Let's say we have a number, `num`. Find the number of values of `n` such that: there exists `n` consecutive **positive** values that sum up to `num`. A positive number is `> 0`. `n` can also be 1.
```python
#Examples
num = 1
#1
return 1
num = 15
#15, (7, 8), (4, 5, 6), (1, 2, 3, 4, 5)
return 4
num = 48
#48, (15, 16, 17)
return 2
num = 97
#97, (48, 49)
return 2
```
The upper limit is `$10^8$`
def consecutive_sum(n):
    top = (1+(1+8*n)**.5) / 2
    return sum( not (n - (v-1)*v//2) % v for v in range(1,int(top)))
def consecutive_sum(n):
    top = (1+(1+8*n)**.5) / 2
    return sum( not (n - (v-1)*v//2) % v for v in range(1,int(top)))