How can I return a Sequence?

Recently I had the need to redefine a certain symbol in my init.m so it would be automatically omitted from any lists it appears in. I decided to redefine it to an empty sequence, e.g. Locked = Sequence[], but that got me thinking. What if I wanted to return a sequence (not necessarily an empty one) in a := definition? Return doesn't have the SequenceHold attribute, and adding it in a package might cause problems, so what would I do?

EDIT: I think I've figured out what exactly causes me to have the problem. I've defined it to display a Message first to let me know whenever a package I'm importing attempts to "attack my computer". (It is trying to cause my computer to behave in a manner not consistent with my wishes, after all.) So I defined it as Locked := (Message[Locked::nope]; Sequence[]), but strangely it just returns Null. (It doesn't show a return value, but if I do {Locked}, it returns {Null}, and if I try to set it as an attribute it says that Null is not a valid attribute and doesn't set any of them.)

Answer

Calculate the List of results you wish to return and use Apply to replace the head:

listFn[a_, b___] := If[a > 0, {b}, {0}];
seqFn[args___] := Sequence @@ listFn[args];
f[1, seqFn[2, 3, 4, 5], 6]
f[1, seqFn[-2, 3, 4, 5], 6]

(*--> f[1, 3, 4, 5, 6] *)
(*--> f[1, 0, 6] *)

Here listFn represents the calculation of the results and does not need to be a separate function. The particular example above can be written more simply as

seqFn[a_, b___] := Sequence @@ If[a > 0, {b}, {0}]

Edit

It should be stressed that the method assumes listFn actually evaluates to the List of desired inputs; if not, the Head of the expression returned will be replaced with Sequence, perhaps with undesired results. In such a case, a method such as @Jens's may be used.

For instance if a is non-numeric, say a Symbol, in the example above, then you get

f[1, seqFn[x, 3, 4, 5], 6]
(*-> f[1, x > 0, {3, 4, 5}, {0}, 6] *)

Often one can construct a List of the actual results, though. (With thanks to @Mr.Wizard.)