Question: In python I need a function that takes an integer and returns the absolute value of n minus the nearest [triangle number](http://en.wikipedia.org/wiki/Triangular_number) to n. The way I'm doing it now is by generating a list of all the triangle numbers up to n. 
Then using:

```
def closest(myList, myNumber):
    c = min(myList, key=lambda x:abs(x-myNumber))

```

To find the closest triangle number. 

The end result list should look like:

```
[0, 0, 1 , 0 , 1 ,1 , 0 , 1, 2 , 1, 0 , 1, 2, 2, 1, 0, etc.]

```

If you have another method of generating the same result that's faster go for it.

Answer: 
A little bit of math will bring a more analytical solution:

```
from math import sqrt
def close_triangle2(n):
    m=int(0.5*(-1+sqrt(1+8*n))) #solve it for the explicit formula
    tan0=(m*m+m)/2              #the closest one is either this
    tan1=(m*m+3*m+2)/2          #or this
    if (n-tan0)>(tan1-n):
        return tan1-n
    else:
        return n-tan0

```

It will become slightly faster than the loop version from @perreal when the number is large:

```
In [87]:

%timeit close_triangle(111111111)
100000 loops, best of 3: 5 µs per loop
In [86]:

%timeit close_triangle2(111111111)
100000 loops, best of 3: 4.13 µs per loop

```

If you wonders:

```
In [94]:

close_triangle2((30**10 * (30**10+1)) / 2 + 100)
Out[94]:
100L
In [95]:

close_triangle((30**10 * (30**10+1)) / 2 + 100)
Out[95]:
100L

In [102]:

%timeit close_triangle((30**10 * (30**10+1)) / 2 + 100)
10000 loops, best of 3: 17.9 µs per loop
In [103]:

%timeit close_triangle2((30**10 * (30**10+1)) / 2 + 100)
100000 loops, best of 3: 12 µs per loop

```
May not be the best way but should be faster than your method:

```
def triangle_around(n):
    k = int((2*n)**0.5)
    p = (k**2 + k) / 2 
    while p < n:
        k, p = k + 1, p + k + 1 
    return (p - k, p)

def close_triangle(n):
    t = triangle_around(n)
    return min(t[1] - n, n - t[0])

print close_triangle(8)    # gives 2

```