¿^creative ^commor
ARS MATHEMATICA CONTEMPORANEA
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 19 (2020) 25-35
https://doi.org/10.26493/1855-3974.2028.cb4
(Also available at http://amc-journal.eu)
Results on the domination number and the total domination number of Lucas cubes*
Zülfükar Saygi ©
Department of Mathematics, TOBB University of Economics and Technology, Turkey Received 27 June 2019, accepted 2 Jüly 2020, published online 28 October 2020
Abstract
Lucas cubes are special subgraphs of Fibonacci cubes. For small dimensions, their domination numbers are obtained by direct search or integer linear programming. For larger dimensions some bounds on these numbers are given. In this work, we present the exact values of total domination number of small dimensional Lucas cubes and present optimization problems obtained from the degree information of Lucas cubes, whose solutions give better lower bounds on the domination numbers and total domination numbers of Lucas cubes.
Keywords: Lucas cube, Fibonacci cube, domination number, total domination number, integer linear programming.
Math. Subj. Class. (2020): 05C69, 68R10, 11B39
1 Introduction
Fibonacci cubes and Lucas cubes are special subgraphs of the hypercube graph, which are introduced as an alternative model for interconnection networks [6, 11]. The structural and enumerative properties of these graphs are considered from various point of view in the literature [2, 6, 7, 8, 9, 10, 11, 15].
Let Qn denote the hypercube of dimension n > 1. It is the graph with vertex set represented by all binary strings of length n and two vertices in Qn are adjacent if they differ in one coordinate. For convenience Q0 = K1. Fibonacci strings of length n are defined as the binary strings ^b2.. .bn such that bi •bi+1 = 0 for all i = 0,1,... ,n-1, that is, binary strings of length n not containing two consecutive 1s. Using this representation n dimensional Fibonacci cube rn is defined as the subgraph of Qn induced by the vertices
* The author is grateful to the anonymous reviewers for their valuable comments and careful reading and would like to thank Omer Egecioglu and Elif Saygi for useful discussions and suggestions. E-mail address: zsaygi@etu.edu.tr (Zulfukar Saygi)
©® This work is licensed under https://creativecommons.Org/licenses/by/4.0/
18
Ars Math. Contemp. 19 (2020) 24-23
whose string representations are Fibonacci strings. Lucas strings of length n are defined as the Fibonacci strings b^2... bn such that 6i • bn = 0. Similar to the Fibonacci cubes n dimensional Lucas cube An is defined as the subgraph of rn induced by the vertices whose string representations are Lucas strings.
Let G = (V, E) be a graph with vertex set V and edge set E. D ç V is called a dominating set of G if every vertex in V either belongs to D or is adjacent to some vertex in D. Then the domination number y(G) of G is defined as the minimum cardinality of a dominating set of the graph G. Similarly, D ç V is called a total dominating set of a graph G without isolated vertex if every vertex in V is adjacent to some vertex in D and the total domination number Yt(G) of G is defined as the minimum cardinality of a total dominating set of G. The domination numbers of rn and An are first considered in [2, 12]. Using integer linear programming, domination numbers of rn and An are considered in [7] and total domination number of rn is considered in [1]. Furthermore, upper bounds and lower bounds on Y(rn), Yt(rn), y(A„) are obtained in [1, 2, 13] and they are improved for r„ in [14].
In this work, we present optimization problems obtained from the degree information of Lucas cubes, whose solutions give better lower bounds on the domination numbers and total domination numbers of Lucas cubes. Our aim is to improve the known results on y(A„) and present new results on Yt (An ). Furthermore, we introduce the up-down degree polynomials for An containing the degree information of all vertices V(An) in more detail. Using these polynomials we define optimization problems whose solutions give lower bound on y(A„) and Yt(A„).
2 Preliminaries
For n > 2 we will use the fundamental decomposition of rn (see, [8]):
r„ = or„_i + ior„_2,	(2.1)
where r0 = Q0 and r1 = Q1. Here note that orn-1 is the subgraph of rn induced by the vertices that start with 0 and rn-2 is the subgraph of rn induced by the vertices that start with 10. Furthermore, 0rn-1 has a subgraph isomorphic to 00rn-2, and there exists a perfect matching between 00rn-2 and 10rn-2. Similar to this decomposition for n > 3 Lucas cubes can be written as
a„ = 0r„_1 + i0r„_30,	(2.2)
where 10rn-30 is the subgraph of An induced by the vertices that start with 10 and end with 0. Here, there exists a perfect matching between 10rn-30 and 00rn-30 c 0rn-1. By convention, A1 = r0 and A2 = r2.
The number of vertices of the rn is /n+2, where fn are the Fibonacci numbers defined as /0 =0, /1 = 1 and /n = /n-1 + /n-2 for n > 2. Similarly, the number of vertices of the An is Ln, where Ln are the Lucas numbers defined as L0 = 2, L1 = 1 and Ln = Ln-1 + Ln-2 for n > 2.
Let x, y be two binary strings of length n. Then the Hamming distance between x and y, dH (x, y) is the number of coordinates in which they differ. The Hamming weight of x, w(x) is the number of nonzero coordinates in x. Note that Hamming distance is the usual graph distance in Qn.
In Figure 1 we present small dimensional Lucas cubes and a minimal total dominating set with circled vertices for 2 < n < 5.
Z. Saygi: Results on the domination number and the total domination number of Lucas cubes 27
100
010101 01000 01001
Figure 1: Lucas cubes and their minimal total dominating sets for 2 < n < 5.
3 Integer linear programming for domination numbers
In this section, we describe a linear programming problem used in [7] for finding the domination number of rn and An. A similar approach is used in [1] for finding the total domination number of rn. The main difficulty for these methods are the number of variables and the number of constraints which are equal to the number of vertices in rn and An. Using this approach we obtain the total domination number of An for n < 12.
Let N(v) denote the set of vertices adjacent to v and N[v] = N(v) u {v}. Suppose each vertex v e V(An) is associated with a binary variable xv. The problems of determining Y(An) and 71 (An) can be expressed as a problem of minimizing the objective function
^ xv	(3.1)
veV (An )
subject to the following constraints for every v e V(An):
y^ xa > 1 (for domination number),
aeN [v]
xa > 1 (for total domination number).
aeN (v)
The value of the objective function gives 7(An) and Yt(An) respectively. Note that this problem has Ln variables and Ln constraints. In [7] 7(An) is obtained up to n = 11 and for larger values of n as the number of variables increases no results are presented.
We implemented the integer linear programming problem (3.1) using CPLEX in NEOS Server [3, 4, 5] for n < 12 and obtain the values of 7t(An) for n < 12 and obtain the estimates 49 < 7(A12) < 54 (takes approximately 2 hours). We collect the known values of 7(An) for n < 11 (see, [7]) and the new values of 7t(An) that we obtained from (3.1) for n < 12 in Table 1.
The fundamental decompositions (2.1) and (2.2) of rn and An are used to obtain the following relations between 7(An) and 7(rn). The main idea in the proof is to partition the set of vertices into disjoint subsets.
18
Ars Math. Contemp. 19 (2020) 26-23
Table 1: Values of y(A„) and Yt(A„) for n < 12.
n	2	3	4	5	6	7	8	9	10	11	12
|V (An)|	3	4	7	11	18	29	47	76	123	199	322
Y (An)	1	1	3	4	5	7	11	16	23	35	
Yt (An)	2	2	3	4	7	9	13	19	27	41	58
Proposition 3.1 ([2, Proposition 3.1]). Let n > 4, then
(i)	Y (An) < Y(r„-l)+ 7(rn-s),
(ii)	Y (An) < Y(r„) < y(A„) + Y (r„-4).
Using a similar idea we obtain the following result. Proposition 3.2. Let n > 4, then
(i)	Yt (An) < Yt(rn-l)+ Yt(rn-3),
(ii)	Yt (An) < Yt(rn) < Yt(An) + Yt(r n—4 ).
Proof. The proof mimics the proof of [2, Proposition 3.1].
(i):	The vertices of An can be partitioned into vertices that start with 0 and vertices that start with 1. The subgraphs induced by these vertices are isomorphic to rn-1 and rn-3 respectively, hence we infer that Yt(AJ < Yt(rn-i) + Yt(rn-3).
(ii):	Let DT be a minimal total dominating set of rn and set
D't = {a | a is a Lucas string from DT} U {0a0 | 1a1 G DT}.
Note that |DT | < |DT | and a vertex of the form 1a 1 dominates two Lucas vertices of the form 0a1 and 1a0. Since these two vertices are dominated by 0a0, we say that D'T is a dominating set of An. Then we need to show that it is also a total dominating set. We know that every vertex in v G V(An) Ç V(rn) is adjacent to some vertex fi G DT. Then if fi G D'T we are done. Otherwise, fi must be of the form 1a1 G DT. In this case v G V(An ) must be of the form 1a0 or 0a1, which means that v is also adjacent to a vertex of the form 0a0 G DT. It follows that Yt(An) < Yt(rn). On the other hand, a total dominating set of An dominates all vertices of rn but the vertices of the form 10a01 where the subgraph induced by these vertices is isomorphic to rn-4. Hence we have Yt(rn) < Yt(An)+ Yt(rn-4).	□
Considering the vertices of high degrees a lower bound on y(Ak) is obtained in [2,
Theorem 3.5] as Y(An) > [
L„— 2n
] where n > 7. Combining this result with the fact that
Yt(An) > Y(An ) we have the following lower bound on Yt(An ). Proposition 3.3. For any n > 7, we have
Yt(An) > Y(An) >
n — 3
Z. Saygi: Results on the domination number and the total domination number of Lucas cubes 27
4 Up-down degree enumerator polynomial
In this section we present the up-down degree enumerator polynomial for An similar to the one for rn given in [14]. Using this polynomial we write optimization problems whose solutions give lower bounds on y(A„) and Yt(An).
By the definition of the edge set E(An), (v, v') G E(An) if and only if the number of different coordinates of v and v' is 1, that is, the Hamming distance dH(v, v') = 1. Here we have at most two kinds of neighbor v' for a vertex in v G V(An), whose weights can take the values w(v) ± 1. If w(v') = w(v) + 1 we call v' is an up neighbor of v and if w(v') = w(v) - 1 we call v' is a down neighbor of v. We denote the number of up neighbors of v by u and the number of down neighbors of v by d which is equal to the w(v) by the definition of An. Note that if the degree of v is k then we have u = k — d. For each fixed v G V(An) having degree k = deg(v), we write a monomial xuyd where d = w(v) is the Hamming weight of v and u is k — d. We call the polynomial
PA„ (x,y) =	xdeS(v)-W(v)yW(v) = £ xV
veY(An)	»ev (An)
as the up-down degree enumerator polynomial of An.
We need the following useful result given in [10] to obtain the recursive structure of PAn (x, y). Let £n,k,w be the number of vertices in An of degree k and weight w.
Theorem 4.1 ([10, Theorem 5.2]). For all n, k, w such that n > 2, 1 < k < n and
0 < w < n,
w — 1 \ /n — 2w\ ^ / w \ /n — 2w — 1 2w + k — n k — w	2w + k — n k — w
Let be the number of vertices in An whose number of up neighbors are u and number of down neighbors are d. Setting k = u + d and w = d in Theorem 4.1 we have
/ d — 1 \/n — 2d\ / d \/n — 2d — A
= 3d+	+2 3d +	. (4.1)
, , \3a + u — n) \ u ) \dd + u — n/\ u y Then using (4.1) we can write the up-down degree enumerator polynomial of An as
PAn (x,y) = £ £'niU,d xuyd,	(4.2)
u,d
where 0 < u, d < n. Furthermore, using (4.1) and (4.2) we obtain the following recursive relation which is very useful to calculate PAn (x, y).
Theorem 4.2. Let PAn (x, y) be the up-down degree enumerator polynomial of An. Then for n > 5 we have
pAn (x,y) = xpAn-i (x,y) + ypAn-2 (x,y) + (y — xy)pAn- (x,y), (4.3)
where
PA2 (x, y) = x2 +2y, PA3 (x, y) = x3 + 3y and Pa4 (x,y)= x4 + 4xy + 2y2.
18
Ars Math. Contemp. 19 (2020) 28-23
Proof. The initial conditions are clear from the definition of An. For fixed integers 1 < u < n and 2 < d < |_nJ, the coefficient of the monomial xuyd in the right hand side of
the equation (4.3) is the sum of t
n— 1,u — 1,d
coming from iPa„_i (x, y), t
n—2,u,d— 1
coming
from ypA„_2 (x,y^ C—3,u,d—1 coming from ypA„_3 (x,y) and —C—3,u—i,d—1 coming
from —xyPAf-3 (x, y). Then we need to show that
^n,u,d = ^n-l,u-l,d + ^n-2,u,d-l + ^n-3,u,d-l ^n-3,u-l,d-l.
By setting X = 3d + u — n and Y = n — 2d in (4.1) and using the binomial identities
m/m — 1\ m + 1 — k
k
k V k - 1
k1
k
m1
we have
^n—1,u—1,d + tn — 2,u,d— 1 + ^
d — 1\/Y — 1\ / d X u — 1 +2 X
- t
n—3,u,d— 1 n—3,u— 1,d— 1
+
+
d — 2 A /Y X - 1
+ 2
d — (Y — 1 X
+2
Y — 2 u — 1
d — 1 \/Y — 1 X — J V u d — 1\/Y — 2 Xu

d — 2 X — 1 d — 1\/Y
Xu
Y — 1\ (d — 1 Yu -11 + 2 Xd-11
u X
Y + d-1+ d- 1
Y — 2^ u — 1 d- 1 - X Y -u

Y
X u d~ ^ Y
+2
d1
X
Y — 1
u
M +2( d
^U + V Xy \ u
u X d — X Y — 1 — u Y - 1 + "d + _d Y - 1
Y1

X
d Y - 1
t:
In particular, the case d = 0 corresponds to the all 0 vertex in An and we have tn n 0 = 1, which means that the coefficient of the terms xny0 in both sides of (4.3) are 1. Similarly, the case u = 0 corresponds to the vertices in An whose weights are | f J and we have (see, Remark 5.1)
n if n is odd,
t
n,0 , L f J
2 if n is even.
Then one can easily see that the coefficient of the terms x0yLf J in both sides of (4.3) are equal to each other. The only remaining particular case is d = 1. For PAf (x, y) this case corresponds to the vertices in An whose weights are 1. We know that there are n such vertices in An and their number of up neighbors are n — 3. That is, the coefficient of the term xn—3y in PAf (x, y) is n. On the other hand the coefficient of the term xn—3y is n — 1 in xPAf-1 (x, y); 0 in yP\f-2 (x, y) and 1 in (y — xy)PAf-3 (x, y) respectively. Hence the coefficient of the terms xuyd in both sides of (4.3) are equal to each other for all cases. □
m
m
m
k
u
Z. Saygi: Results on the domination number and the total domination number of Lucas cubes 27
Remark 4.3. The recursive relation for the up-down degree enumerator polynomial of An in Theorem 4.2 is the same with the recursive relation for the up-down degree enumerator polynomial of rn, which is proved using the fundamental decomposition rn = 0rn-1 + 10rn-2. The only differences are the initial polynomials. For the proof we directly used the degree information of An obtained in [10], since An do not have a decomposition like 0A„-1 + 10A„-2.
5 Lower bounds on domination numbers using optimization problems
In this section, we present optimization problems giving lower bounds on y(A„) and Yt(An), whose number of variables and number of constraints are fewer than the general optimization problem described in Section 3.
We use the up-down degree enumerator polynomial PAn (x, y) to construct an optimization problem, which is similar to the optimization problem given in [14]. Let D and DT be a dominating set and a total dominating set of An respectively. Let vD e D (vD e Dt respectively) and xUyd be its corresponding monomial in PAn (x, y). Then vd dominates u distinct vertices v g V(An) having weight w(v) = w(vD) + 1 and d distinct vertices v e V(An) having weight w(v) = w(vD) - 1. Note that for all vd g D (vD e DT respectively) some of the vertices of An may be dominated more than one times. Note that for every vertex v e V(An) there must exist at least one vertex vd e N[v] n DT with w(vD) = w(v) + 1 or vD = v for the dominating set D and vD e N(v) n DT with w(vD) = w(v) + 1 for the total dominating set DT.
Now we write the up-down degree enumerator polynomial of An (see, 4.2) as
PAn (x, y) = £ cUxV,	(5.1)
u,d
where cU = i'nu d. For each pair (u, d) in the monomials of the up-down degree enumerator polynomial PAn (x, y) we associate an integer variable zU which counts the number of vertices in D or DT having d down neighbors and u up neighbors. For any fixed value of d, the number of vertices having weight d gives the bounds 0 < zU < cU. Our aim is to minimize |D| for domination number and to minimize |DT| for total domination number. Hence our objective function is to minimize
£ zU.
u , d
To dominate all the vertices having a fixed weight d such that 1 < d < |_ n J - 1 we must have the following constraints rd for domination number and r'd for the total domination number.
rd :	£(u • zU-i + zU + (d +1) • zU+i) > £ cU
U	U
rd :	£(u • zU-i + (d +1) • zU+i) > £ cU
since any vertex corresponding to the monomial xUyd 1 can dominate u distinct vertices (u up neighbors) having weight d and any vertex corresponding to the monomial xU yd+1
18
Ars Math. Contemp. 19 (2020) 30-23
can dominate d +1 distinct vertices (d +1 down neighbors) having weight d. By the same argument, for d = 0 we must have
ro:
E zu + zu > E
1
and
ro :
E zu > E-
and for d = |_ 2 J we must have
l n j :
Eu ■ z-f j-1 + z"nJ > E
cL f J =
E
u■zLfJ-1
>E
L f J
In if n is odd,
12	if n is even.
n if n is odd,
2	if n is even.
Now subject to the above constraints r0,..., rLnj (constraints r0,..., r'nj) the value of the objective function will be a lower bound on 7(An) (Yt(An), respectively).
Remark 5.1. The number of vertices of An having weight d is equal to the right hand side of the above constraints rd and rd. By setting k = u + d and w = d in [10, Corollary 5.3] we have

n-d
E ^
n — d d
+
n — d— 1 n 2d
Remark 5.2. The number of variables z- in our optimization problem is equal to the number of monomials in PAf (x, y). Assume that n is even. By the string representation of the vertices in An we have n — 3d < u < n — 2d — 1. The bounds come from the maximum number of the sub-strings 010 and 10 in the representation of the vertices. That is, u can take n — 2d — 1 — (n — 3d) + 1 = d distinct values when d ranges from 1 up to |_ n J and u can take n — 2d distinct values when n + 1 < d < |_ n J. Furthermore, u can take only one values for d = 0 and d = |_ n J. Therefore, the number of variables z- becomes
LfJ LfJ-1
2 + E d + E (n — 2d)
d=LfJ+1
d=1
which is equal to
3	n	(	n
2	.3.		.3.
2n
+ 1 — T
+
+ n — n.
(5.2)
For n > 2 this sequence starts as 2, 2, 3,3,5,5, 7, 8,10,11,14,15,18, 20, 23, 25, 29,... Note that in (3.1) the number of variables is Ln, which exhibit exponential growth. In our case, if we omit the floor functions in (5.2) then the number of variables z- is approximately
2
equals to 2 + 12.
u
u
c
0
0
rL f J
u
u
d
n
For n = 12 we illustrate our optimization problem as follows. First we obtain PAl2 (x, y)
Z. Saygi: Results on the domination number and the total domination number of Lucas cubes 27
by using the recursion in Theorem 4.2 as
PAl2 (x,y) = 2y6+
12y5x + 24y5 +
12y4xf + 54y4 x2 + 36y4x + 3y4+
12yfx5 + 60yf x4 + 40yfxf+
12y2x2 + 42y2x6+
12yx9+
x12
Then using PAl2 (x, y) we have the corresponding optimization problem:
Objective function:
minimize : z^ + zj9 + 4 + + zf + zf + zf + z| + z| + z4 + z^ + z5 + z0 +
Constraints for 7 (A 12):
r6 :	z1 + zo z5 + z6	>	2;
r5 :	3zf + 2z4 + z1 + z5 + z0 + 6z0	>	36;
r4 :	5zf + 4zf + 3zf + zf + z42 + z4 + z40 + 5zt° + 5z5°	>	105
rf:	7z2 + 6z6 + zf + zf + zf + 4z| + 4z4 + 4z°1 + 4z°	>	112
r2 :	9z9 + z2 + z6 + 3zf + 3zf + 3zf	>	54;
r1 :	12z12 + z9 + 2z2 + 2z6	>	12;
r0 :	z12 + z9 z0 + z1	>	1;
Constraints for jt (A 12):
z5 > 2;
3z3 + 2z2 + z| + 6z0 > 36; 5zf + 4zf + 3zf + 5z5 + 5z5 > 105; 7z27 + 6z| + 4zf + 4z| + 4z4 + 4z4 > 112;
9z9 + 3zf + 3zf + 3zf > 54; 12z012 + 2z2 + 2z| > 12;
z9 > 1;
Bounds:
z^ < 1;
0 - 1; z9 < 12; z2 < 12; z26 < 42; zf < 12; zf < 60; zf < 40;
zf < 12; z42 < 54; z4 < 36; z40 < 3;
z51 < 12;
z50 < 24; z0 < 2.
Depending on the constraints rd and r'd (d = 0,1,..., 6) the value of the objective function gives a lower bound on y(A12) and Yt(A12) respectively. The above problem has
r
r
r
r
r
r
r
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Ars Math. Contemp. 19 (2020) 32-23
only 14 variables and 7 constraints (instead of having L12 = 322 variables and 322 constraints as in (3.1)). To find lower bounds on y(A„) and Yî(A„) one can use the up-down degree enumerator polynomial PAn (x, y) of An in Theorem 4.2 and one can write an optimization problem having fewer number of variables z^ (see Remark 5.2) and |_ÇJ + 1 constraints rd or r'd. The solutions of the optimization problems give lower bounds on Y (An) and yt(An). It is easy to see that the number of variables and the number of constraints in our optimization problems are very smaller than the ones in the optimization problem (3.1).
For illustration we implemented the above integer linear programming problem using CPLEX in NEOS Server [3, 4, 5] for 12 < n < 26 and immediately (less than 0.02 seconds) obtain the lower bounds on y (An) and yt(An) presented in Table 2 and Table 3 (better than the ones in Proposition 3.3). Note that for n = 26, the number of variables in our optimization problem is 58 by Remark 5.2 and the number of constraints is 14, on the other hand, these numbers are equal to L26 = 271443 for the general optimization problem (3.1). In addition, the upper bounds in these tables are obtained by Proposition 3.1 and Proposition 3.2 by using the upper bounds on the values of Y(rn) and Yt(rn) given in [14] for n > 14.
Table 2: Current best bounds on Y(An), 12 < n < 26.
n	Y (An)	n	Y(An )	n	Y(An)
12	49* - 54	17	310 - 555	22	2686 - 6140
13	61* - 86	18	471 - 895	23	4184 - 9935
14	89 -132	19	725 -1450	24	6519 -16075
15	134 - 215	20	1114 - 2345	25	10163 - 26010
16	203 - 340	21	1724 - 3795	26	15835 - 42085
Table 3: Current best bounds on yt(An), 12 < n < 26.
n	Yt (An)	n	Yt (An )	n	Yt (An )
12	58*	17	340 - 567	22	2893 - 6140
13	77* - 95	18	514 - 909	23	4490 - 9935
14	101 -145	19	787 -1450	24	6974 -16075
15	151 - 231	20	1205 - 2345	25	10839 - 26010
16	225 - 362	21	1862 - 3795	26	16838 - 42085
Remark 5.3. It is shown in [1, 7] that Y(r9) = 17, Y(r10) = 25, 54 < Y(r12) < 61 and 78 < Y(ri3) < 93 (shown in [14]). Substituting these results in Proposition 3.1 we obtain the bounds for n = 13 in Table 2.
Similarly, it is shown in [1, 7] that Yt(r) = 20, Yt(rio) = 30, Yt(ri2) = 65 and 97 < Yt(r13) < 101. Substituting these results in Proposition 3.2 we obtain the bounds for n = 13 in Table 3.
Note that our optimization problems obtained from up-down degree enumerator polynomial give y(A12) > 39, yt(A12) > 45 and y(A13) > 59, Yt(A13) > 68. Furthermore,
Z. Saygi: Results on the domination number and the total domination number of Lucas cubes 27
using (3.1) we obtain that 49 < y(Ai2) < 54 and y1(A12) = 58. For these reasons we put
a * to the lower bounds for the cases n =12 and n = 13 in Table 2 and Table 3.
ORCID iD
ZUlfUkar Saygi © https://orcid.org/0000-0002-7575-3272 References
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