Three Romanian tennis players, Cezar Cretu, Filip Jianu and Radu Papoe, have progressed to the round of 16 of the Sekyra Group Prague Open 2026 by Advantage Cars challenger tournament, prize money EUR 97,640 euros.
Cretu, 25, ranked 266th, sixth seed, on Monday prevailed over France's Thomas Faurel, 20, ranked 331st, 2-6 6-2 6-4, in two hours and 10 minutes.
Cretu's next opponent will be Czech Maxim Mrva, 19, ranked 309th.
Romanian Radu Mihai Papoe, 23, ranked 300th, came from the qualifiers, after victories against Czechs Jakub Vrtilka, 6-0 6-1 and Matyas Cerny, 6-2 6-3, and, on Tuesday in the first round, he defeated Spain's Oriol Roca Batalla, 33, ranked 351st, 6-7 (4/7) 7-5 7-5.
In the round of 16, Papoe will meet another Spaniard, qualifier Javier Barranco Cosano, 27, ranked 466th.
Also on Tuesday, Filip Cristian Jianu, 24, ranked 284th , beat Slovakian Andrej Martin, 36, ranked 402nd, 6-2 6-2, in an hour and 16 minutes. Jianu is now 2-1 in the head-to-head count against Martin.
In the second round, Jianu will face German Rudolf Molleker, 25, ranked, 481st, whom he defeated in all their four previous meets.
All three Romanians won EUR 1,590 euros, Cretu and Jianu 6 points each, and Papoe 10 points.
In men's doubles, Bogdan Pavel and Ukrainian Denis Molceanov reached the quarterfinals, after 6-3 6-2 against the Nick Hardt (Dominican Republic)/Cristian Rodriguez (Colombia) in 62 minutes. The Romanian-Ukrainian duo won EUR 920 euros and 16 ATP doubles points.
In the quarterfinals, they will play either Swedes Erik Grevelius/Adam Heinonen or Spain's Mario Mansilla Diez/Benjamin Winter Lopez. AGERPRES
