actually not the same

I've added a translation to Ruby. Please check and approve.

@bkaes: Wow, thank you so much. Here was the point: up to f(68) (aka 69 numbers)

Thanks, will be waiting for the answer. But I guess this issue is applicable to Python too, because the test case is the same: test.assert_equals(count_odd_pentaFib(68), 23).

Hi, Haskell

My solution passes the tests, but not this one: it "should return 23 on 68" $ countOddPentaFib 68 shouldBe 23.

Due to my function I have 22, not 23. I checked the sequence and counted 23 numbers with two times '1', so the output should be 22. What point have I missed?

Just to verify: the 68th element of the sequence is 12778498417777343135.

Added example test cases.

As I mentioned earlier, I have already done it. You can see and check JS translation if you press '+' button in the top of the page near the Python logo.

I translated this kata into JS, but it should be renamed. Is there any possibility to do it?

Expected: Hello World!, instead got: Helo World!

Couldn't pass the test. What does it mean?

Expected: 0, instead got: 1
10 Passed
1 Failed

I wonder, what is for '0', as it is expected by the 11th test?

You can create arrays like that: new Array(n), where n is quantity of elements. For example, new Array(3) will produce [undefined, undefined, undefined]. Besides, every array has join method, which creates a string from the array's elements and join's argument will play as separator, so that [1,2,3].join('anyStringYouLike') will be evaluated in '1anyStringYouLike2anyStringYouLike3'. And [undefined, undefined, undefined].join('String') will be evaluated in 'StringString'.

Hope this explanation will help to undestand this solution.

very helpful, thanks

You're right, _sort will be called every iteration, so you can try to sort array2 outside the loop for a better performance (see bellow). Or you have another vision for completing this task?

function comp(array1, array2){
function _sort(arr) { return arr.sort(function(a,b) { return a-b; }); }
array2 = _sort(array2);
return !array1 || !array2 ? false : _sort(array1).every(function(elm,i) {
return elm === Math.sqrt(array2[i]);
});
}