The Buffalo Bills have released former Tampa Bay Buccaneers tight end O.J. Howard, per multiple reports.
Howard, the No. 19 overall pick by the Bucs in the 2017 NFL draft, signed a one-year deal with the Bills in free agency, most of which was guaranteed.
Even with his release, the Bucs are still expected to receive a seventh-round compensatory pick in the 2023 NFL draft for losing him in free agency this offseason.
Despite the loss of Rob Gronkowski to retirement, the Bucs are unlikely to pursue a reunion with Howard, having signed veteran Kyle Rudolph and drafted a pair of promising rookies in Cade Otton and Ko Kieft.