是否可以在普通的JPA或JPA + Hibernate扩展中声明复合键，而复合键的元素是一个序列？

这是我的复合课程：

@Embeddable
public class IntegrationEJBPk implements Serializable {

    //...


    @ManyToOne(cascade = {}, fetch = FetchType.EAGER)
    @JoinColumn(name = "APPLICATION")
    public ApplicationEJB getApplication() {
        return application;
    }


    @Column(name = "ENTITY", unique = false, nullable = false, insertable = true, updatable = true)
    public String getEntity() {
        return entity;
    }

    @GeneratedValue(strategy = GenerationType.AUTO, generator = "INTEGRATION_ID_GEN")
    @SequenceGenerator(name = "INTEGRATION_ID_GEN", sequenceName = "OMP_INTEGRATION_CANONICAL_SEQ")
    @Column(name = "CANONICAL_ID", unique = false, nullable = false, insertable = true, updatable = true)
    public String getCanonicalId() {
        return canonicalId;
    }

    @Column(name = "NATIVE_ID", unique = false, nullable = false, insertable = true, updatable = true)
    public String getNativeId() {
        return nativeId;
    }

    @Column(name = "NATIVE_KEY", unique = false, nullable = false, insertable = true, updatable = true)
    public String getNativeKey() {
        return nativeKey;
    }

    //...
}

我已经提供的值application，entity，nativeId和nativeKey。我想构造一个像下面这样的实体：

IntegrationEJB i1 = new IntegrationEJB();
i1.setIntegrationId(new IntegrationEJBPk());
i1.getIntegrationId().setApplication(app1);
i1.getIntegrationId().setEntity("Entity");
i1.getIntegrationId().setNativeId("Nid");
i1.getIntegrationId().setNativeKey("NK");

当我调用em.persist(i1）时，我希望canonicalId生成并插入集成。

这可能吗？如果是这样，简单的方法是什么？（我不想使用应用程序提供的密钥或本机sql）。