# Tell a sequence in which there comes three consecutive even numbers after three odd numbers indefinitely? [closed]

For example, in triangular numbers $1, 3, 6, 10, 15, 21, 28, 36, 45, 55\dots$, two evens come after two odds. I want something like it for three odds and three evens.

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## closed as too broad by BaronVT, vadim123, Nick Peterson, Jesko Hüttenhain, Jack D'AurizioJan 10 '14 at 18:36

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

Does '1, 3, 5, 2, 4, 6, 7, 9, 11, 8, 10, 12, ...' not satisfy your constraint? Without knowing more about what you need it's impossible to give a good answer to this question. – Steven Stadnicki Jan 10 '14 at 17:31
well, 3,5,7,4,6,8,2,2,2,2,2 etc. (all 2s) This is just to say: Any sequence is mathematically logically, see comments here: math.stackexchange.com/questions/632321/… – Bernd Jan 10 '14 at 17:31

The sequence $a_n = \lceil n/3\rceil$ works.
The notation $\lceil m \rceil$ indicates the smallest integer not greater than $m$ (the "ceiling"). So the sequence for $n = 1, 2, 3, ...$ is $a_n = 1, 1, 1, 2, 2, 2, 3, 3, 3, ...$
Actually I just changed it to ceiling because $n$ are the natural numbers (usually). – John Jan 10 '14 at 17:39