can be also defined recurrently, namely \(a_0(\delta )=1,\ b_0(\delta )=0\) and

$$\begin{aligned} \left[ \begin{array}{c} a_{n+1}(\delta )\\ b_{n+1}(\delta ) \end{array}\right] =\left[ \begin{array}{cc} 1&{}\delta \\ \delta &{} 1-\delta \end{array}\right] \left[ \begin{array}{c} a_{n}(\delta )\\ b_{n}(\delta ) \end{array}\right] , \quad n\in {\mathbb {N}}_0. \end{aligned}$$

One of the most important properties of \(\delta \)-Fibonacci numbers is the fact that they represent the binomials transformation of the scaled Fibonacci numbers, that is the following identities hold

$$\begin{aligned} a_n(\delta )&=\sum _{k=0}^n{n \atopwithdelims ()k}F_{k-1}(-\delta )^k=\sum _{k=0}^n {n \atopwithdelims ()k}F_{k+1}(1-\delta )^{n-k}\delta ^k,\\ b_n(\delta )&=\sum _{k=1}^n {n \atopwithdelims ()k}(-1)^{k-1}F_k\delta ^k=\sum _{k=1}^n{n \atopwithdelims ()k}F_k(1-\delta )^{n-k}\delta ^k, \end{aligned}$$

for \( n\in {\mathbb {N}}\). Also, interesting sequences were obtained for some values of \(\delta \):

$$\begin{aligned} \begin{array}{ll} a_n(1)=F_{n+1},&{}b_n(1)=F_n,\\ a_n(-1)=F_{2n-1}, &{}b_n(-1)=-F_{2n},\\ a_{2n+2}(-i)=(2i-1)^{n+1}F_n, &{}b_{2n}(-i)=(2i-1)^nF_n,\\ \displaystyle {a_{2n}\left( \frac{1+i}{2}\right) =\frac{1}{5}\left( 1+\frac{1}{2}i\right) ^nL_{n+2}}, &{}\displaystyle {b_{2n}\left( \frac{1+i}{2}\right) =\left( 1+\frac{1}{2}i\right) ^nF_n}.\end{array} \end{aligned}$$

Moreover, for odd natural n we have

$$\begin{aligned} a_n(\delta )=\left( \frac{\sqrt{5}}{2\cos \varphi }\right) ^n \left( \frac{\sqrt{5}}{5}L_{n-1}T_n(\cos \varphi )+iF_n\sin \varphi U_{n-1}(\cos \varphi )\right) \end{aligned}$$

and for even natural n we have

$$\begin{aligned} a_n(\delta )=\left( \frac{\sqrt{5}}{2\cos \varphi }\right) ^n \left( F_{n-1}T_n(\cos \varphi )+i\frac{\sqrt{5}}{5}L_{n-1}\sin \varphi U_{n-1}(\cos \varphi )\right) , \end{aligned}$$

where \(L_n\) denotes the n-th Lucas numberFootnote 1 whereas \(T_n\) and \(U_n\) denote the Chebyshev polynomials of the first and second kind, respectivelyFootnote 2 (see [8]).

It turned out that it is even more natural and proper to consider generalizations of Fibonacci numbers depending on more parameters [17]. These are quasi-Fibonacci numbers of order n (which is assumed to be odd) and depend on \(\frac{1}{2}\varphi (n)-1\) parameters, where \(\varphi =\varphi (n)\) is the Euler’s totient function. In [17] the first nontrivial cases were discussed in details, namely quasi-Fibonacci numbers of order 7 and 9 (these are the only two-parametric numbers). For instance, two-parametric quasi-Fibonacci numbers of 9-th order are members of sequences \((A_{n,9}(\delta ,\gamma )),\ (B_{n,9}(\delta ,\gamma ))\) and \((C_{n,9}(\delta ,\gamma ))\) defined by the following relations

$$\begin{aligned}&(1+\delta (\zeta ^k +\zeta ^{8k})+ \gamma (\zeta ^{2k} +\zeta ^{7k}))^n \\&\quad =A_{n,9}(\delta ,\gamma )+B_{n,9}(\delta ,\gamma )(\zeta ^k +\zeta ^{8k})+C_{n,9}(\delta ,\gamma )(\zeta ^{2k} +\zeta ^{7k}), \end{aligned}$$

where \( n \in {\mathbb {N}}_0,\ k \in {\mathbb {N}}{\setminus } 9{\mathbb {N}}, \ \delta , \gamma \in {\mathbb {C}},\ \zeta \in {\mathbb {C}}\) is a primitive 9-th root of unity.

They can be also defined recurrently, namely we have \(A_{0,9}(\delta ,\gamma )=1,\) \(B_{0,9}(\delta ,\gamma )=C_{0,9}(\delta ,\gamma )=0\) and for all \(n\in {\mathbb {N}}_0\)

$$\begin{aligned} \left[ \begin{array}{c} A_{n+1,9}(\delta ,\gamma )\\ B_{n+1,9}(\delta ,\gamma )\\ C_{n+1,9}(\delta ,\gamma ) \end{array}\right] = \left[ \begin{array}{ccc} 1&{}2\delta -\gamma &{}2\gamma -\delta \\ \delta &{}1+\gamma &{} \delta -\gamma \\ \gamma &{}\delta &{}1-\gamma \end{array}\right] \left[ \begin{array}{c} A_{n,9}(\delta ,\gamma )\\ B_{n,9}(\delta ,\gamma )\\ C_{n,9}(\delta ,\gamma ) \end{array}\right] . \end{aligned}$$

Note that for \(\gamma =0\) we get the (one-parameter) quasi-Fibonacci numbers (this type of numbers was discussed e.g. in [21, 22]).

From one side n-th roots of unity form a cyclic group under multiplication, and from the other some sums of these roots are linearly independent (like \(1,\ \zeta ^k +\zeta ^{8k}\) and \(\zeta ^{2k} +\zeta ^{7k}\) used in the above definition). That made us extend the idea of quasi-Fibonacci numbers to real quaternions. During our investigations we obtained a lot of interesting results, also on quaternions structure itself, and they will be presented in this paper.

We first remind the definition and basic properties of real quaternions. They arose as a natural extension of complex numbers [15].

Definition 1

Let \({\mathbb {H}}\) be a 4-dimensional vector space over \({\mathbb {R}}\) with the basis \(\{1,i,j,k\}\). Then a (real) quaternion q is an element of \({\mathbb {H}}\) written with respect to this basis, that is

$$\begin{aligned} q:=a_0+a_1i+a_2j+a_3k, \end{aligned}$$

where \(a_0\;,a_1,\;a_2,\;a_3 \in {\mathbb {R}}\) with i, j, k called imaginary units.

Now, let the multiplication be defined on imaginary units by the following rules, called Hamilton formulae,

$$\begin{aligned} i^2 = j^2 = k^2 =ijk=-1 \end{aligned}$$
(2)

and extended to \({\mathbb {H}}\) by requirements that it is associative, distributive over addition and commutative with respect to scalar multiplication. The resulting division ring is also denoted by \({\mathbb {H}}\).

Moreover, a unary operation \({\overline{\omega }}:{\mathbb {H}}\rightarrow {\mathbb {H}}\) called conjugation is defined in the following way

$$\begin{aligned} {\overline{q}}:=\overline{a_0+a_1i+a_2j+a_3k}=a_0-a_1i-a_2j-a_3k, \end{aligned}$$

and the norm of a quaternion q is defined as

$$\begin{aligned} |q|:=\sqrt{a_0^2+a_1^2+a_2^2+a_3^2}. \end{aligned}$$

From now on i, j, k will be used only as imaginary units.

Note that quaternions can be also represented by matrices in such a manner that multiplication of quaternions is compatible with that of matrices.Footnote 3 Namely, let I be the identity matrix of order 4 and let H, J, K be matrices of order 4 representing i, j, k, respectively, i.e. satisfying the relations like in (2). Then \(q\in {\mathbb {H}}\) can be written in a matrix form Q as

$$\begin{aligned} Q=a_0I+a_1H+a_2J+a_3K. \end{aligned}$$
(3)

It was proven in [4] that there are 48 distinct representations of quaternions by skew-symmetric signed permutation matrices. More precisely, H, J, K can be chosen from two sets \(\{\pm X_s,\pm Y_s,\pm Z_s\},\ s=1,2\) under assumption that we take either a matrix or its negation and formulae (2) are satisfied where

$$\begin{aligned} X_1&=\left[ \begin{array}{cccc} 0&{}0&{}1&{}0\\ 0&{}0&{}0&{}-1\\ -1&{}0&{}0&{}0\\ 0&{}1&{}0&{}0 \end{array}\right] , \ \ Y_1=\left[ \begin{array}{cccc} 0&{}1&{}0&{}0\\ -1&{}0&{}0&{}0\\ 0&{}0&{}0&{}1\\ 0&{}0&{}-1&{}0 \end{array}\right] , \ \ Z_1=\left[ \begin{array}{cccc} 0&{}0&{}0&{}1\\ 0&{}0&{}1&{}0\\ 0&{}-1&{}0&{}0\\ -1&{}0&{}0&{}0 \end{array}\right] ,\\ X_2&=\left[ \begin{array}{cccc} 0&{}1&{}0&{}0\\ -1&{}0&{}0&{}0\\ 0&{}0&{}0&{}-1\\ 0&{}0&{}1&{}0 \end{array}\right] , \ \ Y_2=\left[ \begin{array}{cccc} 0&{}0&{}1&{}0\\ 0&{}0&{}0&{}1\\ -1&{}0&{}0&{}0\\ 0&{}-1&{}0&{}0 \end{array}\right] ,\ \ Z_2=\left[ \begin{array}{cccc} 0&{}0&{}0&{}1\\ 0&{}0&{}-1&{}0\\ 0&{}1&{}0&{}0\\ -1&{}0&{}0&{}0 \end{array}\right] . \end{aligned}$$

Now, let \(S^3\) denote the set of unit quaternions, that is

$$\begin{aligned} S^3:=\{q\in {\mathbb {H}};\quad |q|=1\}. \end{aligned}$$

Note that \(S^3\) is a subgroup of the multiplicative group of \({\mathbb {H}}\). Moreover, in \({\mathbb {H}}\simeq {\mathbb {R}}^4\) it is a surface of a unit sphere with the centre at the origin (see [13]).

Next, each quaternion \(q\in S^3{\setminus }\{\pm 1\}\) can be written in a polar form

$$\begin{aligned} q=\cos \theta +\omega \sin \theta , \end{aligned}$$

where \(\cos \theta = a_0\) and \(\omega =\frac{a_1 i+a_2j +a_3 k}{\sqrt{a_1^2+a_2^2+a_3^2}}=\frac{a_1 i+a_2j +a_3 k}{\sqrt{1-a_0^2}}\), from which Cho in [2] obtained an analogue of de Moivre’s formula, i.e.

$$\begin{aligned} q^n =\cos n\theta +\omega \sin n\theta . \end{aligned}$$
(4)

Actually, each nonreal quaternion \(q\in {\mathbb {H}}\), i.e. such that \(a_1^2+a_2^2+a_3^2\ne 0\), can be written in a polar form

$$\begin{aligned} q=|q|(\cos \theta +\varepsilon \sin \theta ), \end{aligned}$$
(5)

where

$$\begin{aligned} \varepsilon =\frac{a_1i+a_2j+a_3k}{\sqrt{a_1^2+a_2^2+a_3^2}},\quad \cos \theta =\frac{a_0}{|q|}, \quad \sin \theta =\frac{\sqrt{a_1^2+a_2^2+a_3^2}}{|q|}, \end{aligned}$$

such that the generalization of (4), still called de Moirve’s formula, holds

$$\begin{aligned} q^n=|q|^n(\cos n \theta +\varepsilon \sin n \theta ). \end{aligned}$$
(6)

2 Quaternaccis

Definition 2

Quaternion equivalents for quasi-Fibonacci numbers (shortly Quaternaccis) are members of sequences \((A_{n}(b,c,d)), (B_{n}(b,c,d)),(C_{n}(b,c,d)),\) \((D_{n}(b,c,d))\) defined by the following relation

$$\begin{aligned} (1+bi+cj+dk)^n= A_{n}(b,c,d)+B_{n}(b,c,d) i+ C_{n}(b,c,d)j+D_{n}(b,c,d) k, \end{aligned}$$

where \(n \in {\mathbb {N}}_0\), \(b,c,d\in {\mathbb {R}}\).

Example

Values of Quaternaccis for \(n\leqslant 6\) are given at the end of this paper.

Notation From now on, we will use (b, c, d) only to denote arguments of Quaternaccis. Moreover, while describing properties valid for any argument (b, c, d) we shall omit them in formulation of results, in particular we will use the following abbreviations:

$$\begin{aligned} \begin{array}{cc} A_{n}=A_{n}(b,c,d),&{} B_{n}=B_{n}(b,c,d),\\ C_{n}=C_{n}(b,c,d),&{} D_{n}=D_{n}(b,c,d). \end{array} \end{aligned}$$

Theorem 1

For all \(n\in {\mathbb {N}}_0\) the following recurrence relation holds

$$\begin{aligned} \left[ \begin{array}{c} A_{n+1}\\ B_{n+1}\\ C_{n+1}\\ D_{n+1} \end{array}\right] = \left[ \begin{array}{cccc} 1&{}-b&{}-c&{}-d\\ b&{}1&{}d&{}-c\\ c&{}-d&{}1&{}b\\ d&{}c&{}-b&{}1 \end{array}\right] \left[ \begin{array}{c} A_{n}\\ B_{n}\\ C_{n}\\ D_{n} \end{array}\right] .\end{aligned}$$
(7)

Proof

It is obviously true for \(n=0\). For \(N\in {\mathbb {N}}\) from Definition 2 we get

$$\begin{aligned} (1+bi+cj+dk)^{N+1} =A_{N+1}+B_{N+1} i+ C_{N+1}j+D_{N+1} k, \end{aligned}$$
(8)

whereas from the properties of exponentiation of quaternions it follows that

$$\begin{aligned}&(1+bi+cj+dk)^{N+1}=(1+bi+cj+dk)^N(1+bi+cj+dk)\nonumber \\&\quad =\bigl (A_{N}+B_{N}i+ C_{N}j+D_{N} k\bigr )(1+bi+cj+dk)\nonumber \\&\quad =A_{N}+bA_{N}i+cA_{N}j+dA_{N}k+\cdots \end{aligned}$$
(9)

Since \(\{1,i,j,k\}\) is the basis of \({\mathbb {H}}\), we obtain the statement by comparison of coefficients of 1, i, j, k in (8) and (9).\(\square \)

Proposition 1

For all \(m,n\in {\mathbb {N}}_0\) the following relations hold

$$\begin{aligned} cD_{n}=dC_{n},\quad bC_{n}=cB_{n}, \quad dB_{n}=bD_{n}, \end{aligned}$$
(10)

and

$$\begin{aligned} C_{n}D_{m}=C_{m}D_{n},\quad B_{n}C_{m}=B_{m}C_{n},\quad D_{n}B_{m}=D_{m}B_{n}. \end{aligned}$$
(11)

Proof

From associativity of powers for \(N\in {\mathbb {N}}\) we also have

$$\begin{aligned} (1+bi+cj+dk)^{N+1}=(1+bi+cj+dk)(1+bi+cj+dk)^N \end{aligned}$$

Using the definition of Quaternaccis for the later factor and comparing with (9) we get (10). The remaining three formulae are derived in an analogous way from \(q^nq^m=q^{m+n}=q^mq^n\) for \(q=1+bi+cj+dk\).\(\square \)

Formulae (7) and (10) immediately imply the following relations:

Corollary 1

For all \(n\in {\mathbb {N}}_0\) the following recurrence relations are satisfied

$$\begin{aligned} \left[ \begin{array}{c} A_{n+1}\\ B_{n+1}\\ C_{n+1}\\ D_{n+1} \end{array}\right] = \left[ \begin{array}{cccc} 1&{}-b&{}-c&{}-d\\ b&{}1&{}-d&{}c\\ c&{}d&{}1&{}-b\\ d&{}-c&{}b&{}1 \end{array}\right] \left[ \begin{array}{c} A_{n}\\ B_{n}\\ C_{n}\\ D_{n} \end{array}\right] \end{aligned}$$
(12)

and

$$\begin{aligned} \left[ \begin{array}{c} A_{n+1}\\ B_{n+1}\\ C_{n+1}\\ D_{n+1} \end{array}\right] = \left[ \begin{array}{cccc} 1&{}-b&{}-c&{}-d\\ b&{}1&{}0&{}0\\ c&{}0&{}1&{}0\\ d&{}0&{}0&{}1 \end{array}\right] \left[ \begin{array}{c} A_{n}\\ B_{n}\\ C_{n}\\ D_{n} \end{array}\right] . \end{aligned}$$
(13)

Remark 1

Note that formulae (10) actually allow to write a form of the recurrence relations for Quaternaccis which generalize (7), (12) and (13), by introducing parameters \(a_1,a_2,a_3\in {\mathbb {R}}\) as follows

$$\begin{aligned} \left[ \begin{array}{c} A_{n+1}\\ B_{n+1}\\ C_{n+1}\\ D_{n+1} \end{array}\right] = \left[ \begin{array}{cccc} 1&{}-b&{}-c&{}-d\\ b&{}1&{}-a_1&{}a_2\\ c&{}a_1&{}1&{}-a_3\\ d&{}-a_2&{}a_3&{}1 \end{array}\right] \left[ \begin{array}{c} A_{n}\\ B_{n}\\ C_{n}\\ D_{n} \end{array}\right] . \end{aligned}$$
(14)

Remark 2

The matrix given in Theorem 1 is one of the basic matrix representations of quaternion \(1+bi+cj+dk\), obtained from (3) for \(H=-X_2,\) \(J=-Y_2,K=-Z_2\), similarly as the matrix in (12), which is obtained for \(H=-Y_1,J=-X_1,K=-Z_1\).

Our next results provide more properties of transition matrices of recurrence relations (7), (12) and (13). We first focus on determining the n-th power of these matrices. For instance, after n successive applications of formula (7) we obtain

$$\begin{aligned} \left[ \begin{array}{c} A_{n+m}\\ B_{n+m}\\ C_{n+m}\\ D_{n+m} \end{array}\right] = \left[ \begin{array}{cccc} 1&{}-b&{}-c&{}-d\\ b&{}1&{}d&{}-c\\ c&{}-d&{}1&{}b\\ d&{}c&{}-b&{}1 \end{array}\right] ^n\cdot \left[ \begin{array}{c} A_{m}\\ B_{m}\\ C_{m}\\ D_{m} \end{array}\right] . \end{aligned}$$

Note that since \(A_{0}=1,\ B_{0}=C_{0}=D_{0}=0\), then the first column of the above matrix is \([A_{n},B_{n},C_{n},D_{n}]^T\). Analogous properties are true for (12), (13) and (14). But it turns out that the situation is even more interesting.

Theorem 2

For all \(n\in {\mathbb {N}}_0\) the following identities hold

$$\begin{aligned}&\left[ \begin{array}{ccccc} 1&{}-b&{}-c&{}-d\\ b&{}1&{}d&{}-c\\ c&{}-d&{}1&{}b\\ d&{}c&{}-b&{}1 \end{array}\right] ^n=\left[ \begin{array}{ccccc} A_{n}&{}-B_{n}&{} -C_{n}&{}-D_{n}\\ B_{n}&{}A_{n}&{}D_{n}&{}-C_{n}\\ C_{n}&{}-D_{n}&{}A_{n}&{}B_{n}\\ D_{n}&{}C_{n}&{} -B_{n}&{}A_{n}\\ \end{array}\right] , \end{aligned}$$
(15)
$$\begin{aligned}&\left[ \begin{array}{ccccc} 1&{}-b&{}-c&{}-d\\ b&{}1&{}-d&{}c\\ c&{}d&{}1&{}-b\\ d&{}-c&{}b&{}1 \end{array}\right] ^n=\left[ \begin{array}{ccccc} A_{n}&{}-B_{n}&{} -C_{n}&{}-D_{n}\\ B_{n}&{}A_{n}&{}-D_{n}&{}C_{n}\\ C_{n}&{}D_{n}&{}A_{n}&{}-B_{n}\\ D_{n}&{}-C_{n}&{} B_{n}&{}A_{n}\\ \end{array}\right] \end{aligned}$$
(16)

and

$$\begin{aligned} \left[ \begin{array}{ccccc} 1&{}-b&{}-c&{}-d\\ b&{}1&{}0&{}0\\ c&{}0&{}1&{}0\\ d&{}0&{}0&{}1 \end{array}\right] ^n =\left[ \begin{array}{cccc} A_{n}&{}-B_{n}&{}-C_{n}&{}-D_{n}\\ B_{n}&{}1+b^2E_{n}&{}bcE_{n}&{}bdE_{n}\\ C_{n}&{}bcE_{n}&{}1+c^2E_{n}&{}cdE_{n}\\ D_{n}&{}bdE_{n}&{}cdE_{n}&{}1+d^2E_{n} \end{array}\right] . \end{aligned}$$
(17)

where \(E_{n}:={\left\{ \begin{array}{ll}0&{} b=c=d=0\\ \frac{A_{n}-1}{b^2+c^2+d^2} &{} \text {otherwise}\end{array}\right. }\).

Proof

We shall prove (15) by induction on n. By definition of Quaternaccis the statement follows for \(n=0\) and moreover, for \(n=1\) we have

$$\begin{aligned} \left[ \begin{array}{ccccc} A_{1}&{}-B_{1}&{} -C_{1}&{}-D_{1}\\ B_{1}&{}A_{1}&{}D_{1}&{}-C_{1}\\ C_{1}&{}-D_{1}&{}A_{1}&{}B_{1}\\ D_{1}&{}C_{1}&{} -B_{1}&{}A_{1}\\ \end{array}\right] =\left[ \begin{array}{ccccc} 1&{}-b&{}-c&{}-d\\ b&{}1&{}d&{}-c\\ c&{}-d&{}1&{}b\\ d&{}c&{}-b&{}1 \end{array}\right] . \end{aligned}$$

Next, assume that (15) is true for some \(n\in {\mathbb {N}}\). Then

Therefore by the induction principle formula (15) is valid for all \(n\in {\mathbb {N}}_0\).

Formula (16) is proven analogously. To prove (17) we need to justify auxiliary identities, namely that \(E_0=E_1=0\) and for \(n\geqslant 2,\ bcd\ne 0\)

$$\begin{aligned} E_{n}=\frac{A_{n}-1}{b^2+c^2+d^2}=-\frac{1}{b}\sum _{k=1}^{n-1}B_{k}=-\frac{1}{c}\sum _{k=1}^{n-1}C_{k}=-\frac{1}{d}\sum _{k=1}^{n-1}D_{k}, \end{aligned}$$
(18)

whereas for \(b=0\) or \(c=0\) or \(d=0\) the corresponding sum is 0.

These identities can be proven by induction, however a more elegant and interesting proof will be given later (see Corollary 2). \(\square \)

Note that if in the proof we use the inductive assumption as follows

$$\begin{aligned} \left[ \begin{array}{cccc} 1&{}-b&{}-c&{}-d\\ b&{}1&{}d&{}-c\\ c&{}-d&{}1&{}b\\ d&{}c&{}-b&{}1 \end{array}\right] \cdot \left[ \begin{array}{cccc} A_{n}&{}-B_{n}&{} -C_{n}&{}-D_{n}\\ B_{n}&{}A_{n}&{}D_{n}&{}-C_{n}\\ C_{n}&{}-D_{n}&{}A_{n}&{}B_{n}\\ D_{n}&{}C_{n}&{} -B_{n}&{}A_{n}\\ \end{array}\right] , \end{aligned}$$

then we obtain a matrix that is equal to the desired one by means of (10).

Remark 3

In Appendix 1 we give two general methods for exponentiation of so-called arrowhead matrices which could be also used for proving (17).

Remark 4

Note that matrices in (15) and (16) are actually of the very special form (mind the colors)—we point it out below for (15).

figure a

More precisely, for every \(n\in {\mathbb {N}}\) the matrix on the right results form the one on the left by substituting 1, b, c, d by \(A_{n}, B_{n},C_{n}\) and \(D_{n}\), respectively. Or, on the other hand, the matrix on the right is obtained from the middle one by replacing the index by the exponent.

An analogous property is true for two-parameter quasi-Fibonacci numbers of order 7 (see [17]) and 9 (the formula presented below is new)

figure b

Notation From now on for \(b,c,d \in {\mathbb {R}}\) we set \(\lambda :=\sqrt{b^2+c^2+d^2}\).

Straightforward calculations lead to the following

Proposition 2

Transition matrices in (7), (12) and (13) are diagonalizable and their Jordan decompositions are as follows:

a)

$$\begin{aligned} \left[ \begin{array}{cccc} 1&{}-b&{}-c&{}-d\\ b&{}1&{}0&{}0\\ c&{}0&{}1&{}0\\ d&{}0&{}0&{}1 \end{array}\right] =P_1\text {diag}(1,1,1-i\lambda ,1+i\lambda )P_1^{-1}, \end{aligned}$$
(20)

where

$$\begin{aligned} P_1^{-1}=\left[ \begin{array}{cccc} 0&{}0&{}-\displaystyle {\frac{i\lambda }{c}}&{}\displaystyle {\frac{i\lambda }{c}}\\ -\displaystyle {\frac{c}{a}}&{}-\displaystyle {\frac{b}{a}}&{}\displaystyle {\frac{a}{c}}&{}\displaystyle {\frac{a}{c}}\\ 0&{}1&{}\displaystyle {\frac{b}{c}}&{}\displaystyle {\frac{b}{c}}\\ 1&{}0&{}1&{}1 \end{array}\right] ^{-1}=\displaystyle {\frac{1}{2\lambda ^2}}\left[ \begin{array}{cccc} 0&{}-2bd&{}-2cd&{}2(b^2+c^2)\\ 0&{}-2bc&{}2(b^2+d^2)&{}-2cd\\ id\lambda &{}bd&{}cd&{}d^2\\ -id\lambda &{}bd&{}cd&{}d^2 \end{array}\right] ; \end{aligned}$$

b)

$$\begin{aligned} \left[ \begin{array}{cccc} 1&{}-b&{}-c&{}-d\\ b&{}1&{}d&{}-c\\ c&{}-d&{}1&{}b\\ d&{}c&{}-b&{}1 \end{array}\right] =P_2\text {diag}(1-i\lambda ,1-i\lambda ,1+i\lambda ,1+i\lambda )P_2^{-1}, \end{aligned}$$
(21)

where

$$\begin{aligned} P_2= & {} \left[ \begin{array}{cccc} \displaystyle {-\frac{bc+di\lambda }{c^2+d^2}}&{}\displaystyle {\frac{bd-ci\lambda }{c^2+d^2}}&{}\displaystyle {\frac{-bc+di\lambda }{c^2+d^2}}&{}\displaystyle {\frac{bd+ci\lambda }{c^2+d^2}}\\ \displaystyle {\frac{bd-ci\lambda }{c^2+d^2}}&{}\displaystyle {\frac{bc+di\lambda }{c^2+d^2}}&{}\displaystyle {\frac{bc+di\lambda }{c^2+d^2}}&{}\displaystyle {\frac{bc-di\lambda }{c^2+d^2}}\\ 0&{}1&{}0&{}1\\ 1&{}0&{}1&{}0 \end{array}\right] , \\ P_2^{-1}= & {} -\displaystyle {\frac{i}{2\lambda }}\left[ \begin{array}{cccc} -d&{}-c&{}b&{}i\lambda \\ -c&{}d&{}i\lambda &{}-b\\ d&{}c&{}-b&{}i\lambda \\ c&{}-d&{}i\lambda &{}b \end{array}\right] ; \end{aligned}$$

c)

$$\begin{aligned} \left[ \begin{array}{cccc} 1&{}-b&{}-c&{}-d\\ b&{}1&{}-d&{}c\\ c&{}d&{}1&{}-b\\ d&{}-c&{}b&{}1 \end{array}\right] =P_3\text {diag}(1-i\lambda ,1-i\lambda ,1+i\lambda ,1+i\lambda )P_3^{-1}, \end{aligned}$$
(22)

where

$$\begin{aligned} P_3= & {} \left[ \begin{array}{cccc} \displaystyle {\frac{bc-di\lambda }{c^2+d^2}}&{}\displaystyle {-\frac{bd+ci\lambda }{c^2+d^2}}&{}\displaystyle {\frac{bc+di\lambda }{c^2+d^2}}&{}\displaystyle {\frac{-bd+ci\lambda }{c^2+d^2}}\\ \displaystyle {\frac{bd+ci\lambda }{c^2+d^2}}&{}\displaystyle {\frac{bc-di\lambda }{c^2+d^2}}&{}\displaystyle {\frac{bd-ci\lambda }{c^2+d^2}}&{}\displaystyle {\frac{bc+di\lambda }{c^2+d^2}}\\ 0&{}1&{}0&{}1\\ 1&{}0&{}1&{}0 \end{array}\right] , \\ P_3^{-1}= & {} -\displaystyle {\frac{i}{2\lambda }}\left[ \begin{array}{cccc} -d&{}c&{}-b&{}i\lambda \\ -c&{}-d&{}i\lambda &{}b\\ d&{}-c&{}b&{}i\lambda \\ c&{}d&{}i\lambda &{}-b \end{array}\right] . \end{aligned}$$

As a consequence of Theorem 2 we get the following

Theorem 3

Quaternacci sequences \((A_{n}), (B_{n}),(C_{n})\) and \((D_{n})\) satisfy the following recurrence relation

$$\begin{aligned} {\mathbb {X}}_{n}=2{\mathbb {X}}_{n-1}-(1+\lambda ^2){\mathbb {X}}_{n-2},\quad n\geqslant 2, \end{aligned}$$
(23)

with initial conditions

$$\begin{aligned} \begin{array}{cc} A_{0}=1, \ A_{1}=1,\ &{}\ B_{0}=0, \ B_{1}=b,\\ C_{0}=0,\ C_{1}=c, \ &{}\ D_{0}=0, \ D_{1}=d. \end{array} \end{aligned}$$

Proof

Let \({\mathcal {W}}^n=[w_{ij}(n)]_{4\times 4}\) for all \(n \in {\mathbb {N}}\). If the matrix \({\mathcal {W}}\) is diagonalizable, then its minimal polynomial \(\psi \) is a product of linear factors for distinct eigenvalues. Since \(\psi ({\mathcal {W}})=0\) and \(\psi ({\mathcal {W}})\) is a linear combination of some powers of \({\mathcal {W}}\), for each pair of indices i, j we obtain a linear equation involving \(w_{ij}(n)\) for different n. Hence each sequence \((w_{ij}(n)), \; n\in {\mathbb {N}}\) can be defined by a linear recurrence relation with characteristic polynomial equal to the minimal polynomial for \({\mathcal {W}}\).

So if we take a transition matrix in formula (7) as \({\mathcal {W}}\), then from (15) in particular we have \(w_{11}(n)=A_{n}, w_{21}(n)=B_{n}, w_{31}(n)=C_{n}, w_{41}(n)=D_{n}\). Therefore (23) follows from b) in Proposition 2. \(\square \)

Remark 5

There is a natural connection between Quaternaccis and Lucas numbers of the first and second kind, i.e. \(U_n(P,Q)\) and \(V_n(P,Q)\), respectively, which are given by a recurrence relation

$$\begin{aligned} {\mathbb {X}}_n(P,Q)=P{\mathbb {X}}_{n-1}(P,Q)-Q{\mathbb {X}}_{n-2}(P,Q),\quad n\geqslant 2 \end{aligned}$$
(24)

with \(P,Q\in {\mathbb {Z}},\ U_0(P,Q)=0,U_1(P,Q)=1\) and \(V_0(P,Q)=2, V_1(P,Q)=P\).

For instance, for all \(n\in {\mathbb {N}}_0\) we have \(B_{n}(1,c,d)=U_n(2,2+c^2+d^2)\) and \(2A_{n}(b,c,d)=V_n(2,1+b^2+c^2+d^2)\).

Corollary 2

Formulae (18) are valid, that is for \(n\geqslant 2,\ bcd\ne 0\) we have

$$\begin{aligned} -\frac{1}{b}\sum _{k=1}^{n-1}B_{k}=-\frac{1}{c}\sum _{k=1}^{n-1}C_{k}=-\frac{1}{d}\sum _{k=1}^{n-1}D_{k}=\frac{A_{n}-1}{b^2+c^2+d^2}=\frac{A_{n}-1}{\lambda ^2} \end{aligned}$$

and for \(b=0\) or \(c=0\) or \(d=0\) the corresponding sum equals 0.

Proof

Let \((x,X)\in \{(b,B),(c,C),(d,D)\}\). If we fix \(n\in {\mathbb {N}}\), then from (23) we get

$$\begin{aligned} X_{n+1}-X_{n}&=X_{n}-X_{n-1}-\lambda ^2X_{n-1}\\ X_{n}-X_{n-1}&=X_{n-1}-X_{n-2}-\lambda ^2X_{n-2}\\&\;\;\vdots \\ X_{2}-X_{1}&=X_{1}-X_{0}-\lambda ^2X_{0} \end{aligned}$$

Adding the above (a telescoping summing) and substituting \(X_0=0,\) \(X_1=x\) we obtain

$$\begin{aligned} X_{n+1}-X_{n}=X_{1}-X_0-\lambda ^2\sum _{k=0}^{n-1}X_{k}=x-\lambda ^2\sum _{k=1}^{n-1}X_{k}. \end{aligned}$$

Now, if we take \(X_{n}=x{\tilde{X}}_{n}\), where

$$\begin{aligned} {\tilde{X}}_{n}=2{\tilde{X}}_{n-1}-(1+\lambda ^2){\tilde{X}}_{n-2},\quad {\tilde{X}}_{0}=0, \quad {\tilde{X}}_{1}=1, \end{aligned}$$

and \(Y_n={\tilde{X}}_{n}-{\tilde{X}}_{n-1}\) for \(n\in {\mathbb {N}}\), then from above we get

$$\begin{aligned} Y_n=2Y_{n-1}-(1+\lambda ^2)Y_{n-2}, \quad Y_1=1, \quad Y_2=1. \end{aligned}$$

Hence by Theorem 3, \(Y_n=A_{n-1}\) and we have

$$\begin{aligned} xA_n=xY_{n+1}=x{\tilde{X}}_{n+1}-x{\tilde{X}}_{n}=X_{n+1}-xX_{n}=x-\lambda ^2\sum _{k=1}^{n-1}X_{k}, \end{aligned}$$

which implies that the sum is 0 if \(x=0\) and for \(x\ne 0\) finally gives

$$\begin{aligned} \frac{A_{n}-1}{\lambda ^2}=-\frac{1}{x}\sum _{k=1}^{n-1}X_{k}. \end{aligned}$$

\(\square \)

A consequence of Theorem 3 are Binet’s formulae for Quaternaccis

Corollary 3

If \(n\in {\mathbb {N}}_0,\ bcd\ne 0\) then Binet’s formulae for Quaternaccis are the following

$$\begin{aligned} A_{n}&=\frac{(1-i\lambda )^n+(1+i\lambda )^n}{(1-i\lambda )+(1+i\lambda )}=\frac{1}{2}\Bigl ((1-i\lambda )^n+(1+i\lambda )^n\Bigr ),\\ \frac{1}{b}B_{n}= \frac{1}{c}C_{n}= \frac{1}{d}D_{n}&=\frac{(1-i\lambda )^n-(1+i\lambda )^n}{(1-\lambda )-(1+i\lambda )}=\frac{i}{2\lambda }\Bigl ((1-i\lambda )^n-(1+i\lambda )^n\Bigr ), \end{aligned}$$

which is equivalent to

$$\begin{aligned} A_{n}= & {} \sum _{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor } {n \atopwithdelims ()2k}(-1)^k\lambda ^{2k}, \end{aligned}$$
(25)
$$\begin{aligned} \frac{1}{b}B_{n}=\frac{1}{c} C_{n}=\frac{1}{d} D_{n}= & {} \sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }{n \atopwithdelims ()2k+1}(-1)^{k+1}\lambda ^{2k} . \end{aligned}$$
(26)

If \(x=0\), where \(x=b,c,d\), then \(X_{n}\equiv 0\) for \(X=B,C,D\), respectively.

Note that the proof of Corollary 2 can be also derived from Binet’s formulae. In particular, from (25) and (26) we can obtain

$$\begin{aligned} A_{n}(F_m,F_{m+1},F_{m+2})= & {} \sum _{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor }\sum _{r=0}^k{n \atopwithdelims ()2k}{k \atopwithdelims ()r}(-1)^{k+r}(3F_{2m+3})^{k-r}F_{m+3}^{2k},\end{aligned}$$
(27)
$$\begin{aligned} \frac{B_{n}(F_m,F_{m+1},F_{m+2})}{F_m}= & {} \frac{C_{n}(F_m,F_{m+1},F_{m+2})}{F_{m+1}}= \frac{D_{n}(F_m,F_{m+1},F_{m+2})}{F_{m+2}}\nonumber \\= & {} \sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\sum _{r=0}^k{n \atopwithdelims ()2k+1}{k \atopwithdelims ()r}(-1)^{k+r+1}(3F_{2m+3})^{k-r}F_{m+3}^{2k},\end{aligned}$$
(28)
$$\begin{aligned} A_{n}(\varphi ,{\overline{\varphi }},0)= & {} \sum _{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor }{n \atopwithdelims ()2k}(-3)^k=(-2)^n \cos \frac{2\pi n }{3}, \end{aligned}$$
(29)
$$\begin{aligned} \frac{1}{\varphi }B_{n}(\varphi ,{\overline{\varphi }},0)= & {} \frac{1}{{\overline{\varphi }}}C_{n}(\varphi ,{\overline{\varphi }},0)\end{aligned}$$
(30)
$$\begin{aligned}= & {} \sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }{n \atopwithdelims ()2k+1}(-3)^k=\frac{(-1)^{n+1}\cdot 2^n}{\sqrt{3}}\sin \frac{2n\pi }{3}, \end{aligned}$$
(31)

where \(\varphi =\frac{1+\sqrt{5}}{2}\) and \({\overline{\varphi }}=\frac{1-\sqrt{5}}{2}\).

Another immediate consequence of Binet’s formulae is the following

Corollary 4

For each \(n\in {\mathbb {N}}_0\) the following cyclic relations for b, c, d holds

$$\begin{aligned} B_{n}(b,c,d)=D_{n}(d,c,b)=C_{n}(c,b,d) =D_{n}(c,d,b)=C_{n}(d,b,c)=B_{n}(b,d,c) \end{aligned}$$

and \((A_{n}(b,c,d))\) is invariant with respect to permutations of b, c, d.

Assume now that \(\alpha ,\beta ,\gamma ,\delta \in {\mathbb {R}},\ \alpha \ne 0\) are such that \(q=\alpha +\beta i+\gamma j+\delta k \in S^3\). Then de Moivre’s fomula (4) and the following application of Definition 2

$$\begin{aligned}&{ q^n=\alpha ^n\left( 1+\frac{\beta }{\alpha } i+\frac{\gamma }{\alpha } j+\frac{\delta }{\alpha }k\right) ^n=\alpha ^n\left( A_{n}\left( \frac{\beta }{\alpha }, \frac{\gamma }{\alpha },\frac{\delta }{\alpha }\right) \right. } \nonumber \\&\quad {\left. +B_{n}\left( \frac{\beta }{\alpha }, \frac{\gamma }{\alpha },\frac{\delta }{\alpha }\right) i+C_{n}\left( \frac{\beta }{\alpha }, \frac{\gamma }{\alpha },\frac{\delta }{\alpha }\right) j+D_{n}\left( \frac{\beta }{\alpha }, \frac{\gamma }{\alpha },\frac{\delta }{\alpha }\right) k\right) ,} \end{aligned}$$
(32)

yields

Corollary 5

Let \(\alpha =\cos \theta \) and \(\beta ,\gamma ,\delta \in {\mathbb {R}}\) be such that \(\alpha \beta \gamma \delta \ne 0\) and \(\alpha ^2+\beta ^2+\gamma ^2+\delta ^2=1\). Then for every \(n\in {\mathbb {N}}\), Quaternaccis and Chebyshev polynomials are related by the following identities

$$\begin{aligned} A_{n}\left( \frac{\beta }{\alpha }, \frac{\gamma }{\alpha },\frac{\delta }{\alpha }\right)= & {} \frac{T_n(\alpha )}{\alpha ^n}, \end{aligned}$$
(33)
$$\begin{aligned} B_{n}\left( \frac{\beta }{\alpha }, \frac{\gamma }{\alpha },\frac{\delta }{\alpha }\right)= & {} \frac{\beta {\tilde{U}}_{n-1}(\alpha )}{\alpha ^n}, \end{aligned}$$
(34)
$$\begin{aligned} C_{n}\left( \frac{\beta }{\alpha }, \frac{\gamma }{\alpha },\frac{\delta }{\alpha }\right)= & {} \frac{\gamma {\tilde{U}}_{n-1}(\alpha )}{\alpha ^n}, \end{aligned}$$
(35)
$$\begin{aligned} D_{n}\left( \frac{\beta }{\alpha }, \frac{\gamma }{\alpha },\frac{\delta }{\alpha }\right)= & {} \frac{\delta {\tilde{U}}_{n-1}(\alpha )}{\alpha ^n}, \end{aligned}$$
(36)

where \({\tilde{U}}_n(\alpha )\) is a scaled Chebyshev polynomial of the second kind, that is

$$\begin{aligned} {\tilde{U}}_n(\alpha )=\frac{\sin (n+1)\theta }{|\sin \theta |}={{\,\mathrm{sgn}\,}}(\sin \theta )U_n(\alpha ). \end{aligned}$$

Finally, Horadam in [12] introduced the number systems called Fibonacci quaternions \(Q_n\) and Lucas quaternions \(K_n\), defined recurrently by

$$\begin{aligned} Q_n= & {} F_n+F_{n+1}i+F_{n+2}j+F_{n+3}k, \end{aligned}$$
(37)
$$\begin{aligned} K_n= & {} L_n+L_{n+1}i+L_{n+2}j+L_{n+3}k, \end{aligned}$$
(38)

where \(n\in {\mathbb {N}}_0\) and \(F_n,L_n\) denote the n-th Fibonacci and Lucas number, respectively (see also [5, 6, 18]).

He also proved equivalents of identities for Fibonacci numbers, like e.g.

$$\begin{aligned} Q_n{\overline{Q}}_n= & {} \sum _{i=0}^3F_{n+i}^2=3F_{2n+3}, \end{aligned}$$
(39)
$$\begin{aligned} Q_n^2= & {} 2F_nQ_n-Q_n{\overline{Q}}_n. \end{aligned}$$
(40)

Note that Quaternaccis and Fibonacci quaternions are related by the following identity, valid for all \(m,n \in {\mathbb {N}}_0\):

$$\begin{aligned} Q_n^m= & {} F_n^m\left( 1+\frac{F_{n+1}}{F_n}i+\frac{F_{n+2}}{F_n}j+\frac{F_{n+3}}{F_n}k \right) ^m \nonumber \\= & {} F_n^m\left[ A_{m}\left( \frac{F_{n+1}}{F_n},\frac{F_{n+2}}{F_n},\frac{F_{n+3}}{F_n}\right) +B_{m}\left( \frac{F_{n+1}}{F_n},\frac{F_{n+2}}{F_n},\frac{F_{n+3}}{F_n}\right) i\right. \nonumber \\&\left. +\,C_{m}\left( \frac{F_{n+1}}{F_n},\frac{F_{n+2}}{F_n},\frac{F_{n+3}}{F_n}\right) j+D_{m}\left( \frac{F_{n+1}}{F_n},\frac{F_{n+2}}{F_n},\frac{F_{n+3}}{F_n}\right) k \right] .\nonumber \\ \end{aligned}$$
(41)

Moreover, from (40) and (41) we obtain

Corollary 6

For all \(m,n\in {\mathbb {N}}_0\) the following identities holds

$$\begin{aligned}&A_{2m}\left( \frac{F_{n+1}}{F_n},\frac{F_{n+2}}{F_n},\frac{F_{n+3}}{F_n}\right) +B_{2m}\left( \frac{F_{n+1}}{F_n},\frac{F_{n+2}}{F_n},\frac{F_{n+3}}{F_n}\right) i\\&\qquad +C_{2m}\left( \frac{F_{n+1}}{F_n},\frac{F_{n+2}}{F_n},\frac{F_{n+3}}{F_n}\right) j+D_{2m}\left( \frac{F_{n+1}}{F_n},\frac{F_{n+2}}{F_n},\frac{F_{n+3}}{F_n}\right) k\\&\quad =\frac{1}{F_n^{2m}}\sum _{r=0}^m {m \atopwithdelims ()r} (2F_nQ_n)^{m-r}(-3F_{2n+3})^r\\&\quad =\sum _{r=0}^m {m \atopwithdelims ()r} 2^{m-r}F_n^{-m-r}(-3F_{2n+3})^r\left( A_{m-r}\left( \frac{F_{n+1}}{F_n},\frac{F_{n+2}}{F_n},\frac{F_{n+3}}{F_n}\right) \right. \\&\qquad \left. +\,B_{m-r}\left( \frac{F_{n+1}}{F_n},\frac{F_{n+2}}{F_n}, \frac{F_{n+3}}{F_n}\right) i+C_{m-r}\left( \frac{F_{n+1}}{F_n}, \frac{F_{n+2}}{F_n},\frac{F_{n+3}}{F_n}\right) j\right. \\&\qquad +\,\left. D_{m-r}\left( \frac{F_{n+1}}{F_n},\frac{F_{n+2}}{F_n},\frac{F_{n+3}}{F_n}\right) k\right) . \end{aligned}$$

3 Reduction Formulae

By the properties of exponentiation and from Proposition 1 we obtain the basic reduction formulae for Quaternaccis, that is

Theorem 4

Let \(X\in \{B,C,D\},\ m,n\in {\mathbb {N}}_0\). Then

$$\begin{aligned} A_{n+m}&=A_{n}A_{m}-B_{n}B_{m} -C_{n}C_{m}-D_{n}D_{m},\\ X_{n+m}&=A_{n}X_{m}+A_{m}X_{n}. \end{aligned}$$

And the special cases are the following

Corollary 7

Let \(X\in \{B,C,D\},\ n\in {\mathbb {N}}_0\). Then

a)

$$\begin{aligned} A_{2n}&=A_{n}^2-B_{n}^2-C_{n}^2-D_{n}^2=\Delta ^nA_{n}\left( \frac{2b}{\Delta },\frac{2c}{\Delta },\frac{2d}{\Delta }\right) ,\\ X_{2n}&=2 A_{n}X_{n}=\Delta ^nX_{n}\left( \frac{2b}{\Delta },\frac{2c}{\Delta },\frac{2d}{\Delta }\right) , \end{aligned}$$

where \(\Delta =1-\lambda ^2\),

b)

$$\begin{aligned} A_{3n}&=A_{n}^3-3A_{n}B_{n}^2-3A_{n}C_{n}^2-3A_{n}D_{n}^2=\bigl (\Delta '\bigr )^nA_{n}({\tilde{b}},{\tilde{c}},{\tilde{d}}),\\ X_{3n}&=X_{n}\bigl (3A_{n}^2-B_{n}^2-C_{n}^2-D_{n}^2\bigr )=\bigl (\Delta '\bigr )^nX_{n}({\tilde{b}},{\tilde{c}},{\tilde{d}}), \end{aligned}$$

where

$$\begin{aligned} \begin{array}{c} \displaystyle {{\tilde{b}}=\frac{3b-b^3-bc^2-bd^2}{\Delta '}, \quad {\tilde{c}}=\frac{3c-c^3-b^2c-cd^2}{\Delta '},}\\ \displaystyle {{\tilde{d}}=\frac{3d-d^3-b^2d-c^2d}{\Delta '}\quad \text {and} \quad \Delta '=1-3\lambda ^2}. \end{array} \end{aligned}$$

Quaternaccis can be used for obtaining decompositions of polynomials.

Theorem 5

a) For all \(n\in {\mathbb {N}}_0\) the following identity holds

$$\begin{aligned}&\bigl ({\mathbb {X}}-(1+bi)^n\bigr )\bigl ({\mathbb {X}}-(1+bj)^n\bigr ) \bigl ({\mathbb {X}}-(1+bk)^n\bigr )= {\mathbb {X}}^3-{\mathbb {X}}^2\Bigl (3A_{b}+iB_{b}+jB_{b} +kB_{b}\Bigr )\\&\quad +\,{\mathbb {X}}\Bigl [3A_{b}^2+i B_{b}\Bigl (2A_{b}+B_{b}\Bigr )+jB_{b}\Bigl (2A_{b}-B_{b}\Bigr ) +kB_{b}\Bigl (2A_{b}+B_{b}\Bigr )\Bigr ]\\&\quad -\,\Bigl [A_{b}^3-B_{b}^3+iA_{b}B_{b}\Big (A_{b}-B_{b}\Bigr )+ jA_{b}B_{b}\Big (A_{b}-B_{b}\Bigr )+kA_{b}B_{b}\Big (A_{b}+B_{b}\Bigr )\bigr ], \end{aligned}$$

where \(X_b=X_n(b,0,0)\) for \(X\in \{A,B,C,D\}\).

b) Let \(\alpha _n=(1+bi+cj)^n,\ \beta _n=(1+bj+ck)^n,\ \gamma _n=(1+bi+ck)^n,\ n\in {\mathbb {N}}_0 \). Then Vieta’s formulae for \(({\mathbb {X}}-\alpha _n)({\mathbb {X}}-\beta _n)({\mathbb {X}}-\gamma _n)\) are the following

$$\begin{aligned}&\alpha _n+\beta _n+\gamma _n=3A_{bc}+2iB_{bc}+j\bigl (B_{bc}+C_{bc}\bigr )+2kC_{bc},\\&\quad \alpha _n\beta _n+\alpha _n\gamma _n+\beta _n\gamma _n= 2A_{bc}^2+A_{2bc}-B_{bc}C_{bc} +i\Bigl (4A_{bc}B_{bc}+B_{bc}C_{bc}+2C_{bc}^2\Bigr )\\&\quad +j\Bigl (2A_{bc}B_{bc}-B_{bc}C_{bc}+4A_{bc}C_{bc}\Bigr )+kC_{bc}\Bigl (4A_{bc}-B_{bc}\Bigr ), \end{aligned}$$

andFootnote 4

$$\begin{aligned} \alpha _n \beta _n \gamma _n= (A_{bc}+B_{bc})(A_{bc}-B_{bc}-C_{bc})(A_{bc}+C_{bc})+i(2A_{bc}^2B_{bc}+\cdots \end{aligned}$$

where \(X_{bc}=X_n(b,c,0)\) for \(X\in \{A,B,C,D\}\).

Proof

We will prove the case a), b) is analogous. After multiplying and rearranging the left side we have

$$\begin{aligned}&\bigl ({\mathbb {X}}-(1+bi)^n\bigr )\bigl ({\mathbb {X}}-(1+bj)^n\bigr )\bigl ({\mathbb {X}}-(1+bk)^n\bigr ) ={\mathbb {X}}^3-{\mathbb {X}}^2\bigl ((1+bi)^n+(1+bj)^n\\&\quad +\,(1+bk)^n\bigr )+ {\mathbb {X}}\bigl ((1+bi)^n(1+bj)^n+(1+bi)^n(1+bk)^n +(1+bj)^n(1+bk)^n\bigr )\\&\quad +\,(1+bi)^n(1+bj)^n(1+bk)^n. \end{aligned}$$

Then the statement follows from Definition 2 and Corollary 4. \(\square \)

From Binet’s formulae we obtain further reduction formulae.

Theorem 6

For \(X \in \{B,C,D\}\) and \(n\in {\mathbb {N}}_0\) the following identities hold

a) for any real a:

$$\begin{aligned} A_{n}(a,a,a)&=A_{n}(\sqrt{3} a,0,0),\\ X_{n}(a,a,a)&=\frac{1}{\sqrt{3}}X_{n}(\sqrt{3} a,0,0)=\frac{1}{\sqrt{3}}B_{n}(\sqrt{3} a,0,0), \end{aligned}$$

b) for \(bcd \ne 0\) such that \(b\ne c\):

$$\begin{aligned} A_{n}(c-b,c+b,d)&=A_{n}(\sqrt{2}c,\sqrt{2}b,d),\\ B_{n}(c-b,c+b,d)&=\frac{c-b}{\sqrt{2} c}B_{n}(\sqrt{2}c,\sqrt{2}b,d),\\ C_{n}(c-b,c+b,d))&=\frac{c+b}{\sqrt{2} c}C_{n}(\sqrt{2}c,\sqrt{2}b,d),\\ D_{n}(c-b,c+b,d))&=D_{n}(\sqrt{2}c,\sqrt{2}b,d), \end{aligned}$$

c) for \(bcd\ne 0\) such that \(b^2+c^2=d^2\):

$$\begin{aligned} A_{n}&=A_{n}(\sqrt{2}d,0,0)=A_{n}(d,d,0),\\ B_{n}&=\frac{b}{\sqrt{2}d}B_{n}(\sqrt{2}d,0,0)=\frac{b}{d}B_{n}(d,d,0),\\ C_{n}&=\frac{c}{\sqrt{2}d}C_{n}(0,\sqrt{2}d,0)=\frac{c}{d}C_{n}(d,d,0),\\ D_{n}&=\frac{1}{\sqrt{2}}D_{n}(0,0,\sqrt{2}d)=D_{n}(0,d,d), \end{aligned}$$

Some index-parameter reduction formulae were also obtained.

Theorem 7

For \(X\in \{A,B,C,D\}\) and \(m,n \in {\mathbb {N}}\) we have

$$\begin{aligned} \displaystyle {X_{m\cdot n}=A^n_{m} X_{n}\left( \frac{B_{m}}{A_{m}},\frac{C_{m}}{A_{m}},\frac{D_{m}}{A_{m}}\right) }. \end{aligned}$$

Proof

Definition of Quaternaccis and properties of exponentiation yield

$$\begin{aligned}&{A_{m\cdot n}+B_{m\cdot n}i+C_{m\cdot n}j+D_{m\cdot n}k=(1+bi+cj+dk)^{m\cdot n}}\nonumber \\&\quad {=\left( A_{m}+B_{m}i+C_{m}j+D_{m}k\right) ^n =A^n_{m}\left( 1+\frac{B_{m}}{A_{m}}i+\frac{C_{m}}{A_{m}}j+\frac{D_{m}}{A_{m}}k\right) ^n}\nonumber \\&\quad {=A^n_{m}\left[ A_{n}\left( \frac{B_{m}}{A_{m}},\frac{C_{m}}{A_{m}},\frac{D_{m}}{A_{m}}\right) \right. +\left. B_{n}\left( \frac{B_{m}}{A_{m}},\frac{C_{m}}{A_{m}},\frac{D_{m}}{A_{m}}\right) i\right. }\nonumber \\&\qquad {+\,\left. C_{n}\left( \frac{B_{m}}{A_{m}},\frac{C_{m}}{A_{m}},\frac{D_{m}}{A_{m}}\right) j\right. + \left. D_{n}\left( \frac{B_{m}}{A_{m}},\frac{C_{m}}{A_{m}},\frac{D_{m}}{A_{m}}\right) k\right] .} \end{aligned}$$
(42)

Comparison of coefficients of 1, i, j, k finishes the proof. \(\square \)

Example

From (25), that is Binet’s formula for \(A_{n}\), we get for instance

$$\begin{aligned} A_{4}(1,2,0)={4\atopwithdelims ()0}-5{4\atopwithdelims ()2}+25{4\atopwithdelims ()4}=-4 \end{aligned}$$

and (26) gives \(B_{4}(1,2,0)=-16, C_{4}(1,2,0)=-32, D_{4}(1,2,0)=0.\) Thus from Theorem 7 we obtain

$$\begin{aligned}&A_{4n}(1,2,0)=A_{4}^n(1,2,0)A_{n} \left( \frac{B_{4}(1,2,0)}{A_{4}(1,2,0)}, \frac{C_{4}(1,2,0)}{A_{4}(1,2,0)},\frac{D_{4}(1,2,0)}{A_{4}(1,2,0)}\right) \\&\quad =(-4)^{n}A_{n}(4,8,0). \end{aligned}$$

Similarly we have

$$\begin{aligned}&\Bigl (A_{4}(-1,1,2)=1 \; \wedge \;B_{4}(-1,1,2)=20\;\wedge \; C_{4}(-1,1,2)=-20 \\&\quad \wedge \; D_{4}(-1,1,2)=-40\Bigr )\Longrightarrow A_{4n}(-1,1,2)=A_{n}(20,-20,-40),\\&\Bigl (A_{5}(-1,-1,0)=1\; \wedge \;B_{5}(-1,-1,0)=11\;\wedge \; C_{5}(-1,-1,0)=11 \\&\quad \wedge \; D_{5}(-1,-1,0)=0\Bigr )\Longrightarrow A_{5n}(-1,-1,0)=A_{n}(11,11,0). \end{aligned}$$

An interesting trigonometric case follows from Theorem 7 and Corollary 5:

Corollary 8

Let \((x,X)\in \{(\beta , B),(\gamma , C),(\delta ,D)\}\). If \(\alpha ^2+\beta ^2+\gamma ^2+\delta ^2=1\) and \( \beta \gamma \delta \ne 0\) then for every \(m,n\in {\mathbb {N}}\) we have

$$\begin{aligned} A_{n}\left( \frac{\beta \tan (m\theta )}{\sqrt{1-\alpha ^2}},\frac{\gamma \tan (m\theta )}{\sqrt{1-\alpha ^2}},\frac{\delta \tan (m\theta )}{\sqrt{1-\alpha ^2}}\right)&=\frac{\cos (mn\theta )}{\cos ^n (m\theta )},\\ xX_{n}\left( \frac{\beta \tan (m\theta )}{\sqrt{1-\alpha ^2}},\frac{\gamma \tan (m\theta )}{\sqrt{1-\alpha ^2}},\frac{\delta \tan (m\theta )}{\sqrt{1-\alpha ^2}}\right)&=\frac{\sqrt{1-\alpha ^2}\cos ^n(m\theta )}{ \sin (mn\theta )}. \end{aligned}$$

Remark 6

Formulae in Theorem 7 significantly simplify the evaluation of Quaternaccis for high indices, e.g. for \(X\in \{A,B,C,D\},\ n\in {\mathbb {N}}\) we have

$$\begin{aligned}&{X_{n^3}}{=A^{n^2}_{n}A^n_{n}\left( \frac{B_{n}}{A_{n}},\frac{C_{n}}{A_{n}},\frac{D_{n}}{A_{n}}\right) }\\&\qquad {\times \, X_{n}\left( \frac{B_{n}\left( \frac{B_{n}}{A_{n}},\frac{C_{n}}{A_{n}},\frac{D_{n}}{A_{n}}\right) }{A_{n}\left( \frac{B_{n}}{A_{n}},\frac{C_{n}}{A_{n}},\frac{D_{n}}{A_{n}}\right) }, \frac{C_{n}\left( \frac{B_{n}}{A_{n}},\frac{C_{n}}{A_{n}},\frac{D_{n}}{A_{n}}\right) }{A_{n}\left( \frac{B_{n}}{A_{n}},\frac{C_{n}}{A_{n}},\frac{D_{n}}{A_{n}}\right) }, \frac{D_{n}\left( \frac{B_{n}}{A_{n}},\frac{C_{n}}{A_{n}},\frac{D_{n}}{A_{n}}\right) }{A_{n}\left( \frac{B_{n}}{A_{n}},\frac{C_{n}}{A_{n}},\frac{D_{n}}{A_{n}}\right) }\right) .} \end{aligned}$$

That means that if we want to evaluate Quaternaccis at some point (b, c, d) for index 1000, say, it is enough to determine their values for index 10, although the calculations has to be done at some other points (specified above).

4 Applications

1. Professor Webb in 2008, during the 13th International Conference on Fibonacci Numbers and Their Applications in Patras, posed a question of finding compact formulae for sums of the form

$$\begin{aligned} \sum _{k=1}^N F_{k^r}, \end{aligned}$$

where \(r\in {\mathbb {N}}\). Thanks to properties of \(\delta \)-Fibonacci numbers (recalled in Sect. 1) this problem was reduced to finding sums of powers of Fibonacci numbers with indices not greater than N. More precisely, by reduction formulae for \(\delta \)-Fibonacci numbers (similar to those in Theorem 7) we can get (see [23])

$$\begin{aligned}&F_{mn}=F_m^nb_n\left( \frac{F_m}{F_{m+1}}\right) =\sum _{k=0}^n{n \atopwithdelims ()k}(-1)^{k-1}F_{k}F_m^kF_{m+1}^{n-k}, \end{aligned}$$
(43)
$$\begin{aligned}&F_{mnp}=F_m^{np}a_n^p\left( \frac{F_m}{F_{m+1}}\right) b_p\left( \frac{b_n\left( \frac{F_m}{F_{m+1}}\right) }{a_n\left( \frac{F_m}{F_{m+1}}\right) }\right) , \end{aligned}$$
(44)

whence for instance

$$\begin{aligned} F_{n^3}=F_n^{n^2}a_{n}^n\left( \frac{F_{n}}{F_{n+1}}\right) b_n\left( \frac{b_n\left( \frac{F_{n}}{F_{n+1}}\right) }{a_n\left( \frac{F_{n}}{F_{n+1}}\right) }\right) . \end{aligned}$$

Similarly, reduction formulae given in Theorem 7 can be used for solving analogous problems in which Quaternaccis can be involved. For instance, high powers of Fibonacci quaternions \(Q_n\) can be replaced by high powers of Quaternaccis (due to formula (41)) which in turn can be lowered significantly by Theorem 7 and Corollary 6, which can be lowered even further by (43) and (44).

For instance, for \(s,r,n,N,M\in {\mathbb {N}},\ M\leqslant N\) we have

$$\begin{aligned}&\sum _{\tau =M}^N Q_{n^s}^{2r\tau }=\sum _{\tau =M}^NF_{n^s}^{2r\tau }A_{2\tau ,F}^r\nonumber \\&\quad \times \Biggl [A_{r}\left( \frac{B_{2\tau ,F}}{A_{2\tau ,F}},\frac{C_{2\tau ,F}}{A_{2\tau ,F}},\frac{D_{2\tau ,F}}{A_{2\tau ,F}}\right) +B_{r}\left( \frac{B_{2\tau ,F}}{A_{2\tau ,F}},\frac{C_{2\tau ,F}}{A_{2\tau ,F}},\frac{D_{2\tau ,F}}{A_{2\tau ,F}}\right) i \Biggl .\nonumber \\&\quad +\Biggl .C_{r}\left( \frac{B_{2\tau ,F}}{A_{2\tau ,F}},\frac{C_{2\tau ,F}}{A_{2\tau ,F}},\frac{D_{2\tau ,F}}{A_{2\tau ,F}}\right) j +D_{r}\left( \frac{B_{2\tau ,F}}{A_{2\tau ,F}},\frac{C_{2\tau ,F}}{A_{2\tau ,F}},\frac{D_{2\tau ,F}}{A_{2\tau ,F}}\right) k\Biggr ]\nonumber \\&\quad {\mathop {=}\limits ^{\text {Cor.}6}}\sum _{\tau =M}^N\left( \sum _{\omega =0}^\tau {\tau \atopwithdelims ()\omega }2^{\tau -\omega }F_{n^s}^{-\tau -\omega }\bigl (-3F_{2{n^s}+3}\bigr )^\omega A_{\tau -\omega ,F}\right) ^r \nonumber \\&\quad \times \Biggl [A_{r}\left( \frac{B_{2\tau ,F}}{A_{2\tau ,F}},\frac{C_{2\tau ,F}}{A_{2\tau ,F}},\frac{D_{2\tau ,F}}{A_{2\tau ,F}}\right) +B_{r}\left( \frac{B_{2\tau ,F}}{A_{2\tau ,F}},\frac{C_{2\tau ,F}}{A_{2\tau ,F}},\frac{D_{2\tau ,F}}{A_{2\tau ,F}}\right) i \Biggr .\nonumber \\&\quad +\Biggl .C_{r}\left( \frac{B_{2\tau ,F}}{A_{2\tau ,F}},\frac{C_{2\tau ,F}}{A_{2\tau ,F}},\frac{D_{2\tau ,F}}{A_{2\tau ,F}}\right) j +D_{r}\left( \frac{B_{2\tau ,F}}{A_{2\tau ,F}},\frac{C_{2\tau ,F}}{A_{2\tau ,F}},\frac{D_{2\tau ,F}}{A_{2\tau ,F}}\right) k\Biggr ],\qquad \end{aligned}$$
(45)

where \(X_{n,F}=X_{n}\left( \frac{F_{{n^s}+1}}{F_{n^s}},\frac{F_{{n^s}+2}}{F_{n^s}},\frac{F_{{n^s}+3}}{F_{n^s}}\right) \) for \(X\in \{A,B,C,D\}\) and indices of Fibonacci numbers \(F_{n^s}\) can be reduced in a way mentioned above.

Remark 7

Since every nonconstant quaternion can be written in a polar form (see (5)), then sums of the form \({\sum _{k=M}^NQ_t^{2m}}\) can be evaluated by determining some trigonometric sums of the form \({\sum _{l=1}^t \alpha ^l \cos l\phi }\) and \({\sum _{l=1}^t \alpha ^l \sin l\phi }\), where \(\alpha ,\phi \in {\mathbb {R}}\). Formulae for these sums are known (see e.g. [20, p.13]). Note however, that our approach, like in (45), allows to reduce arguments of cosine, sine and powers of \(\alpha \), unlike the other known formulae.

2. Quaternaccis can be used also to determine some trigonometric sums. What is interesting in this approach is that we do not use known trigonometric identities but we use only those we obtained from properties of exponentiation of quaternions, i.e. (33)–(36).

Theorem 8

Set \(\alpha =\cos \theta \), where \(\theta \in {\mathbb {R}}{\setminus } \{\frac{k\pi }{2},\ k\in {\mathbb {Z}}\}\). Then for \(N\in {\mathbb {N}}\)

(46)
(47)
(48)

Proof

We will prove case a), the remaining cases are proven analogously.

Since by assumption \(\alpha ^2< 1\) and \(\alpha \ne 0\), there exist \(\beta ,\gamma ,\delta \in {\mathbb {R}},\ \beta \ne 0\) such that \(\alpha ^2+\beta ^2+\gamma ^2+\delta ^2=1\) and \(q=\left( 1+\frac{\beta }{\alpha }i+\frac{\gamma }{\alpha }j+\frac{\delta }{\alpha }k\right) \in {\mathbb {H}}\). Let \({{\,\mathrm{Im}\,}}_i(q)\) denote the imaginary part of q corresponding to the imaginary unit i and—in this proof only—we denote

$$\begin{aligned} \displaystyle {X_{n}=X_{n}\left( \frac{\beta }{\alpha },\frac{\gamma }{\alpha },\frac{\delta }{\alpha }\right) },\quad X\in \{A,B,C,D\}. \end{aligned}$$

Then

Now, from (33) and (34) we finally get (46). \(\square \)

Remark 8

In view of (5), \(\alpha , \beta ,\gamma ,\delta \) can be any real numbers such that \(\alpha \beta \gamma \delta \ne 0\) and the above statements remain true if \(\displaystyle {\cos \theta =\frac{\alpha }{|q|}},\) \(\displaystyle {\sin \theta =\frac{\sqrt{\beta ^2+\gamma ^2+\delta ^2}}{|q|}}\).

5 OEIS

Note that some Quaternacci sequences can be found in OEIS [25], e.g.:

  • \(A_{n}(1,0,0)=A146559(n)\) is a difference counterpart of cosine (see [16]);

  • For \(\varphi =\arccos \left( \frac{1}{3}\right) \) we have \(\cos (n \varphi )=\frac{A_{n}(2,2,0)}{3^n}=\frac{A025172(n)}{3^n}\). The sequence appears in the negative answer to the third Hilbert Problem (see [1]);

  • \(B_{n}(1,0,1)=D_{n} (1,0,1)=A088137(n),\ B_{n}(1,1,1)=C_{n} (1,1,1)=D_{n}(1,1,1) = A_{n-1}(1,-1,1)+B_{n-1}(1,-1,1)+C_{n-1}(1,-1,1)+D_{n-1}(1,-1,1) =A088138(n)\) are Generalized Gaussian Fibonacci integers;

  • \(A_{n}(1,-1,-1)+B_{n}(1,-1, -1)+C_{n}(1,-1,-1) +D_{n}(1,-1,-1)=A_{n}(1,1,1)- C_{n}(1,1,1)=A120580(n)\) is the Hankel transform of the sequence of sums \({\sum _{k=0}^n {2k \atopwithdelims ()k}}\);

  • \(|A_{n}(0,0,-1)|=|A009116(n)|\) and members of A009116(n) are coefficients in the expansion of \(\frac{\cos x}{e^x}\);

  • We have \(2^nA_{n}\left( \frac{1}{2},\frac{1}{2},\frac{1}{2}\right) =A213421(n)=\text {Re}\bigl ((2+i+j+k)^n\bigr )\) and also \(2^n\left( A_{n}\left( 0,\frac{1}{2},\frac{1}{2}\right) +C_{n}\left( 0,\frac{1}{2},\frac{1}{2}\right) +D_{n}\left( 0,\frac{1}{2},\frac{1}{2}\right) \right) =A190965(n+1) ={{\,\mathrm{Im}\,}}_j\bigl ((2+j+k)^{n +1}\bigr )\), where \({{\,\mathrm{Im}\,}}_j\) denotes the imaginary part of a quaternion corresponding to the imaginary unit j. Moreover, \(2^nA_{n}\left( 0,\frac{1}{2},\frac{1}{2}\right) = A266046(n)=\text {Re}\bigl ((2+j+k)^n\bigr )\). Also, in the description of the later sequence we have found a remark by Stanislav Sykora in which he suggests to consider the coefficients of 1, i, j, k in powers of a quaternion as recurrent sequences of coefficients of 1, i, j, k in that quaternion;

  • \(\displaystyle {\frac{B_{n}(1,0,1)}{A_{n}(1,0, 1)}=\frac{D_{n}(1,0,1)}{A_{n}(1,0, 1)}=\frac{A088137(n)}{A087455(n)}=\underbrace{1 \circ 1\circ \cdots \circ 1}_{n \text { times}}},\) where \(a\circ b=\displaystyle {\frac{a+b}{1-2ab}}\).

In OEIS there is a number of Quaternacci sequences connected with Lucas sequences of the first and second kind, e.g.:

  • \(A_{n}(1,0,1)=A_{n}(0,1, -1)=A_{n}(1,0,-1)=A087455(n) =\frac{1}{2}(-1)^n V_n(-2,3) \);

  • \(A_{n}(2,0,0)=A006495(n)=\frac{(-1)^n}{2} V_n(-2,5)\);

  • \(B_{n}(1,1,1)=C_{n} (1,1,1)=D_{n} (1,1,1)=A088138(n)=(-1)^{n+1}U_{n}(-2,4) =U_{n}(2,4)\);

  • \(B_{n}(1,0,1)=D_{n}(1,0,1)=A088137(n)=U_n(2,3)\);

  • \(2^n A_{n}\left( \frac{1}{2},\frac{1}{2},\frac{1}{2}\right) =A213421(n)=V_n(4,7)\);

  • \(2^nA_{n}\left( \frac{1}{2},0,0\right) =A139011(n)=V_n(4,5)\);

  • \(2^nB_{n}\left( \frac{1}{2},0,0\right) =A099456(n)=U_n(4,5)\);

  • From (23) we get

    $$\begin{aligned} 2^nB_{n}\left( \frac{1}{2},\frac{1}{2},\frac{1}{2}\right) =2^nC_{n}\left( \frac{1}{2},\frac{1}{2},\frac{1}{2}\right) =2^nD_{n}\left( \frac{1}{2},\frac{1}{2},\frac{1}{2}\right) =U_n(4,7). \end{aligned}$$

    Hence from Theorem 7 we obtain the following reduction formulae

    $$\begin{aligned} V_{n^2}(4,7)=V_n^n(4,7)A_{n}\left( \frac{U_n(4,7)}{V_n(4,7)},\frac{U_n(4,7)}{V_n(4,7)},\frac{U_n(4,7)}{V_n(4,7)}\right) . \end{aligned}$$

    A similar argument can be used for \(A_{n}(1,1,1)=A138230(n)={\mathcal {U}}_n\), where \({\mathcal {U}}_n=2{\mathcal {U}}_{n-1}-4{\mathcal {U}}_{n-2}\), \({\mathcal {U}}_0={\mathcal {U}}_1=1\), which gives

    $$\begin{aligned} {\mathcal {U}}_{n^2}={\mathcal {U}}_{n}^nA_{n}\left( \frac{(-1)^{n+1}U_n(2,4)}{{\mathcal {U}}_n},\frac{(-1)^{n+1}U_n(2,4)}{{\mathcal {U}}_n},\frac{(-1)^{n+1}U_n(2,4)}{{\mathcal {U}}_n}\right) . \end{aligned}$$

We have also found some other Quaternaccis (we omit their descriptions):

  • \(A_{n}(1,2,0)=A_{n}(2,1,0)=A_{n}(2,0,1)=A138229(n);\)

  • \(A_{n}(1,1,1)=A138230(n);\)

  • \(A_{n}(0,0,-1)=A146559(n);\)

  • \(B_{n}(1,2,0)=A088139(n);\)

  • \(A_{n}(1,1,1)+B_{n}(1,1,1)+C_{n}(1,1,1)+D_{n}(1,1,1)=A137717(n);\)

  • \(3^n A_{n}\left( \frac{2}{3},0,0\right) =A121622(n);\)

  • \(3^nB_{n}\left( \frac{1}{3},0,0\right) =A106392(n+1);\)

  • \(2^nA_{n}\left( \frac{1}{2},1,1\right) =2^nA_{n}\left( \frac{3}{2},0,0\right) =A121621(n);\)

  • \(2^nB_{n}\left( \frac{1}{2},1,\frac{1}{2}\right) =A190968(n);\)

  • \(2^nB_{n}\left( \frac{1}{2},1,0 \right) =A190967(n);\)

  • \(2^n\left( A_{n}\left( \frac{1}{2},1,0\right) +B_{n}\left( \frac{1}{2},1,0\right) -C_{n}\left( \frac{1}{2},1,0\right) \right) =A176333(n).\)

Note that many Quaternacci sequences for small integer values of b, c, d does not appear in OEIS in any form, e.g. \(B_{n}(2,1,0),C_{n}(2,1,0),B_{n}(2,2,0),\) \(C_{n}(2,2,0),\ A_{n}(1,-1,1),\ B_{n}(1,-1,1),\ C_{n}(1,-1,1),\ D_{n}(1,-1,1),A_{n}(1,0,3),\) \(B_{n}(1,0,3),\ D_{n}(1,0,3),\ A_{n}(-1,3,1),\ B_{n}(-1,3,1),\ C_{n}(-1,3,1),\ A_{n}(3,1,0),\) \(B_{n}(3,1,0),C_{n}(3,1,0), A_{n}(3,0,0)\).