Why do ints require three times as much memory in Python?

Remember that the Python int type does not have a limited range like C int has; the only limit is the available memory.

Memory goes to storing the value, the current size of the integer storage (the storage size is variable to support arbitrary sizes), and the standard Python object bookkeeping (a reference to the relevant object and a reference count).

You can look up the longintrepr.h source (the Python 3 int type was traditionally known as the long type in Python 2); it makes effective use of the PyVarObject C type to track integer size:

struct _longobject {
        PyObject_VAR_HEAD
        digit ob_digit[1];
};

The ob_digit array stores ‘digits’ of either 15 or 30 bits wide (depending on your platform); so on my 64-bit OS X system, an integer up to (2 ^ 30) – 1 uses 1 ‘digit’:

>>> sys.getsizeof((1 << 30) - 1)
28

but if you use 2 30-bit digits in the number an additional 4 bytes are needed, etc:

>>> sys.getsizeof(1 << 30)
32
>>> sys.getsizeof(1 << 60)
36
>>> sys.getsizeof(1 << 90)
40

The base 24 bytes then are the PyObject_VAR_HEAD structure, holding the object size, the reference count and the type pointer (each 8 bytes / 64 bits on my 64-bit OS X platform).

On Python 2, integers <= sys.maxint but >= -sys.maxint - 1 are stored using a simpler structure storing just the single value:

typedef struct {
    PyObject_HEAD
    long ob_ival;
} PyIntObject;

because this uses PyObject instead of PyVarObject there is no ob_size field in the struct and the memory size is limited to just 24 bytes; 8 for the long value, 8 for the reference count and 8 for the type object pointer.