What is the pattern in the sequence 1 2 4 3 6 8 7 14 16?

What is the pattern in the sequence 1 2 4 3 6 8 7 14 16?
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Karla Bautista
Not a very mathematically significant sequence, but can you express it algebraically with a single formula?
Consider $\omega =-\frac{1}{2}+i\frac{\sqrt{3}}{2}$
This has the property that ${\omega }^{3}=1$
Then we can write:
${a}_{0}=1$
${a}_{i+1}=\frac{\left({\omega }^{i}-\omega \right)\left({\omega }^{i}-{\omega }^{2}\right)}{\left(1-\omega \right)\left(1-{\omega }^{2}\right)}2{a}_{i}+\frac{\left({\omega }^{i}-{\omega }^{2}\right)\left({\omega }^{i}-1\right)}{\left(\omega -{\omega }^{2}\right)\left(\omega -1\right)}\left({a}_{i}+2\right)+\frac{\left({\omega }^{i}-1\right)\left({\omega }^{i}-\omega \right)}{\left({\omega }^{2}-1\right)\left({\omega }^{2}-\omega \right)}\left({a}_{i}-1\right)$
This can be simplified, but it helps to have it in this formulation so you can understand how it works.
When i=0 modulo 3, then:
$\frac{\left({\omega }^{i}-\omega \right)\left({\omega }^{i}-{\omega }^{2}\right)}{\left(1-\omega \right)\left(1-{\omega }^{2}\right)}=\frac{\left(1-\omega \right)\left(1-{\omega }^{2}\right)}{\left(1-\omega \right)\left(1-{\omega }^{2}\right)}=1$
$\frac{\left({\omega }^{i}-{\omega }^{2}\right)\left({\omega }^{i}-1\right)}{\left(\omega -{\omega }^{2}\right)\left(\omega -1\right)}=\frac{\left(1-{\omega }^{2}\right)\left(1-1\right)}{\left(1-{\omega }^{2}\right)\left(\omega -1\right)}=0$
$\frac{\left({\omega }^{i}-1\right)\left({\omega }^{i}-\omega \right)}{\left({\omega }^{2}-1\right)\left({\omega }^{2}-\omega \right)}=\frac{\left(1-1\right)\left(1-\omega \right)}{\left({\omega }^{2}-1\right)\left({\omega }^{2}-\omega \right)}=0$
When i=1 modulo 3, then these coefficient expressions evaluate as 0, 1 and 0.
When i=2 modulo 3, then these coefficient expressions evaluate as 0, 0 and 1.
So we use these to pick out each of the three rules cyclically.