2000
DOI: 10.1002/(sici)1097-0118(200005)34:1<89::aid-jgt8>3.0.co;2-k
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Half-transitive graphs of valency 4 with prescribed attachment numbers

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Cited by 24 publications
(28 citation statements)
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“…, d m − 1} for which t = cm and m = ct, claims (i), (iii) and (iv) follow from (20), (21), (24) Table 1). In particular 3 + r 2 = t, and so equation (26) implies that 2 + 2r = t. Note also that (26) and (30) combined together imply that 2(r − 1) = 0. By (27) we thus have 2(r − 1) = n 2 and t = n 2 + 4.…”
Section: Lemma 52mentioning
confidence: 99%
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“…, d m − 1} for which t = cm and m = ct, claims (i), (iii) and (iv) follow from (20), (21), (24) Table 1). In particular 3 + r 2 = t, and so equation (26) implies that 2 + 2r = t. Note also that (26) and (30) combined together imply that 2(r − 1) = 0. By (27) we thus have 2(r − 1) = n 2 and t = n 2 + 4.…”
Section: Lemma 52mentioning
confidence: 99%
“…The nature of the action of ϕ will contradict half-arctransitivity of X, which thus proves that (AutX) v ∼ = Z 2 , as claimed. Note that (26) and (33) imply that r 2 = 7 u To see that ϕ is in fact a permutation of V (X), we only need to observe that it is injective. Consider the vertices of Table 2, which are of the form u Table 2 for each i = 1, 2, 3, and that the corresponding congruencies modulo 8 of their superscripts are again the eight possibilities 0, 1, .…”
Section: Lemma 52mentioning
confidence: 99%
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