Compare squares.md

Given two arrays a and b write a function comp(a, b) that checks whether the two arrays have the "same" elements, with the same multiplicities. "Same" means, here, that the elements in b are the elements in a squared, regardless of the order.

Examples

Valid arrays

a = [121, 144, 19, 161, 19, 144, 19, 11]  
b = [121, 14641, 20736, 361, 25921, 361, 20736, 361]

comp(a, b) returns true because in b 121 is the square of 11, 14641 is the square of 121, 20736 the square of 144, 361 the square of 19, 25921 the square of 161, and so on. It gets obvious if we write b's elements in terms of squares:

a = [121, 144, 19, 161, 19, 144, 19, 11] 
b = [11*11, 121*121, 144*144, 19*19, 161*161, 19*19, 144*144, 19*19]

Invalid arrays

If we change the first number to something else, comp may not return true anymore:

a = [121, 144, 19, 161, 19, 144, 19, 11]  
b = [132, 14641, 20736, 361, 25921, 361, 20736, 361]

comp(a,b) returns false because in b 132 is not the square of any number of a.

a = [121, 144, 19, 161, 19, 144, 19, 11]  
b = [121, 14641, 20736, 36100, 25921, 361, 20736, 361]

comp(a,b) returns false because in b 36100 is not the square of any number of a.

Remarks

a or b might be []. a or b might be null.

If a or b are null, the problem doesn't make sense so return false.

If a or b are empty the result is evident by itself.

odd number triangle.md

Given the triangle of consecutive odd numbers:

             1
          3     5
       7     9    11
   13    15    17    19
21    23    25    27    29
...

Calculate the row sums of this triangle from the row index (starting at index 1) e.g.:

rowSumOddNumbers(1); // 1
rowSumOddNumbers(2); // 3 + 5 = 8

Prime decomposition - Opcional.md

Given a positive number n > 1 find the prime factor decomposition of n. The result will be a string with the following form :

"(p**n1)(p2**n2)...(pk**nk)"

Remarks

1. the prime numbers are in increasing order and
2. n(i) == '' if n(i) is 1.

Example

n(86240) should return (2**5)(5)(7**2)(11)