Journal of Combinatorial Mathematics and Combinatorial Computing, 49 (2004) pp.9-31.
In a previous article on caps in PG(5,3) and PG(6,3), it was proven that every 53-cap in PG(5,3) is contained in the 56-cap of Hill and that there exist complete 48-caps in PG(5,3). The first result was used to lower the upper bound on m2(6,3) from 164 to 154. Presently, the known upper bound on m2(6,3) is 148. In this article, using computer searches, we prove that every 49-cap in PG(5,3) is contained in a 56-cap, and that every 48-cap, having a 20-hyperplane with at most 8-solids, is also contained in a 56-cap. These computer results enable us to present a geometrical proof that m2(6,3) = 147. A computer search for caps in PG(6,3) which uses the computer results of PG(5,3) then lowers this upper bound to m2(6,3) = 136. So now we know that 112 = m2(6,3) = 136.
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