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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.

Martin Sleziak
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Apurva
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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 $3(K+1)>n$? – HorizonsMaths Nov 03 '12 at 22:48
  • 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 typo – Apurva Nov 03 '12 at 22:51
  • There's a duplicate somewhere, but I'm too lazy to find it. See this: http://en.wikipedia.org/wiki/Series_multisection – wj32 Nov 03 '12 at 22:52
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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 $b > a$. – Peter Taylor Nov 03 '12 at 22:59
  • @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. – HorizonsMaths Nov 03 '12 at 23:28
  • 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. – JMoravitz Dec 02 '16 at 21:13

2 Answers2

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Let $\omega$ be a third root of 1.

Then

$$(1+1)^n = \binom{n}{0} +\binom{n}{1}+ \binom{n}{2}+ \binom{n}{3}+ ...+\binom{n}{n} \,.$$ $$ (1+\omega)^n = \binom{n}{0} + \binom{n}{1}\omega+ \binom{n}{2} \omega^2+ \binom{n}{3}+ ...+ \binom{n}{n} \omega^n \,.$$

$$ (1+\omega^2)^n =\binom{n}{0} + \binom{n}{1} \omega^2+ \binom{n}{2} \omega+ \binom{n}{3}+ ...+ \binom{n}{n}\omega^{2n} \,.$$

Now, since $1+ \omega +\omega^2=0$, adding them only every third column remains.

Thus

$$2^n+ (1+\omega)^n+(1+\omega^2)^n =3 \left( \binom{n}{0} +\binom{n}{3}+ \binom{n}{6}+ \binom{n}{9}+ ...+\binom{n}{3k} \right)$$

All you have left is to calculate $(1+\omega)^n$ and $(1+\omega^2)^n$ by writing them in polar/trig form.

P.S. Same trick with $\omega(1+\omega)^n$ and $\omega^2 (1+\omega^2)^n$ yields $\left( \binom{n}{2} +\binom{n}{5}+ \binom{n}{8}+ \binom{n}{11}+ ... \right)$ while $\omega^2(1+\omega)^n$ and $\omega (1+\omega^2)^n$ yields $\left( \binom{n}{1} +\binom{n}{4}+ \binom{n}{7}+ \binom{n}{10}+ ... \right)$.

N. S.
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A closed form is given by $\ \displaystyle \frac{2^n+2\cos(n\pi/3)}3$.

See this entry of OEIS for more information.

Raymond Manzoni
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