Concatinating stream in it's own forEach loop : Example 1: 
   

     public TailCall<TestClass> useOfStream(Stream<Test> streamL) {
            ArrayList<Test> testList2 = new ArrayList<>();
            Stream<Test> streamL2 = testList2.stream();
    
            streamL.forEach(test -> {
                for (int i = 1; i < 14; i++) {
                    if (/*insert if statement*/) {
                        Test test2 = new Test();
                        Stream<Test> streamT = stream.Of(test2);
                        **streamL2.concat(streamL2, streamT);**
                    } else {
                        //do something with TestClass
                    }
                }
            });
            if (streamL2.findAny().isPresent()) {
                return call(() -> useOfStream(streamL2));
            } else {
                return TailCalls.done(TestClass);
            }
        }


So, for a certain element in streamL I can possibly make up to 13 same-class elements. These newly made elements (that I've added to streamL2) should be iterated over the same way streamL was iterated over. Is there a possibility of adding those new elements to streamL?
Doing:

        streamL.forEach(test -> {
            for (int i = 1; i < 14; i++) {
                if (/*insert if statement*/) {
                    Test test2 = new Test();
                    Stream<Test> streamT = stream.Of(test2);
                    **streamL.concat(streamL, streamT);**
                } else {
                    //do something with TestClass
                }
            }
        });
If it's even possible to concat a stream within it's own for-Each loop, would the forEach loop also go through those newly added elements? That would eliminate my need of a recursive method.
Another question is, will an object made within a stream (for ex. `Test test2 = new Test();` in the forEach loop of streamL) be processed lazily by the program? With other words, would this object take any place in the memory heap? (but this question is not that important, primarly my first question)
