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I have recently been interested in the problem of summing Combinatorials. I have been beating my brain for the past days to figure out how to find an explicit closed form of:

$n \choose 0 $+$ n \choose 3 $+$ n \choose 6$ + $\dots$ + $n \choose 3K$, where $3K$ is the largest multiple of $3$ less than or equal to $n$.

I have tried the expansion of $(1+1+1)^n$ which got nowhere, and I dont know how to proceed. I figure the answer will be conditional on $n \pmod 6$.

Can someone lend help? Thank you.

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    Since the set $\mathbb{Z}\cap(n,\infty)$ is not bounded above, what do you mean by $3K$ being the largest number such that $3(K+1)>n$?2012-11-03
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    Ah right. $n$ is a constant- a fixed number. Thus, if we had-say- $n=100$, $K$ would be $33$. What i was trying to get at is pretty much to just have all the multiples of $3$ that are less then $n$ being in the combinatorics. I will fix the typo2012-11-03
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    There's a duplicate somewhere, but I'm too lazy to find it. See this: http://en.wikipedia.org/wiki/Series_multisection2012-11-03
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    You don't need to worry about defining K precisely: $$\binom{a}{b} = 0$$ when $a,b \in \mathbb{N}$ and $b > a$.2012-11-03
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    @PeterTaylor I'm not worried about defining $K$, but it simply makes no sense to say that $3K$ is the largest the largest number satisfying $3(K+1)>n$ since such a number doesn't actually exist.2012-11-03
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    Reading this title and question hurts my eyes. For anyone who visits this page in the future, note: Combinatorics is the genre of mathematics which studies discrete objects, their properties, and how many of them there are. Combinatorial is an adjective used to describe something having to do with combinatorics. Neither are what you should call $\binom{n}{k}$. That is worded as a *Combination*. There is no such thing as a sum of combinatorics neither is there such a thing as a sum of combinatorials.2016-12-02

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