H proves (i). Suppose that ui , ui+2 , ui+5 J for some
H proves (i). Suppose that ui , ui+2 , ui+5 J for some i. Then, ui+1 , ui+3 , ui+4 , vi , vi+2 , vi+5 J. / four. Because ui+3 , ui+4 have to be 2-dominated, vi+3 , vi+4 J. 1st, we prove that vi+1 J. Suppose / on contrary that vi+1 J, as in Figure 16. Then, vi+1+k , vi+3+k , vi+4+k J. Because vi / has to be 2-dominated, vi-k J or vi+k J. With no loss of generality, assume that vi+k J. Thus, ui+k J, ui+1+k J and ui+2+k J. Additionally, ui+3+k J, ui+4+k J, / / / ui+5+k J, vi+5+k J, and vi+2+k J. Proving analogously as in subcase 3.three., we receive a / contradiction using the assumption that J is usually a (2-d)-kernel.+kFigure 16. The case when ui , ui+2 , ui+5 , vi+1 J.Symmetry 2021, 13,7 ofHence, vi+1 J (see Figure 17). (Z)-Semaxanib Epigenetics Moreover, vi+1+k J and ui+1+k , vi+3+k , vi+4+k J. / / We consider two subcases.+kFigure 17. The case when ui , ui+2 , ui+5 J, and vi+1 J. /4.1. vi+2+k J for some i (see Figure 18). / Then, ui+2+k J, ui+3+k J, ui+4+k J, ui+5+k J, and v5+i+k J. Furthermore, vi+k J / / and ui+k J; otherwise, we acquire precisely the same configuration as in subcase 3.three. /+kFigure 18. The case when ui , ui+2 , ui+5 J (the very first subcase).Hence, n has to be divisible by 5, and from the definition of P(n, k), we conclude that k two(mod 5), which proves (ii). four.2. vi+2+k J for some i (see Figure 19). Then, ui+2+k J, ui+k , ui+3+k J and vi+k , ui+4+k J. Additionally, ui+5+k J and / / v i +5+ k J /+kFigure 19. The case when ui , ui+2 , ui+5 J (the second subcase).Therefore, n has to be divisible by 5, and in the definition of P(n, k), we conclude that k three(mod five), which proves (iii), which ends the proof. Basing around the proof of Theorem 1, the following corollaries are obtained. They concern the number of (2-d)-kernels within the generalized Petersen graph also as the reduced and upper (2-d)-BI-0115 In stock kernel numbers. By a rotation of configurations shown on Figure ten, condition (i) of Theorem 1 gives two (2-d)-kernels in generalized Petersen graph and situations (ii) and (iii) give five (2-d)-kernels. Therefore, if n and k satisfy a lot more than 1 of these circumstances, we receive more (2-d)-kernels. Additionally, the proof on the Theorem 1 presents the constructions from the (2-d)-kernels within the generalized Petersen graph P(n, k). Figure 20 shows the smallest along with the largest (2-d)-kernel within the graph P(20, 7).P (20, 7)P (20, 7)Figure 20. The biggest (left side) as well as the smallest (correct side) (2-d)-kernel within the graph P(20, 7).Symmetry 2021, 13,8 ofCorollary 1. Let n 3, k 7 five ( P(n, k)) =nbe integers. Then,for n 0(mod ten) and k a(mod ten), a = 3, 7 for n five(mod 10) and k a(mod five), a = 2, three or for n 0(mod 10) and k a(mod 10), a = 2, eight for n 0(mod 10) and k a(mod ten), a = 1, five, 9 or for even n, n 0(mod ten) and odd k.Corollary two. Let n three, k n be integers. If n 0(mod ten) and k a(mod 10), a = three, 7, two then four (2-d) ( P(n, k )) = n and (2-d) ( P(n, k )) = n. five Corollary 3. Let n 3, k n be integers. If n five(mod ten) and k a(mod five), a = two, 3 or 2 n 0(mod ten) and k a(mod ten), a = two, eight, then (2-d) ( P(n, k)) = (2-d) ( P(n, k)) = 4 n.Corollary four. Let n 3, k n be integers. If n 0(mod 10) and k a(mod ten), a = 1, five, 9 2 or n is even, n 0(mod 10) and k is odd, then (2-d) ( P(n, k)) = (2-d) ( P(n, k)) = n. The above corollaries characterize all doable graphs P(n, k), which have the (2-d)kernel. two.2. The Second Generalization of your Petersen Graph Now, we contemplate another generalization of your Petersen graph. Let n five be an integer. Let Cn be a cycle and Cn its complement.
