By Dileepkumar R
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Extra info for Arithmetic Graphs
K + (a + b − 2)d}. Hence f (ui ) = f (vj ) yields k2 − k1 = (i − j)d (16) Therefore, if d | (k2 − k1 ), then (16) is a contradiction, and on the other hand, if k2 − k1 = rd, r ≥ a, then (16) yields rd = k2 − k1 = (i − j)d ⇒ i−j =r ≥a ⇒ i≥a+j a contradiction to the fact that 1 ≤ i ≤ a and 1 ≤ j ≤ b. Thus f must be a required (k, d)-arithmetic numbering of Ca,b . 2. 3 displays two such instance. 3 It seems a hard problem to determine in general the full range of values of k and d and possible partitions (k1 , k2 ) of k, 0 ≤ k1 < k2 , with k2 − k1 = rd and r < a such that Ca,b , 2 ≤ a ≤ b, has a (k, d)-arithmetic numbering f for which k1 , k2 ∈ f (Ca,b ).
Then by equation (14) we get k1 + (i − 1)d = k2 + (j − 1)ad ⇒ k2 − k1 = [(i − 1) − (j − 1)a]d. The last eqality contradicts C1. Thus f must be injective and hence is a required (k,d)arithmetic numbering of Ka,b in this case. On other hand, if under C2 we have f (ui ) = f (vj ) we get rd = k2 − k1 = [(i − 1) − (j − 1)a]d ⇒ (i − 1) − (j − 1)a = r ≥ a ⇒ (i − 1) ≥ ja which is a contradiction as 1 ≤ j ≤ b. Hence f is a (k, d)-arithmetic numbering of Ka,b when C2 holds. One can in fact prove that each of C1 and C2 is also neccessary for Ka,b , 2 ≤ a ≤ b, to have a (k, d)-arithmetic numbering f with k1 , k2 ∈ f (Ka,b ) and the proof, which is rather tedious, essentially makes use of the following general property of arithmetic numbers of Ka,b .
X4t+1 + x4t+2 )} = (4t + 2)k + (2t + 1)(4t + 1)d ⇒ 2k + 2(2tk + md) = 2k + 4tk + (2t + 1)(4t + 1)d where m is a positive integer. Hence 2md = (2t + 1)(4t + 1)d ⇒ 2m = (2t + 1)(4t + 1)d This is a preposterous identity sice both sides are positive integers. 6 For any nonnegative integer r and positive integer d, C4t+3 is ((2t+1)d+2r, d)-arithmetic. Proof Under the hypotheses, the map f : V (C4t+3 ) −→ N defined by 46 f (ui ) = r + ( i−1 )d, if i is odd 2 1 (2t + 1)d + r + 2 id, if i is even (19) is a required arithmetic numbering of C4t+3 .