IntStream iterate in steps

Actually range is ideal for this.

IntStream.range(0, 100).filter(x -> x % 3 == 0); //107,566 ns/op [Average]

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Edit: Holgers's solution is the fastest performing solution.

Since the following lines of code

IntStream.range(0, 100).filter(x -> x % 3 == 0).forEach((x) -> x = x + 2); 

IntStream.range(0, 100 / 3).map(x -> x * 3).forEach((x) -> x = x + 2); 

int limit = ( 100 / 3 ) + 1; 
IntStream.iterate(0, n -> n + 3).limit(limit).forEach((x) -> x = x + 2);

show these benchmark results

Benchmark                 Mode  Cnt    Score    Error  Units
Benchmark.intStreamTest   avgt    5  485,473 ± 58,402  ns/op
Benchmark.intStreamTest2  avgt    5  202,135 ±  7,237  ns/op
Benchmark.intStreamTest3  avgt    5  280,307 ± 41,772  ns/op

Actually you can also achieve the same results with a combination of peek and allMatch:

IntStream.iterate(0, n -> n + 3).peek(n -> System.out.printf("%d,", n)).allMatch(n -> n < 100 - 3);

This prints

0,3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99,

But nevertheless, this one is faster:

IntStream.range(0, 100 / 3 + 1).map(x -> x * 3).forEach((x) -> System.out.printf("%d,", x));

Java 9 Update:

Now the same iteration easier to achieve with Java 9:

Stream.iterate(0, i -> i <= 100, i -> 3 + i).forEach(i -> System.out.printf("%d,", i));

In JDK9 there's takeWhile 1

IntStream
  .iterate(0, n -> n + 3)
  .takeWhile(n -> n < 100)
  .forEach(System.out::println);