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

```
You can leverage the *triangle-root* property to quickly calculate the nearest triangle number to a number, and from there easily calculate your value

```
from math import sqrt

def triangle_number(tnumber):
    '''
    returns the tnumberth triangle number (I.E 3 = 6)
    '''
    return (tnumber*(tnumber+1))/2

def triangle_root(number):
    '''
    returns the triangle root of a number (I.E 6 = 3)
    '''
    return (sqrt(8*number+1)-1)/2

def nearest_root(number):
    '''
    Calculates the nearest whole triangle root to a number (I.E. 5 = 3)
    '''
    t_root = triangle_root(number)
    return round(t_root)

def find_delta(number):
    '''
    Given a number, returns abs(n-nearest_triangle)
    '''
    return abs(number - triangle_number(nearest_root(number)))

```

Of course you could condense this into a single function for efficiency, I just spaced it out for clarity.