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

```
I have one solution that doesn't need calculate every triangle number first, but I am not sure whether it's best.

As formula of triange number is

```
    A(n) = n*(n-1)/2       (n is integer)

```

For input x, it should be in middle of A(n) and A(n+1)

```
    A(n) <= x <= A(n+1)

```

For A(n) we have:

```
    n * (n - 1) / 2 <= x
    (n - 1)^2 < n * (n-1) <= 2*x
    n < sqrt(2*x) + 1

```

For A(n+1) we have:

```
    (n + 1)^2 > (n + 1) * n >= 2*x
    n > sqrt(2*x) - 1

```

Now, if we have input number x, we could rapidly find n that makes x in [A(n),A(n+1)], then we only need to calculate distance of [abs(A(n) - x),abs(A(n+1) - x)] and make smallar one as real output.