A formula for a sequence which has three odds and then three evens, alternately We know that triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36...  where we have alternate two odd and two even numbers. This sequence has a simple formula $a_n=n(n+1)/2$.
What would be an example of a sequence, described by a similar algebraic formula, which has three odds and then three evens, alternately?
Ideally, it would be described by a polynomial of low degree. 
 A: The question says "ideally it would be described by a polynomial of low degree," and neither of the other two answers give such a polynomial. So it's worth asking whether there could be a polynomial formula with this parity property. I claim that there is no such formula.
To see this, we can use the method of successive differences and work modulo $2$. Modulo $2$, our sequence is $6$-periodic:
$$1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0,\dots$$
So the first difference of the sequence (modulo $2$) is
$$1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0,\dots$$
and the second difference is
$$1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0,\dots$$
and finally the third difference is
$$1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1,\dots$$
Note that the third difference of the sequence is a rotated version of the second difference! In consequence, all subsequent differences will also be rotated versions of the second difference. In particular, no successive difference of the original sequence can ever be all zero modulo 2, which means that no successive difference of the original sequence can ever be zero in the integers.
But the $(n+1)$st difference of any polynomial of degree $n$ is zero. So no polynomial sequence of any degree can have the desired property.
A: How about the sequence
$$a_n=\frac{1+(-1)^{\lfloor n/3\rfloor}}{2},\qquad \begin{array}{c|c|c|c|c|c|c|c|c|c}
n \strut& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline
a_n \strut& 1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1
\end{array}\;\cdots$$
or even simpler, the sequence
$$a_n=\left\lfloor \frac{n}{3}\right\rfloor+1,\qquad \begin{array}{c|c|c|c|c|c|c|c|c|c}
n \strut& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline
a_n \strut& 1 & 1 & 1 & 2 & 2 & 2 & 3 & 3 & 3
\end{array}\;\cdots$$
A: The sequence $$n \mapsto 4n^6+n^5+6n^3+4n \pmod 7$$ for $n \geq 1$ gives $$1,1,1,0,0,0,1,1,1,0,0,0,\ldots.$$
