Generate a sequence of numbers in Python

Every number from 1,2,5,6,9,10... is divisible by 4 with remainder 1 or 2.

>>> ','.join(str(i) for i in xrange(100) if i % 4 in (1,2))'1,2,5,6,9,10,13,14,...'

>>> ','.join('{},{}'.format(i, i + 1) for i in range(1, 100, 4))'1,2,5,6,9,10,13,14,17,18,21,22,25,26,29,30,33,34,37,38,41,42,45,46,49,50,53,54,57,58,61,62,65,66,69,70,73,74,77,78,81,82,85,86,89,90,93,94,97,98'

That was a quick and quite dirty solution.

Now, for a solution that is suitable for different kinds of progression problems:

def deltas():    while True:        yield 1        yield 3def numbers(start, deltas, max):    i = start    while i <= max:        yield i        i += next(deltas)print(','.join(str(i) for i in numbers(1, deltas(), 100)))

And here are similar ideas implemented using itertools:

from itertools import cycle, takewhile, accumulate, chaindef numbers(start, deltas, max):    deltas = cycle(deltas)    numbers = accumulate(chain([start], deltas))    return takewhile(lambda x: x <= max, numbers)print(','.join(str(x) for x in numbers(1, [1, 3], 100)))

Includes some guessing on the exact sequence you are expecting:

>>> l = list(range(1, 100, 4)) + list(range(2, 100, 4))>>> l.sort()>>> ','.join(map(str, l))'1,2,5,6,9,10,13,14,17,18,21,22,25,26,29,30,33,34,37,38,41,42,45,46,49,50,53,54,57,58,61,62,65,66,69,70,73,74,77,78,81,82,85,86,89,90,93,94,97,98'

As one-liner:

>>> ','.join(map(str, sorted(list(range(1, 100, 4))) + list(range(2, 100, 4))))

(btw. this is Python 3 compatible)