### Question:
Retryer { static double fib(int n) { double phi = 1.6180339887; return (Math.pow(phi, n) - Math.pow(-phi, -n)) / (2 * phi - 1); } Retryer(); private  Retryer(ExceptionPlan<Delay<?>> plan); final Retryer upon(
      Class<E> exceptionType, List<? extends Delay<? super E>> delays); final Retryer upon(
      Class<E> exceptionType, Stream<? extends Delay<? super E>> delays); Retryer upon(
      Class<E> exceptionType, Predicate<? super E> condition,
      List<? extends Delay<? super E>> delays); Retryer upon(
      Class<E> exceptionType, Predicate<? super E> condition,
      Stream<? extends Delay<? super E>> delays); T retryBlockingly(CheckedSupplier<T, E> supplier); CompletionStage<T> retry(
      CheckedSupplier<T, ?> supplier, ScheduledExecutorService executor); CompletionStage<T> retryAsync(
      CheckedSupplier<? extends CompletionStage<T>, ?> asyncSupplier,
      ScheduledExecutorService executor); ForReturnValue<T> ifReturns(
      Predicate<T> condition, List<? extends Delay<? super T>> delays); ForReturnValue<T> ifReturns(
      Predicate<T> condition, Stream<? extends Delay<? super T>> delays); ForReturnValue<T> uponReturn(
      T returnValue, Stream<? extends Delay<? super T>> delays); ForReturnValue<T> uponReturn(
      T returnValue, List<? extends Delay<? super T>> delays);  }### Answer:
@Test public void testFibonacci() { assertThat(Math.round(Retryer.fib(0))).isEqualTo(0); assertThat(Math.round(Retryer.fib(1))).isEqualTo(1); List<Long> results = new ArrayList<>(); results.add(0L); results.add(1L); for (int i = 2; i < 93; i++) { long f = Math.round(Retryer.fib(i)); assertThat(f).isLessThan(Long.MAX_VALUE); assertThat((double) f) .isWithin(f / 1000).of(results.get(i - 2).doubleValue() + results.get(i - 1).doubleValue()); results.add(f); } }