Arithmetic Graphs by Dileepkumar R

By Dileepkumar R

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Extra info for Arithmetic Graphs

Sample text

A, A Dynamic Survey of Graph Labeling, Electron. J. Combinatorics, 10, 2007, DS6. M and Shetty, Sudhakar, On Arithmetic Graphs, Indian J. , 33(8), 2002, 1275-1283.

5 For the class of complete graphs in general we have the following conjecture to propose: Conjecture 2 For any integer n ≥ 5, Kn is not arithmetic. 2 there exists a (k, d)-arithmetic numbering f of C4t , t ≥ 1, such that k1 , k2 ∈ f (C4t ). 2 is satisfied. 6 shows a (1, 2)-arithmetic numbering f of C8 (under C1) and an (8,2)-arithmetic numbering of C8 (under C2). 4 For any nonnegative integer r and positive integer d, C4t+1 is (2dt+2r, d)-arithmetic. Proof Under the hypotheses the map f : V (C4t+1 ) −→ N defined by f (ui ) = )d, if i is odd r + ( i−1 2 1 2dt + r + 2 id, if i is even can be easily verified to be a required arithmetic numbering of C4t+1 .

Then d | (k2 − k1 ) for each of these partitions (k1 , K2 ), but only for the first three of these do we have r ≥ 3 = a, k2 − k1 = 2r. 1 As another example, (a, b, k, d)=(3, 4, 12, 3), then the partitions (k1 , k2 ) of k with 0 ≤ k1 < k2 are (0, 12), (1, 11), (2, 10), (3, 9), (4, 8), (5, 7). Then only (0, 12) satisfies C2 and hence when (k1 , k2 ) is any of these partitions of k, the graph K3,4 has a (k,d)arithmetic numbering f with k1 , k2 ∈ f (K3,4 ). The partition (3,9) does not satisfy either C1 or C2, hence K3,4 has no (k, d)-arithmetic numbering f for which 3, 9 ∈ f (K3,4 ).

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